Classify the following matrices :
(i) [ 2 − 1 5 1 ] \text{(i) }\begin{bmatrix*}[r] 2 & -1 \\ 5 & 1 \end{bmatrix*} \\[0.5em] (i) [ 2 5 − 1 1 ]
(ii) [ 2 3 − 7 ] \text{(ii) } \begin{bmatrix} 2 & 3 & -7 \end{bmatrix} \\[0.5em] (ii) [ 2 3 − 7 ]
(iii) [ 3 0 − 1 ] \text{(iii) } \begin{bmatrix*}[r] 3 \\ 0 \\ -1 \end{bmatrix*} \\[0.5em] (iii) 3 0 − 1
(iv) [ 2 − 4 0 0 1 7 ] \text{(iv) } \begin{bmatrix} 2 & -4 \\ 0 & 0 \\ 1 & 7 \end{bmatrix} \\[0.5em] (iv) 2 0 1 − 4 0 7
(v) [ 2 7 8 − 1 2 0 ] \text{(v) } \begin{bmatrix*}[r] 2 & 7 & 8 \\ -1 & \sqrt{2} & 0 \end{bmatrix*} \\[0.5em] (v) [ 2 − 1 7 2 8 0 ]
(vi) [ 0 0 0 0 0 0 ] . \text{(vi) } \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}. \\[0.5em] (vi) [ 0 0 0 0 0 0 ] .
Answer
(i) It is a square matrix of order 2.
(ii) It is a row matrix of order 1 x 3.
(iii) It is a column matrix of order 3 x 1.
(iv) It is a 3 x 2 matrix.
(v) It is a 2 x 3 matrix.
(vi) It is a zero matrix of order 2 x 3.
If a matrix has 4 elements, what are the possible orders it can have ?
Answer
If a matrix has 4 elements the possible orders are,
1 x 4, 2 x 2, 4 x 1.
If a matrix has 8 elements, what are the possible orders it can have ?
Answer
If a matrix has 8 elements the possible orders are,
1 x 8, 4 x 2, 2 x 4, 8 x 1.
Construct a 2 × \times × ij are given by
(i) aij = 2i - j
(ii) aij = i.j
Answer
(i) Given aij = 2i - j,
∴ a11 = 2(1) - 1 = 1, a12 = 2(1) - 2 = 0,21 = 2(2) - 1 = 3, a22 = 2(2) - 2 = 2.
Hence, the required matrix =[ 1 0 3 2 ] \begin{bmatrix} 1 & 0 \\ 3 & 2 \end{bmatrix} [ 1 3 0 2 ]
(ii) Given aij = i.j,
∴ a11 = 1.1 = 1, a12 = 1.2 = 2,21 = 2.1 = 2, a22 = 2.2 = 4.
Hence, the required matrix =[ 1 2 2 4 ] \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix} [ 1 2 2 4 ]
Find the values of x and y if [ 2 x + y 3 x − 2 y ] = [ 5 4 ] \begin{bmatrix*}[r] 2x + y \\ 3x - 2y \end{bmatrix*} = \begin{bmatrix} 5 \\ 4 \end{bmatrix} [ 2 x + y 3 x − 2 y ] = [ 5 4 ]
Answer
Given [ 2 x + y 3 x − 2 y ] = [ 5 4 ] \begin{bmatrix*}[r] 2x + y \\ 3x - 2y \end{bmatrix*} = \begin{bmatrix} 5 \\ 4 \end{bmatrix} [ 2 x + y 3 x − 2 y ] = [ 5 4 ]
By definition of equality of matrices, we get
2x + y = 5 [....Eq 1]
2x + y = 5
Putting value of y in Eq 2,
3x - 2(5 - 2x) = 4
∴ x = 2, y = 5 - 2x = 5 - 4 = 1.
Hence, the value of x = 2 and y = 1.
Find the values of x if [ 3 x + y − y 2 y − x 3 ] = [ 1 2 − 5 3 ] \begin{bmatrix*}[r] 3x + y & -y \\ 2y - x & 3 \end{bmatrix*} = \begin{bmatrix*}[r] 1 & 2 \\ -5 & 3 \end{bmatrix*} [ 3 x + y 2 y − x − y 3 ] = [ 1 − 5 2 3 ]
Answer
Given [ 3 x + y − y 2 y − x 3 ] = [ 1 2 − 5 3 ] \begin{bmatrix*}[r] 3x + y & -y \\ 2y - x & 3 \end{bmatrix*} = \begin{bmatrix*}[r] 1 & 2 \\ -5 & 3 \end{bmatrix*} [ 3 x + y 2 y − x − y 3 ] = [ 1 − 5 2 3 ]
By definition of equality of matrices, we get
-y = 2 or y = -2 and
Putting value of y = -2 in above equation
⇒ 3x - 2 = 1
Hence, the value of x = 1.
If [ x + 3 4 y − 4 x + y ] = [ 5 4 3 9 ] \begin{bmatrix*}[r] x + 3 & 4 \\ y - 4 & x + y \end{bmatrix*} = \begin{bmatrix*}[r] 5 & 4 \\ 3 & 9 \end{bmatrix*} [ x + 3 y − 4 4 x + y ] = [ 5 3 4 9 ]
Answer
Given [ x + 3 4 y − 4 x + y ] = [ 5 4 3 9 ] \begin{bmatrix*}[r] x + 3 & 4 \\ y - 4 & x + y \end{bmatrix*} = \begin{bmatrix} 5 & 4 \\ 3 & 9 \end{bmatrix} [ x + 3 y − 4 4 x + y ] = [ 5 3 4 9 ]
By definition of equality of matrices, we get
x + 3 = 5, y - 4 = 3 and x + y = 9
Note that x = 2 and y = 7 satisfies the equation x + y = 9.
Hence, the value of x = 2 and y = 7.
Find the values of x, y and z if [ x + 2 6 3 5 z ] = [ − 5 y 2 + y 3 − 20 ] . \begin{bmatrix*}[r] x + 2 & 6 \\ 3 & 5z \end{bmatrix*} = \begin{bmatrix*}[r] -5 & y^2 + y \\ 3 & -20 \end{bmatrix*} . [ x + 2 3 6 5 z ] = [ − 5 3 y 2 + y − 20 ] .
Answer
Given [ x + 2 6 3 5 z ] = [ − 5 y 2 + y 3 − 20 ] . \begin{bmatrix*}[r] x + 2 & 6 \\ 3 & 5z \end{bmatrix*} = \begin{bmatrix*}[r] -5 & y^2 + y \\ 3 & -20 \end{bmatrix*} . [ x + 2 3 6 5 z ] = [ − 5 3 y 2 + y − 20 ] .
By definition of equality of matrices, we get
x + 2 = -5 or x = -7,2 + y = 6 [....Eq 1],
From Eq 1 we get,
⇒ y 2 + y − 6 = 0 ⇒ y 2 + 3 y − 2 y − 6 = 0 ⇒ y ( y + 3 ) − 2 ( y + 3 ) = 0 ⇒ ( y − 2 ) ( y + 3 ) = 0 ⇒ y = 2 or y = − 3. \Rightarrow y^2 + y - 6 = 0 \\[0.5em] \Rightarrow y^2 + 3y - 2y - 6 = 0 \\[0.5em] \Rightarrow y(y + 3) - 2(y + 3) = 0 \\[0.5em] \Rightarrow (y - 2)(y + 3) = 0 \\[0.5em] \Rightarrow y = 2 \text{ or } y = -3. \\[0.5em] ⇒ y 2 + y − 6 = 0 ⇒ y 2 + 3 y − 2 y − 6 = 0 ⇒ y ( y + 3 ) − 2 ( y + 3 ) = 0 ⇒ ( y − 2 ) ( y + 3 ) = 0 ⇒ y = 2 or y = − 3.
∴ x = -7, y = 2 or -3 and z = -4.
Hence, the values are:
Find the values of x, y, a and b if [ x − 2 y a + 2 b 3 a − b ] = [ 3 1 5 1 ] \begin{bmatrix*}[r] x - 2 & y \\ a + 2b & 3a - b \end{bmatrix*} = \begin{bmatrix*}[r] 3 & 1 \\ 5 & 1 \end{bmatrix*} [ x − 2 a + 2 b y 3 a − b ] = [ 3 5 1 1 ]
Answer
Given, [ x − 2 y a + 2 b 3 a − b ] = [ 3 1 5 1 ] \begin{bmatrix*}[r] x - 2 & y \\ a + 2b & 3a - b \end{bmatrix*} = \begin{bmatrix*}[r] 3 & 1 \\ 5 & 1 \end{bmatrix*} [ x − 2 a + 2 b y 3 a − b ] = [ 3 5 1 1 ]
By definition of equality of matrices, we get
x - 2 = 3 or x = 5,
From Eq 1 we get,
⇒ a + 2b = 5
Putting this value of a in Eq 2,
⇒ 3(5 - 2b) - b = 1
Using value of b to find value of a,
⇒ a = 5 - 2b
∴ x = 5, y = 1, a = 1 and b = 2.
Hence, the values are:
Find the values of a, b, c and d if [ a + b 3 5 + c a b ] = [ 6 d − 1 8 ] . \begin{bmatrix*}[r] a + b & 3 \\ 5 + c & ab \end{bmatrix*} = \begin{bmatrix*}[r] 6 & d \\ -1 & 8 \end{bmatrix*} . [ a + b 5 + c 3 ab ] = [ 6 − 1 d 8 ] .
Answer
Given [ a + b 3 5 + c a b ] = [ 6 d − 1 8 ] . \begin{bmatrix*}[r] a + b & 3 \\ 5 + c & ab \end{bmatrix*} = \begin{bmatrix*}[r] 6 & d \\ -1 & 8 \end{bmatrix*} . [ a + b 5 + c 3 ab ] = [ 6 − 1 d 8 ] .
By definition of equality of matrices, we get
d = 3,
Putting value of b from Eq 1 in Eq 2,
a + b = 6 or b = 6 − a ⇒ a ( 6 − a ) = 8 ⇒ 6 a − a 2 = 8 ⇒ a 2 − 6 a + 8 = 0 ⇒ a 2 − 4 a − 2 a + 8 = 0 ⇒ a ( a − 4 ) − 2 ( a − 4 ) = 0 ⇒ ( a − 2 ) ( a − 4 ) = 0 ⇒ a = 2 or a = 4. a + b = 6 \text{ or } b = 6 - a \\[1em] \Rightarrow a(6 - a) = 8 \\[0.5em] \Rightarrow 6a - a^2 = 8 \\[0.5em] \Rightarrow a^2 - 6a + 8 = 0 \\[0.5em] \Rightarrow a^2 - 4a - 2a + 8 = 0 \\[0.5em] \Rightarrow a(a - 4) - 2(a - 4) = 0 \\[0.5em] \Rightarrow (a - 2)(a - 4) = 0 \\[0.5em] \Rightarrow a = 2 \text{ or } a = 4. a + b = 6 or b = 6 − a ⇒ a ( 6 − a ) = 8 ⇒ 6 a − a 2 = 8 ⇒ a 2 − 6 a + 8 = 0 ⇒ a 2 − 4 a − 2 a + 8 = 0 ⇒ a ( a − 4 ) − 2 ( a − 4 ) = 0 ⇒ ( a − 2 ) ( a − 4 ) = 0 ⇒ a = 2 or a = 4.
Now, finding value of b = 6 - a,
if, a = 2, b = 6 - 2 = 4.
or, a = 4, b = 6 - 4 = 2.
∴ a = 2 or 4, b = 4 or 2, c = -6 and d = 3.
Hence, the values are
Given that M = [ 2 0 1 2 ] and N = [ 2 0 − 1 2 ] \text{M} = \begin{bmatrix*}[r] 2 & 0 \\ 1 & 2 \end{bmatrix*} \text{ and N }= \begin{bmatrix*}[r] 2 & 0 \\ -1 & 2 \end{bmatrix*} M = [ 2 1 0 2 ] and N = [ 2 − 1 0 2 ]
Answer
M + 2N = [ 2 0 1 2 ] + 2 [ 2 0 − 1 2 ] = [ 2 0 1 2 ] + [ 4 0 − 2 4 ] = [ 2 + 4 0 + 0 1 − 2 2 + 4 ] = [ 6 0 − 1 6 ] \text{M + 2N }= \begin{bmatrix*}[r] 2 & 0 \\ 1 & 2 \end{bmatrix*} + 2\begin{bmatrix*}[r] 2 & 0 \\ -1 & 2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 & 0 \\ 1 & 2 \end{bmatrix*} + \begin{bmatrix*}[r] 4 & 0 \\ -2 & 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 + 4 & 0 + 0 \\ 1 - 2 & 2 + 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 6 & 0 \\ -1 & 6 \end{bmatrix*} \\[1em] M + 2N = [ 2 1 0 2 ] + 2 [ 2 − 1 0 2 ] = [ 2 1 0 2 ] + [ 4 − 2 0 4 ] = [ 2 + 4 1 − 2 0 + 0 2 + 4 ] = [ 6 − 1 0 6 ]
Hence, the matrix M + 2N = [ 6 0 − 1 6 ] \begin{bmatrix*}[r] 6 & 0 \\ -1 & 6 \end{bmatrix*} [ 6 − 1 0 6 ]
If A = [ 2 0 − 3 1 ] and B = [ 0 1 − 2 3 ] \text{If A }= \begin{bmatrix*}[r] 2 & 0 \\ -3 & 1 \end{bmatrix*} \text{ and B }= \begin{bmatrix*}[r] 0 & 1 \\ -2 & 3 \end{bmatrix*} If A = [ 2 − 3 0 1 ] and B = [ 0 − 2 1 3 ]
Answer
2A - 3B = 2 [ 2 0 − 3 1 ] − 3 [ 0 1 − 2 3 ] = [ 4 0 − 6 2 ] − [ 0 3 − 6 9 ] = [ 4 + 0 0 − 3 − 6 − ( − 6 ) 2 − 9 ] = [ 4 − 3 0 − 7 ] \text{2A - 3B }= 2\begin{bmatrix*}[r] 2 & 0 \\ -3 & 1 \end{bmatrix*} - 3\begin{bmatrix*}[r] 0 & 1 \\ -2 & 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 4 & 0 \\ -6 & 2 \end{bmatrix*} - \begin{bmatrix*}[r] 0 & 3 \\ -6 & 9 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 4 + 0 & 0 - 3 \\ -6 - (-6) & 2 - 9 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 4 & -3 \\ 0 & -7 \end{bmatrix*} \\[1em] 2A - 3B = 2 [ 2 − 3 0 1 ] − 3 [ 0 − 2 1 3 ] = [ 4 − 6 0 2 ] − [ 0 − 6 3 9 ] = [ 4 + 0 − 6 − ( − 6 ) 0 − 3 2 − 9 ] = [ 4 0 − 3 − 7 ]
Hence, the matrix 2A - 3B = [ 4 − 3 0 − 7 ] . \begin{bmatrix*}[r] 4 & -3 \\ 0 & -7 \end{bmatrix*}. [ 4 0 − 3 − 7 ] .
Simplify : sin A [ sin A − cos A cos A sin A ] + cos A [ cos A sin A − sin A cos A ] \text{sin A}\begin{bmatrix*}[r] \text{sin A} & -\text{cos A} \\ \text{cos A} & \text{sin A} \end{bmatrix*} + \text{cos A}\begin{bmatrix*}[r] \text{cos A} & \text{sin A} \\ -\text{sin A} & \text{cos A} \end{bmatrix*} sin A [ sin A cos A − cos A sin A ] + cos A [ cos A − sin A sin A cos A ]
Answer
Given,
⇒ sin A [ sin A − cos A cos A sin A ] + cos A [ cos A sin A − sin A cos A ] ⇒ [ sin 2 A − sin A . cos A sin A . cos A sin 2 A ] + [ cos 2 A cos A . sin A − cos A . sin A cos 2 A ] = [ sin 2 A + cos 2 A − sin A . cos A + cos A . sin A sin A . cos A − cos A . sin A sin 2 A + cos 2 A ] ∵ sin 2 A + cos 2 A = 1 = [ 1 0 0 1 ] \Rightarrow \text{sin A}\begin{bmatrix*}[r] \text{sin A} & -\text{cos A} \\ \text{cos A} & \text{sin A} \end{bmatrix*} + \text{cos A}\begin{bmatrix*}[r] \text{cos A} & \text{sin A} \\ -\text{sin A} & \text{cos A} \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] \text{sin}^2\text{ A} & -\text{sin A}.\text{cos A} \\ \text{sin A}.\text{cos A} & \text{sin}^2\text{ A} \end{bmatrix*} + \begin{bmatrix*}[r] \text{cos}^2\text{ A} & \text{cos A}.\text{sin A} \\ -\text{cos A}.\text{sin A} & \text{cos}^2\text{ A} \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] \text{sin}^2\text{ A} + \text{cos}^2\text{ A} & -\text{sin A}.\text{cos A} + \text{cos A}.\text{sin A} \\ \text{sin A}.\text{cos A} - \text{cos A}.\text{sin A} & \text{sin}^2\text{ A} + \text{cos}^2\text{ A} \end{bmatrix*} \\[1em] \because \text{sin}^2\text{ A} + \text{cos}^2\text{ A} = 1 \\[1em] = \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em] ⇒ sin A [ sin A cos A − cos A sin A ] + cos A [ cos A − sin A sin A cos A ] ⇒ [ sin 2 A sin A . cos A − sin A . cos A sin 2 A ] + [ cos 2 A − cos A . sin A cos A . sin A cos 2 A ] = [ sin 2 A + cos 2 A sin A . cos A − cos A . sin A − sin A . cos A + cos A . sin A sin 2 A + cos 2 A ] ∵ sin 2 A + cos 2 A = 1 = [ 1 0 0 1 ]
Hence, on simplifying, the resultant matrix = [ 1 0 0 1 ] . \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*}. [ 1 0 0 1 ] .
If A = [ 1 2 − 2 3 ] , B = [ − 2 − 1 1 2 ] and C = [ 0 3 2 − 1 ] \text{If A } = \begin{bmatrix*}[r] 1 & 2 \\ -2 & 3 \end{bmatrix*}, \text{ B } = \begin{bmatrix*}[r] -2 & -1 \\ 1 & 2 \end{bmatrix*} \text{ and C} = \begin{bmatrix*}[r] 0 & 3 \\ 2 & -1 \end{bmatrix*} If A = [ 1 − 2 2 3 ] , B = [ − 2 1 − 1 2 ] and C = [ 0 2 3 − 1 ]
Answer
A + 2B - 3C = [ 1 2 − 2 3 ] + 2 [ − 2 − 1 1 2 ] − 3 [ 0 3 2 − 1 ] = [ 1 2 − 2 3 ] + [ − 4 − 2 2 4 ] − [ 0 9 6 − 3 ] = [ 1 + ( − 4 ) − 0 2 + ( − 2 ) − 9 − 2 + 2 − 6 3 + 4 − ( − 3 ) ] = [ − 3 − 9 − 6 10 ] \text{A + 2B - 3C }= \begin{bmatrix*}[r] 1 & 2 \\ -2 & 3 \end{bmatrix*} + 2\begin{bmatrix*}[r] -2 & -1 \\ 1 & 2 \end{bmatrix*} - 3\begin{bmatrix*}[r] 0 & 3 \\ 2 & -1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 & 2 \\ -2 & 3 \end{bmatrix*} + \begin{bmatrix*}[r] -4 & -2 \\ 2 & 4 \end{bmatrix*} - \begin{bmatrix*}[r] 0 & 9 \\ 6 & -3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 + (-4) - 0 & 2 + (-2) - 9 \\ -2 + 2 - 6 & 3 + 4 - (-3) \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -3 & -9 \\ -6 & 10 \end{bmatrix*} A + 2B - 3C = [ 1 − 2 2 3 ] + 2 [ − 2 1 − 1 2 ] − 3 [ 0 2 3 − 1 ] = [ 1 − 2 2 3 ] + [ − 4 2 − 2 4 ] − [ 0 6 9 − 3 ] = [ 1 + ( − 4 ) − 0 − 2 + 2 − 6 2 + ( − 2 ) − 9 3 + 4 − ( − 3 ) ] = [ − 3 − 6 − 9 10 ]
Hence, the matrix A + 2B - 3C = [ − 3 − 9 − 6 10 ] . \begin{bmatrix*}[r] -3 & -9 \\ -6 & 10 \end{bmatrix*}. [ − 3 − 6 − 9 10 ] .
If A = [ 0 − 1 1 2 ] and B = [ 1 2 − 1 1 ] \begin{bmatrix*}[r] 0 & -1 \\ 1 & 2 \end{bmatrix*} \text{ and B} = \begin{bmatrix*}[r] 1 & 2 \\ -1 & 1 \end{bmatrix*} [ 0 1 − 1 2 ] and B = [ 1 − 1 2 1 ]
(i) 3A + X = B
(ii) X - 3B = 2A.
Answer
(i) 3A + X = B
⇒ X = B - 3A
⇒ X = [ 1 2 − 1 1 ] − 3 [ 0 − 1 1 2 ] = [ 1 2 − 1 1 ] − [ 0 − 3 3 6 ] = [ 1 − 0 2 − ( − 3 ) − 1 − 3 1 − 6 ] = [ 1 5 − 4 − 5 ] \Rightarrow X = \begin{bmatrix*}[r] 1 & 2 \\ -1 & 1 \end{bmatrix*} - 3\begin{bmatrix*}[r] 0 & -1 \\ 1 & 2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 & 2 \\ -1 & 1 \end{bmatrix*} - \begin{bmatrix*}[r] 0 & -3 \\ 3 & 6 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 - 0 & 2 - (-3) \\ -1 - 3 & 1 - 6 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 & 5 \\ -4 & -5 \end{bmatrix*} ⇒ X = [ 1 − 1 2 1 ] − 3 [ 0 1 − 1 2 ] = [ 1 − 1 2 1 ] − [ 0 3 − 3 6 ] = [ 1 − 0 − 1 − 3 2 − ( − 3 ) 1 − 6 ] = [ 1 − 4 5 − 5 ]
Hence, matrix X = [ 1 5 − 4 − 5 ] . \begin{bmatrix*}[r] 1 & 5 \\ -4 & -5 \end{bmatrix*}. [ 1 − 4 5 − 5 ] .
(ii) X - 3B = 2A
⇒ X = 2A + 3B
X = 2 [ 0 − 1 1 2 ] + 3 [ 1 2 − 1 1 ] = [ 0 − 2 2 4 ] + [ 3 6 − 3 3 ] = [ 0 + 3 − 2 + 6 2 + ( − 3 ) 4 + 3 ] = [ 3 4 − 1 7 ] X = 2\begin{bmatrix*}[r] 0 & -1 \\ 1 & 2 \end{bmatrix*} + 3\begin{bmatrix*}[r] 1 & 2 \\ -1 & 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 0 & -2 \\ 2 & 4 \end{bmatrix*} + \begin{bmatrix*}[r] 3 & 6 \\ -3 & 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 0 + 3 & -2 + 6 \\ 2 + (-3) & 4 + 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 3 & 4 \\ -1 & 7 \end{bmatrix*} X = 2 [ 0 1 − 1 2 ] + 3 [ 1 − 1 2 1 ] = [ 0 2 − 2 4 ] + [ 3 − 3 6 3 ] = [ 0 + 3 2 + ( − 3 ) − 2 + 6 4 + 3 ] = [ 3 − 1 4 7 ]
Hence, matrix X = [ 3 4 − 1 7 ] . \begin{bmatrix*}[r] 3 & 4 \\ -1 & 7 \end{bmatrix*}. [ 3 − 1 4 7 ] .
Solve the matrix equation [ 2 1 5 0 ] − 3 X = [ − 7 4 2 6 ] \begin{bmatrix} 2 & 1 \\ 5 & 0 \end{bmatrix} - 3X = \begin{bmatrix*}[r] -7 & 4 \\ 2 & 6 \end{bmatrix*} [ 2 5 1 0 ] − 3 X = [ − 7 2 4 6 ]
Answer
Given,
⇒ [ 2 1 5 0 ] − 3 X = [ − 7 4 2 6 ] ⇒ 3 X = [ 2 1 5 0 ] − [ − 7 4 2 6 ] ⇒ 3 X = [ 2 − ( − 7 ) 1 − 4 5 − 2 0 − 6 ] ⇒ 3 X = [ 9 − 3 3 − 6 ] ⇒ X = 1 3 [ 9 − 3 3 − 6 ] ⇒ X = [ 3 − 1 1 − 2 ] \Rightarrow \begin{bmatrix} 2 & 1 \\ 5 & 0 \end{bmatrix} - 3X = \begin{bmatrix*}[r] -7 & 4 \\ 2 & 6 \end{bmatrix*} \\[1em] \Rightarrow 3X = \begin{bmatrix} 2 & 1 \\ 5 & 0 \end{bmatrix} - \begin{bmatrix*}[r] -7 & 4 \\ 2 & 6 \end{bmatrix*} \\[1em] \Rightarrow 3X = \begin{bmatrix*}[r] 2 - (-7) & 1 - 4 \\ 5 - 2 & 0 - 6 \end{bmatrix*} \\[1em] \Rightarrow 3X = \begin{bmatrix*}[r] 9 & -3 \\ 3 & -6 \end{bmatrix*} \\[1em] \Rightarrow X = \dfrac{1}{3}\begin{bmatrix*}[r] 9 & -3 \\ 3 & -6 \end{bmatrix*} \\[1em] \Rightarrow X = \begin{bmatrix*}[r] 3 & -1 \\ 1 & -2 \end{bmatrix*} ⇒ [ 2 5 1 0 ] − 3 X = [ − 7 2 4 6 ] ⇒ 3 X = [ 2 5 1 0 ] − [ − 7 2 4 6 ] ⇒ 3 X = [ 2 − ( − 7 ) 5 − 2 1 − 4 0 − 6 ] ⇒ 3 X = [ 9 3 − 3 − 6 ] ⇒ X = 3 1 [ 9 3 − 3 − 6 ] ⇒ X = [ 3 1 − 1 − 2 ]
Hence, matrix X = [ 3 − 1 1 − 2 ] . \begin{bmatrix*}[r] 3 & -1 \\ 1 & -2 \end{bmatrix*} . [ 3 1 − 1 − 2 ] .
If [ 1 4 − 2 3 ] + 2 M = 3 [ 3 2 0 − 3 ] , \begin{bmatrix*}[r] 1 & 4 \\ -2 & 3 \end{bmatrix*} + 2\text{M} = 3\begin{bmatrix*}[r] 3 & 2 \\ 0 & -3 \end{bmatrix*}, \\[1em] [ 1 − 2 4 3 ] + 2 M = 3 [ 3 0 2 − 3 ] ,
Answer
Given,
⇒ [ 1 4 − 2 3 ] + 2 M = 3 [ 3 2 0 − 3 ] ⇒ [ 1 4 − 2 3 ] + 2 M = [ 9 6 0 − 9 ] ⇒ 2 M = [ 9 6 0 − 9 ] − [ 1 4 − 2 3 ] ⇒ 2 M = [ 9 − 1 6 − 4 0 − ( − 2 ) − 9 − 3 ] ⇒ M = 1 2 [ 8 2 2 − 12 ] ⇒ M = [ 4 1 1 − 6 ] \Rightarrow \begin{bmatrix*}[r] 1 & 4 \\ -2 & 3 \end{bmatrix*} + 2\text{M} = 3\begin{bmatrix*}[r] 3 & 2 \\ 0 & -3 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 1 & 4 \\ -2 & 3 \end{bmatrix*} + 2\text{M} = \begin{bmatrix*}[r] 9 & 6 \\ 0 & -9 \end{bmatrix*} \\[1em] \Rightarrow 2\text{M} = \begin{bmatrix*}[r] 9 & 6 \\ 0 & -9 \end{bmatrix*} - \begin{bmatrix*}[r] 1 & 4 \\ -2 & 3 \end{bmatrix*} \\[1em] \Rightarrow 2\text{M} = \begin{bmatrix*}[r] 9 - 1 & 6 - 4 \\ 0 - (-2) & -9 - 3 \end{bmatrix*} \\[1em] \Rightarrow \text{M} = \dfrac{1}{2}\begin{bmatrix*}[r] 8 & 2 \\ 2 & -12 \end{bmatrix*} \\[1em] \Rightarrow \text{M} = \begin{bmatrix*}[r] 4 & 1 \\ 1 & -6 \end{bmatrix*} \\[1em] ⇒ [ 1 − 2 4 3 ] + 2 M = 3 [ 3 0 2 − 3 ] ⇒ [ 1 − 2 4 3 ] + 2 M = [ 9 0 6 − 9 ] ⇒ 2 M = [ 9 0 6 − 9 ] − [ 1 − 2 4 3 ] ⇒ 2 M = [ 9 − 1 0 − ( − 2 ) 6 − 4 − 9 − 3 ] ⇒ M = 2 1 [ 8 2 2 − 12 ] ⇒ M = [ 4 1 1 − 6 ]
Hence, matrix M = [ 4 1 1 − 6 ] . \begin{bmatrix*}[r] 4 & 1 \\ 1 & -6 \end{bmatrix*}. [ 4 1 1 − 6 ] .
Given A = [ 2 − 6 2 0 ] , B = [ − 3 2 4 0 ] and C = [ 4 0 0 2 ] . A = \begin{bmatrix*}[r] 2 & -6 \\ 2 & 0 \end{bmatrix*}, B = \begin{bmatrix*}[r] -3 & 2 \\ 4 & 0 \end{bmatrix*} \text{ and } C = \begin{bmatrix*}[r] 4 & 0 \\ 0 & 2 \end{bmatrix*}. A = [ 2 2 − 6 0 ] , B = [ − 3 4 2 0 ] and C = [ 4 0 0 2 ] .
Find the matrix X such that A + 2X = 2B + C.
Answer
Putting values of A, B and C in A + 2X = 2B + C,
⇒ [ 2 − 6 2 0 ] + 2 X = 2 [ − 3 2 4 0 ] + [ 4 0 0 2 ] ⇒ 2 X = [ − 6 4 8 0 ] + [ 4 0 0 2 ] − [ 2 − 6 2 0 ] ⇒ 2 X = [ − 6 + 4 − 2 4 + 0 − ( − 6 ) 8 + 0 − 2 0 + 2 − 0 ] ⇒ X = 1 2 [ − 4 10 6 2 ] ⇒ X = [ − 2 5 3 1 ] \Rightarrow \begin{bmatrix*}[r] 2 & -6 \\ 2 & 0 \end{bmatrix*} + 2X = 2\begin{bmatrix*}[r] -3 & 2 \\ 4 & 0 \end{bmatrix*} + \begin{bmatrix*}[r] 4 & 0 \\ 0 & 2 \end{bmatrix*} \\[1em] \Rightarrow 2X = \begin{bmatrix*}[r] -6 & 4 \\ 8 & 0 \end{bmatrix*} + \begin{bmatrix*}[r] 4 & 0 \\ 0 & 2 \end{bmatrix*} - \begin{bmatrix*}[r] 2 & -6 \\ 2 & 0 \end{bmatrix*} \\[1em] \Rightarrow 2X = \begin{bmatrix*}[r] -6 + 4 - 2 & 4 + 0 - (-6) \\ 8 + 0 - 2 & 0 + 2 - 0 \end{bmatrix*} \\[1em] \Rightarrow X = \dfrac{1}{2}\begin{bmatrix*}[r] -4 & 10 \\ 6 & 2 \end{bmatrix*} \\[1em] \Rightarrow X = \begin{bmatrix*}[r] -2 & 5 \\ 3 & 1 \end{bmatrix*} ⇒ [ 2 2 − 6 0 ] + 2 X = 2 [ − 3 4 2 0 ] + [ 4 0 0 2 ] ⇒ 2 X = [ − 6 8 4 0 ] + [ 4 0 0 2 ] − [ 2 2 − 6 0 ] ⇒ 2 X = [ − 6 + 4 − 2 8 + 0 − 2 4 + 0 − ( − 6 ) 0 + 2 − 0 ] ⇒ X = 2 1 [ − 4 6 10 2 ] ⇒ X = [ − 2 3 5 1 ]
Hence, matrix X = [ − 2 5 3 1 ] \begin{bmatrix*}[r] -2 & 5 \\ 3 & 1 \end{bmatrix*} [ − 2 3 5 1 ]
Find X and Y if
X + Y = [ 7 0 2 5 ] and X - Y = [ 3 0 0 3 ] . \text{X + Y} = \begin{bmatrix} 7 & 0 \\ 2 & 5 \end{bmatrix} \text{ and X - Y} = \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}. X + Y = [ 7 2 0 5 ] and X - Y = [ 3 0 0 3 ] .
Answer
Given,
X + Y = [ 7 0 2 5 ] \begin{bmatrix} 7 & 0 \\ 2 & 5 \end{bmatrix} [ 7 2 0 5 ]
X - Y = [ 3 0 0 3 ] \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} [ 3 0 0 3 ]
Adding both the equations,
⇒ X + Y + X − Y = [ 7 0 2 5 ] + [ 3 0 0 3 ] ⇒ 2 X = [ 7 + 3 0 + 0 2 + 0 5 + 3 ] ⇒ X = 1 2 [ 10 0 2 8 ] ⇒ X = [ 5 0 1 4 ] \Rightarrow X + Y + X - Y = \begin{bmatrix} 7 & 0 \\ 2 & 5 \end{bmatrix} + \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} \\[1em] \Rightarrow 2X = \begin{bmatrix} 7 + 3 & 0 + 0 \\ 2 + 0 & 5 + 3 \end{bmatrix} \\[1em] \Rightarrow X = \dfrac{1}{2}\begin{bmatrix} 10 & 0 \\ 2 & 8 \end{bmatrix} \\[1em] \Rightarrow X = \begin{bmatrix} 5 & 0 \\ 1 & 4 \end{bmatrix} \\[1em] ⇒ X + Y + X − Y = [ 7 2 0 5 ] + [ 3 0 0 3 ] ⇒ 2 X = [ 7 + 3 2 + 0 0 + 0 5 + 3 ] ⇒ X = 2 1 [ 10 2 0 8 ] ⇒ X = [ 5 1 0 4 ]
From Eq 1 we get,
⇒ Y = [ 7 0 2 5 ] − X = [ 7 0 2 5 ] − [ 5 0 1 4 ] = [ 7 − 5 0 − 0 2 − 1 5 − 4 ] = [ 2 0 1 1 ] ∴ X = [ 5 0 1 4 ] , Y = [ 2 0 1 1 ] . \Rightarrow Y = \begin{bmatrix} 7 & 0 \\ 2 & 5 \end{bmatrix} - X \\[1em] = \begin{bmatrix} 7 & 0 \\ 2 & 5 \end{bmatrix} - \begin{bmatrix} 5 & 0 \\ 1 & 4 \end{bmatrix} \\[1em] = \begin{bmatrix} 7 - 5 & 0 - 0 \\ 2 - 1 & 5 - 4 \end{bmatrix} \\[1em] = \begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix} \\[1.5em] \therefore X = \begin{bmatrix} 5 & 0 \\ 1 & 4 \end{bmatrix} , Y = \begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix}. ⇒ Y = [ 7 2 0 5 ] − X = [ 7 2 0 5 ] − [ 5 1 0 4 ] = [ 7 − 5 2 − 1 0 − 0 5 − 4 ] = [ 2 1 0 1 ] ∴ X = [ 5 1 0 4 ] , Y = [ 2 1 0 1 ] .
Hence, X = [ 5 0 1 4 ] and Y = [ 2 0 1 1 ] . X = \begin{bmatrix} 5 & 0 \\ 1 & 4 \end{bmatrix} \text{ and } Y = \begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix}. X = [ 5 1 0 4 ] and Y = [ 2 1 0 1 ] .
If 2 [ x 7 9 y − 5 ] + [ 6 − 7 4 5 ] = [ 10 7 22 15 ] , 2\begin{bmatrix*}[r] x & 7 \\ 9 & y - 5 \end{bmatrix*} + \begin{bmatrix*}[r] 6 & -7 \\ 4 & 5 \end{bmatrix*} = \begin{bmatrix*}[r] 10 & 7 \\ 22 & 15 \end{bmatrix*}, 2 [ x 9 7 y − 5 ] + [ 6 4 − 7 5 ] = [ 10 22 7 15 ] ,
Answer
Given,
⇒ 2 [ x 7 9 y − 5 ] + [ 6 − 7 4 5 ] = [ 10 7 22 15 ] ⇒ [ 2 x 14 18 2 y − 10 ] + [ 6 − 7 4 5 ] = [ 10 7 22 15 ] ⇒ [ 2 x + 6 14 + ( − 7 ) 18 + 4 2 y − 10 + 5 ] = [ 10 7 22 15 ] ⇒ 2 x + 6 = 10 and 2 y − 5 = 15 ⇒ 2 x = 4 and 2 y = 20 \Rightarrow 2\begin{bmatrix*}[r] x & 7 \\ 9 & y - 5 \end{bmatrix*} + \begin{bmatrix*}[r] 6 & -7 \\ 4 & 5 \end{bmatrix*} = \begin{bmatrix*}[r] 10 & 7 \\ 22 & 15 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2x & 14 \\ 18 & 2y - 10 \end{bmatrix*} + \begin{bmatrix*}[r] 6 & -7 \\ 4 & 5 \end{bmatrix*} = \begin{bmatrix*}[r] 10 & 7 \\ 22 & 15 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2x + 6 & 14 + (-7) \\ 18 + 4 & 2y - 10 + 5 \end{bmatrix*} = \begin{bmatrix*}[r] 10 & 7 \\ 22 & 15 \end{bmatrix*} \\[1em] \Rightarrow 2x + 6 = 10 \text{ and } 2y - 5 = 15 \\[1em] \Rightarrow 2x = 4 \text{ and } 2y = 20 \\[1em] ⇒ 2 [ x 9 7 y − 5 ] + [ 6 4 − 7 5 ] = [ 10 22 7 15 ] ⇒ [ 2 x 18 14 2 y − 10 ] + [ 6 4 − 7 5 ] = [ 10 22 7 15 ] ⇒ [ 2 x + 6 18 + 4 14 + ( − 7 ) 2 y − 10 + 5 ] = [ 10 22 7 15 ] ⇒ 2 x + 6 = 10 and 2 y − 5 = 15 ⇒ 2 x = 4 and 2 y = 20
∴ x = 2 and y = 10.
Hence, the value of x = 2 and y = 10.
If 2 [ 3 4 5 x ] + [ 1 y 0 1 ] = [ z 0 10 5 ] , 2\begin{bmatrix*}[r] 3 & 4 \\ 5 & x \end{bmatrix*} + \begin{bmatrix*}[r] 1 & y \\ 0 & 1 \end{bmatrix*} = \begin{bmatrix*}[r] z & 0 \\ 10 & 5 \end{bmatrix*}, 2 [ 3 5 4 x ] + [ 1 0 y 1 ] = [ z 10 0 5 ] ,
Answer
Given,
⇒ 2 [ 3 4 5 x ] + [ 1 y 0 1 ] = [ z 0 10 5 ] ⇒ [ 6 8 10 2 x ] + [ 1 y 0 1 ] = [ z 0 10 5 ] ⇒ [ 6 + 1 8 + y 10 + 0 2 x + 1 ] = [ z 0 10 5 ] ⇒ [ 7 8 + y 10 2 x + 1 ] = [ z 0 10 5 ] ⇒ 7 = z , 8 + y = 0 and 2 x + 1 = 5 ⇒ z = 7 , y = − 8 and 2 x = 4 ⇒ z = 7 , y = − 8 and x = 2. \Rightarrow 2\begin{bmatrix*}[r] 3 & 4 \\ 5 & x \end{bmatrix*} + \begin{bmatrix*}[r] 1 & y \\ 0 & 1 \end{bmatrix*} = \begin{bmatrix*}[r] z & 0 \\ 10 & 5 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 6 & 8 \\ 10 & 2x \end{bmatrix*} + \begin{bmatrix*}[r] 1 & y \\ 0 & 1 \end{bmatrix*} = \begin{bmatrix*}[r] z & 0 \\ 10 & 5 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 6 + 1 & 8 + y \\ 10 + 0 & 2x + 1 \end{bmatrix*} = \begin{bmatrix*}[r] z & 0 \\ 10 & 5 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 7 & 8 + y \\ 10 & 2x + 1 \end{bmatrix*} = \begin{bmatrix*}[r] z & 0 \\ 10 & 5 \end{bmatrix*} \\[1em] \Rightarrow 7 = z, 8 + y = 0 \text{ and } 2x + 1 = 5 \\[0.5em] \Rightarrow z = 7, y = -8 \text{ and } 2x = 4 \\[0.5em] \Rightarrow z = 7, y = -8 \text{ and } x = 2. ⇒ 2 [ 3 5 4 x ] + [ 1 0 y 1 ] = [ z 10 0 5 ] ⇒ [ 6 10 8 2 x ] + [ 1 0 y 1 ] = [ z 10 0 5 ] ⇒ [ 6 + 1 10 + 0 8 + y 2 x + 1 ] = [ z 10 0 5 ] ⇒ [ 7 10 8 + y 2 x + 1 ] = [ z 10 0 5 ] ⇒ 7 = z , 8 + y = 0 and 2 x + 1 = 5 ⇒ z = 7 , y = − 8 and 2 x = 4 ⇒ z = 7 , y = − 8 and x = 2.
∴ x = 2, y = -8 and z = 7.
Hence, the values are x = 2, y = -8 and z = 7.
If [ 5 2 − 1 y + 1 ] − 2 [ 1 2 x − 1 3 − 2 ] = [ 3 − 8 − 7 2 ] , \begin{bmatrix*}[r] 5 & 2 \\ -1 & y + 1 \end{bmatrix*} - 2\begin{bmatrix*}[r] 1 & 2x - 1 \\ 3 & -2 \end{bmatrix*} = \begin{bmatrix*}[r] 3 & -8 \\ -7 & 2 \end{bmatrix*}, [ 5 − 1 2 y + 1 ] − 2 [ 1 3 2 x − 1 − 2 ] = [ 3 − 7 − 8 2 ] ,
Answer
Given,
⇒ [ 5 2 − 1 y + 1 ] − 2 [ 1 2 x − 1 3 − 2 ] = [ 3 − 8 − 7 2 ] ⇒ [ 5 2 − 1 y + 1 ] − [ 2 4 x − 2 6 − 4 ] = [ 3 − 8 − 7 2 ] ⇒ [ 5 − 2 2 − 4 x + 2 − 1 − 6 y + 1 − ( − 4 ) ] = [ 3 − 8 − 7 2 ] ⇒ [ 3 4 − 4 x − 7 y + 5 ] = [ 3 − 8 − 7 2 ] ⇒ 4 − 4 x = − 8 and y + 5 = 2 ⇒ 4 x = 8 + 4 and y = 2 − 5 ⇒ 4 x = 12 and y = − 3 \Rightarrow \begin{bmatrix*}[r] 5 & 2 \\ -1 & y + 1 \end{bmatrix*} - 2\begin{bmatrix*}[r] 1 & 2x - 1 \\ 3 & -2 \end{bmatrix*} = \begin{bmatrix*}[r] 3 & -8 \\ -7 & 2 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 5 & 2 \\ -1 & y + 1 \end{bmatrix*} - \begin{bmatrix*}[r] 2 & 4x - 2 \\ 6 & -4 \end{bmatrix*} = \begin{bmatrix*}[r] 3 & -8 \\ -7 & 2 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 5 - 2 & 2 - 4x + 2 \\ -1 - 6 & y + 1 - (-4) \end{bmatrix*} = \begin{bmatrix*}[r] 3 & -8 \\ -7 & 2 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 3 & 4 - 4x \\ -7 & y + 5 \end{bmatrix*} = \begin{bmatrix*}[r] 3 & -8 \\ -7 & 2 \end{bmatrix*} \\[1em] \Rightarrow 4 - 4x = -8 \text{ and } y + 5 = 2 \\[0.5em] \Rightarrow 4x = 8 + 4 \text{ and } y = 2 - 5 \\[0.5em] \Rightarrow 4x = 12 \text{ and } y = -3 \\[0.5em] ⇒ [ 5 − 1 2 y + 1 ] − 2 [ 1 3 2 x − 1 − 2 ] = [ 3 − 7 − 8 2 ] ⇒ [ 5 − 1 2 y + 1 ] − [ 2 6 4 x − 2 − 4 ] = [ 3 − 7 − 8 2 ] ⇒ [ 5 − 2 − 1 − 6 2 − 4 x + 2 y + 1 − ( − 4 ) ] = [ 3 − 7 − 8 2 ] ⇒ [ 3 − 7 4 − 4 x y + 5 ] = [ 3 − 7 − 8 2 ] ⇒ 4 − 4 x = − 8 and y + 5 = 2 ⇒ 4 x = 8 + 4 and y = 2 − 5 ⇒ 4 x = 12 and y = − 3
∴ x = 3 and y = -3.
Hence, the value of x = 3 and y = -3.
If [ a 3 4 2 ] + [ 2 b 1 − 2 ] − [ 1 1 − 2 c ] = [ 5 0 7 3 ] , \begin{bmatrix*}[r] a & 3 \\ 4 & 2 \end{bmatrix*} + \begin{bmatrix*}[r] 2 & b \\ 1 & -2 \end{bmatrix*} - \begin{bmatrix*}[r] 1 & 1 \\ -2 & c \end{bmatrix*} = \begin{bmatrix*}[r] 5 & 0 \\ 7 & 3 \end{bmatrix*}, [ a 4 3 2 ] + [ 2 1 b − 2 ] − [ 1 − 2 1 c ] = [ 5 7 0 3 ] ,
Answer
Given,
⇒ [ a 3 4 2 ] + [ 2 b 1 − 2 ] − [ 1 1 − 2 c ] = [ 5 0 7 3 ] ⇒ [ a + 2 − 1 3 + b − 1 4 + 1 − ( − 2 ) 2 + ( − 2 ) − c ] = [ 5 0 7 3 ] ⇒ [ a + 1 b + 2 7 − c ] = [ 5 0 7 3 ] ⇒ a + 1 = 5 , b + 2 = 0 and − c = 3 \Rightarrow \begin{bmatrix*}[r] a & 3 \\ 4 & 2 \end{bmatrix*} + \begin{bmatrix*}[r] 2 & b \\ 1 & -2 \end{bmatrix*} - \begin{bmatrix*}[r] 1 & 1 \\ -2 & c \end{bmatrix*} = \begin{bmatrix*}[r] 5 & 0 \\ 7 & 3 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] a + 2 - 1 & 3 + b - 1 \\ 4 + 1 -(-2) & 2 + (-2) - c \end{bmatrix*} = \begin{bmatrix*}[r] 5 & 0 \\ 7 & 3 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] a + 1 & b + 2 \\ 7 & -c \end{bmatrix*} = \begin{bmatrix*}[r] 5 & 0 \\ 7 & 3 \end{bmatrix*} \\[1em] \Rightarrow a + 1 = 5, b + 2 = 0 \text{ and } -c = 3 \\[1em] ⇒ [ a 4 3 2 ] + [ 2 1 b − 2 ] − [ 1 − 2 1 c ] = [ 5 7 0 3 ] ⇒ [ a + 2 − 1 4 + 1 − ( − 2 ) 3 + b − 1 2 + ( − 2 ) − c ] = [ 5 7 0 3 ] ⇒ [ a + 1 7 b + 2 − c ] = [ 5 7 0 3 ] ⇒ a + 1 = 5 , b + 2 = 0 and − c = 3
∴ a = 4, b = -2 and c = -3.
Hence, the values are a = 4, b = -2 and c = -3.
If A = [ 2 a − 3 5 ] , B = [ − 2 3 7 b ] , C = [ c 9 − 1 − 11 ] A = \begin{bmatrix*}[r] 2 & a \\ -3 & 5 \end{bmatrix*}, B = \begin{bmatrix*}[r] -2 & 3 \\ 7 & b \end{bmatrix*}, C = \begin{bmatrix*}[r] c & 9 \\ -1 & -11 \end{bmatrix*} A = [ 2 − 3 a 5 ] , B = [ − 2 7 3 b ] , C = [ c − 1 9 − 11 ]
Answer
Given, 5A + 2B = C
⇒ 5 [ 2 a − 3 5 ] + 2 [ − 2 3 7 b ] = [ c 9 − 1 − 11 ] ⇒ [ 10 5 a − 15 25 ] + [ − 4 6 14 2 b ] = [ c 9 − 1 − 11 ] ⇒ [ 10 + ( − 4 ) 5 a + 6 − 15 + 14 25 + 2 b ] = [ c 9 − 1 − 11 ] ⇒ [ 6 5 a + 6 − 1 25 + 2 b ] = [ c 9 − 1 − 11 ] ⇒ 6 = c , 5 a + 6 = 9 and 25 + 2 b = − 11 ⇒ c = 6 , 5 a = 3 and 2 b = − 11 − 25 ⇒ c = 6 , a = 3 5 and 2 b = − 36 ⇒ c = 6 , a = 3 5 and b = − 18 ∴ a = 3 5 , b = − 18 and c = 6. \Rightarrow 5\begin{bmatrix*}[r] 2 & a \\ -3 & 5 \end{bmatrix*} + 2\begin{bmatrix*}[r] -2 & 3 \\ 7 & b \end{bmatrix*} = \begin{bmatrix*}[r] c & 9 \\ -1 & -11 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 10 & 5a \\ -15 & 25 \end{bmatrix*} + \begin{bmatrix*}[r] -4 & 6 \\ 14 & 2b \end{bmatrix*} = \begin{bmatrix*}[r] c & 9 \\ -1 & -11 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 10 + (-4) & 5a + 6 \\ -15 + 14 & 25 + 2b \end{bmatrix*} = \begin{bmatrix*}[r] c & 9 \\ -1 & -11 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 6 & 5a + 6 \\ -1 & 25 + 2b \end{bmatrix*} = \begin{bmatrix*}[r] c & 9 \\ -1 & -11 \end{bmatrix*} \\[1em] \Rightarrow 6 = c, 5a + 6 = 9 \text{ and } 25 + 2b = -11 \\[0.5em] \Rightarrow c = 6, 5a = 3 \text{ and } 2b = -11 - 25 \\[0.5em] \Rightarrow c = 6, a = \dfrac{3}{5} \text{ and } 2b = -36 \\[0.5em] \Rightarrow c = 6, a = \dfrac{3}{5} \text{ and } b = -18 \\[0.5em] \therefore a = \dfrac{3}{5}, b = -18 \text{ and } c = 6. ⇒ 5 [ 2 − 3 a 5 ] + 2 [ − 2 7 3 b ] = [ c − 1 9 − 11 ] ⇒ [ 10 − 15 5 a 25 ] + [ − 4 14 6 2 b ] = [ c − 1 9 − 11 ] ⇒ [ 10 + ( − 4 ) − 15 + 14 5 a + 6 25 + 2 b ] = [ c − 1 9 − 11 ] ⇒ [ 6 − 1 5 a + 6 25 + 2 b ] = [ c − 1 9 − 11 ] ⇒ 6 = c , 5 a + 6 = 9 and 25 + 2 b = − 11 ⇒ c = 6 , 5 a = 3 and 2 b = − 11 − 25 ⇒ c = 6 , a = 5 3 and 2 b = − 36 ⇒ c = 6 , a = 5 3 and b = − 18 ∴ a = 5 3 , b = − 18 and c = 6.
Hence, the values are a = 3 5 , \bold{\dfrac{3}{5}}, 5 3 ,
If A = [ 3 5 4 − 2 ] and B = [ 2 4 ] , \begin{bmatrix*}[r] 3 & 5 \\ 4 & -2 \end{bmatrix*} \text{ and B } = \begin{bmatrix*}[r] 2 \\ 4 \end{bmatrix*}, [ 3 4 5 − 2 ] and B = [ 2 4 ] ,
Answer
The product is possible because number of rows in A = number of columns in B =
A B = [ 3 5 4 − 2 ] [ 2 4 ] = [ 3.2 + 5.4 4.2 + ( − 2 ) .4 ] = [ 26 0 ] AB = \begin{bmatrix*}[r] 3 & 5 \\ 4 & -2 \end{bmatrix*} \begin{bmatrix*}[r] 2 \\ 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 3.2 + 5.4 \\ 4.2 + (-2).4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 26 \\ 0 \end{bmatrix*} A B = [ 3 4 5 − 2 ] [ 2 4 ] = [ 3.2 + 5.4 4.2 + ( − 2 ) .4 ] = [ 26 0 ]
Hence, the matrix AB = [ 26 0 ] . \begin{bmatrix*}[r] 26 \\ 0 \end{bmatrix*}. [ 26 0 ] .
If A = [ 2 5 1 3 ] , B = [ 1 − 1 − 3 2 ] , \begin{bmatrix*}[r] 2 & 5 \\ 1 & 3 \end{bmatrix*}, \text{ B } = \begin{bmatrix*}[r] 1 & -1 \\ -3 & 2 \end{bmatrix*}, [ 2 1 5 3 ] , B = [ 1 − 3 − 1 2 ] ,
Answer
AB = [ 2 5 1 3 ] [ 1 − 1 − 3 2 ] = [ 2.1 + 5. ( − 3 ) 2. ( − 1 ) + 5.2 1.1 + 3. ( − 3 ) 1. ( − 1 ) + 3.2 ] = [ − 13 8 − 8 5 ] BA = [ 1 − 1 − 3 2 ] [ 2 5 1 3 ] = [ 1 × 2 + ( − 1 ) × 1 1 × 5 + ( − 1 ) × 3 − 3 × 2 + 2 × 1 − 3 × 5 + 2 × 3 ] = [ 2 − 1 5 − 3 − 6 + 2 − 15 + 6 ] = [ 1 2 − 4 − 9 ] \text{ AB } = \begin{bmatrix*}[r] 2 & 5 \\ 1 & 3 \end{bmatrix*} \begin{bmatrix*}[r] 1 & -1 \\ -3 & 2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2.1 + 5.(-3) & 2.(-1) + 5.2 \\ 1.1 + 3.(-3) & 1.(-1) + 3.2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -13 & 8 \\ -8 & 5 \end{bmatrix*} \\[1em] \text{ BA } = \begin{bmatrix*}[r] 1 & -1 \\ -3 & 2 \end{bmatrix*} \begin{bmatrix*}[r] 2 & 5 \\ 1 & 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 \times 2 + (-1) \times 1 & 1 \times 5 + (-1) \times 3 \\ -3 \times 2 + 2 \times 1 & -3 \times 5 + 2 \times 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 - 1 & 5 - 3 \\ -6 + 2 & -15 + 6 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 & 2 \\ -4 & -9 \end{bmatrix*} AB = [ 2 1 5 3 ] [ 1 − 3 − 1 2 ] = [ 2.1 + 5. ( − 3 ) 1.1 + 3. ( − 3 ) 2. ( − 1 ) + 5.2 1. ( − 1 ) + 3.2 ] = [ − 13 − 8 8 5 ] BA = [ 1 − 3 − 1 2 ] [ 2 1 5 3 ] = [ 1 × 2 + ( − 1 ) × 1 − 3 × 2 + 2 × 1 1 × 5 + ( − 1 ) × 3 − 3 × 5 + 2 × 3 ] = [ 2 − 1 − 6 + 2 5 − 3 − 15 + 6 ] = [ 1 − 4 2 − 9 ]
The matrix AB = [ − 13 8 − 8 5 ] and BA = [ 1 2 − 4 − 9 ] . \begin{bmatrix*}[r] -13 & 8 \\ -8 & 5 \end{bmatrix*} \text{ and BA } = \begin{bmatrix*}[r] 1 & 2 \\ -4 & -9 \end{bmatrix*}. [ − 13 − 8 8 5 ] and BA = [ 1 − 4 2 − 9 ] .
If A = [ 3 7 2 4 ] , B = [ 0 2 5 3 ] and C = [ 1 − 5 − 4 6 ] , \begin{bmatrix*}[r] 3 & 7 \\ 2 & 4 \end{bmatrix*}, \text{ B } = \begin{bmatrix*}[r] 0 & 2 \\ 5 & 3 \end{bmatrix*} \text{ and C } = \begin{bmatrix*}[r] 1 & -5 \\ -4 & 6 \end{bmatrix*}, [ 3 2 7 4 ] , B = [ 0 5 2 3 ] and C = [ 1 − 4 − 5 6 ] ,
Answer
A B − 5 C = [ 3 7 2 4 ] [ 0 2 5 3 ] − 5 [ 1 − 5 − 4 6 ] = [ 3 × 0 + 7 × 5 3 × 2 + 7 × 3 2 × 0 + 4 × 5 2 × 2 + 4 × 3 ] − [ 5 − 25 − 20 30 ] = [ 0 + 35 6 + 21 0 + 20 4 + 12 ] − [ 5 − 25 − 20 30 ] = [ 35 − 5 27 − ( − 25 ) 20 − ( − 20 ) 16 − 30 ] = [ 30 52 40 − 14 ] AB - 5C = \begin{bmatrix*}[r] 3 & 7 \\ 2 & 4 \end{bmatrix*}\begin{bmatrix*}[r] 0 & 2 \\ 5 & 3 \end{bmatrix*} - 5\begin{bmatrix*}[r] 1 & -5 \\ -4 & 6 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 3 \times 0 + 7 \times 5 & 3 \times 2 + 7 \times 3 \\ 2 \times 0 + 4 \times 5 & 2 \times 2 + 4 \times 3 \end{bmatrix*} \\[1em] - \begin{bmatrix*}[r] 5 & -25 \\ -20 & 30 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 0 + 35 & 6 + 21 \\ 0 + 20 & 4 + 12 \end{bmatrix*} - \begin{bmatrix*}[r] 5 & -25 \\ -20 & 30 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 35 - 5 & 27 - (-25) \\ 20 - (-20) & 16 - 30 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 30 & 52 \\ 40 & -14 \end{bmatrix*} A B − 5 C = [ 3 2 7 4 ] [ 0 5 2 3 ] − 5 [ 1 − 4 − 5 6 ] = [ 3 × 0 + 7 × 5 2 × 0 + 4 × 5 3 × 2 + 7 × 3 2 × 2 + 4 × 3 ] − [ 5 − 20 − 25 30 ] = [ 0 + 35 0 + 20 6 + 21 4 + 12 ] − [ 5 − 20 − 25 30 ] = [ 35 − 5 20 − ( − 20 ) 27 − ( − 25 ) 16 − 30 ] = [ 30 40 52 − 14 ]
Hence, the matrix AB - 5C = [ 30 52 40 − 14 ] . \begin{bmatrix*}[r] 30 & 52 \\ 40 & -14 \end{bmatrix*}. [ 30 40 52 − 14 ] .
If A = [ 1 2 2 1 ] and B = [ 2 1 1 2 ] , \begin{bmatrix*}[r] 1 & 2 \\ 2 & 1 \end{bmatrix*} \text{ and B } = \begin{bmatrix*}[r] 2 & 1 \\ 1 & 2 \end{bmatrix*}, [ 1 2 2 1 ] and B = [ 2 1 1 2 ] ,
Answer
BA = [ 2 1 1 2 ] [ 1 2 2 1 ] = [ 2 × 1 + 1 × 2 2 × 2 + 1 × 1 1 × 1 + 2 × 2 1 × 2 + 2 × 1 ] = [ 2 + 2 4 + 1 1 + 4 2 + 2 ] = [ 4 5 5 4 ] ∴ A(BA) = A × BA = [ 1 2 2 1 ] [ 4 5 5 4 ] = [ 1 × 4 + 2 × 5 1 × 5 + 2 × 4 2 × 4 + 1 × 5 2 × 5 + 1 × 4 ] = [ 4 + 10 5 + 8 8 + 5 10 + 4 ] = [ 14 13 13 14 ] \text{BA } = \begin{bmatrix*}[r] 2 & 1 \\ 1 & 2 \end{bmatrix*} \begin{bmatrix*}[r] 1 & 2 \\ 2 & 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 \times 1 + 1 \times 2 & 2 \times 2 + 1 \times 1 \\ 1 \times 1 + 2 \times 2 & 1 \times 2 + 2 \times 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 + 2 & 4 + 1 \\ 1 + 4 & 2 + 2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 4 & 5 \\ 5 & 4 \end{bmatrix*} \\[1.5em] \therefore \text{A(BA) } = \text{A} \times \text{BA} \\[1em] = \begin{bmatrix*}[r] 1 & 2 \\ 2 & 1 \end{bmatrix*} \begin{bmatrix*}[r] 4 & 5 \\ 5 & 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 \times 4 + 2 \times 5 & 1 \times 5 + 2 \times 4 \\ 2 \times 4 + 1 \times 5 & 2 \times 5 + 1 \times 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 4 + 10 & 5 + 8 \\ 8 + 5 & 10 + 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 14 & 13 \\ 13 & 14 \end{bmatrix*} BA = [ 2 1 1 2 ] [ 1 2 2 1 ] = [ 2 × 1 + 1 × 2 1 × 1 + 2 × 2 2 × 2 + 1 × 1 1 × 2 + 2 × 1 ] = [ 2 + 2 1 + 4 4 + 1 2 + 2 ] = [ 4 5 5 4 ] ∴ A(BA) = A × BA = [ 1 2 2 1 ] [ 4 5 5 4 ] = [ 1 × 4 + 2 × 5 2 × 4 + 1 × 5 1 × 5 + 2 × 4 2 × 5 + 1 × 4 ] = [ 4 + 10 8 + 5 5 + 8 10 + 4 ] = [ 14 13 13 14 ]
Hence, the matrix A(BA) = [ 14 13 13 14 ] . \begin{bmatrix*}[r] 14 & 13 \\ 13 & 14 \end{bmatrix*}. [ 14 13 13 14 ] .
Given the matrices :
A = [ 2 1 4 2 ] , B = [ 3 4 − 1 − 2 ] and C = [ − 3 1 0 − 2 ] . \text { A } = \begin{bmatrix*}[r] 2 & 1 \\ 4 & 2 \end{bmatrix*}, \text{ B } = \begin{bmatrix*}[r] 3 & 4 \\ -1 & -2 \end{bmatrix*} \text{ and C } = \begin{bmatrix*}[r] -3 & 1 \\ 0 & -2 \end{bmatrix*} . A = [ 2 4 1 2 ] , B = [ 3 − 1 4 − 2 ] and C = [ − 3 0 1 − 2 ] .
Find the products of (i) ABC (ii) ACB and state whether they are equal.
Answer
(i)
ABC = [ 2 1 4 2 ] [ 3 4 − 1 − 2 ] [ − 3 1 0 − 2 ] = [ 2 × 3 + 1 × ( − 1 ) 2 × 4 + 1 × ( − 2 ) 4 × 3 + 2 × ( − 1 ) 4 × 4 + 2 × ( − 2 ) ] [ − 3 1 0 − 2 ] = [ 6 − 1 8 − 2 12 − 2 16 − 4 ] [ − 3 1 0 − 2 ] = [ 5 6 10 12 ] [ − 3 1 0 − 2 ] = [ 5 × ( − 3 ) + 6 × 0 5 × 1 + 6 × ( − 2 ) 10 × ( − 3 ) + 12 × 0 10 × 1 + 12 × ( − 2 ) ] = [ − 15 + 0 5 − 12 − 30 + 0 10 + ( − 24 ) ] = [ − 15 − 7 − 30 − 14 ] \text{ABC } = \begin{bmatrix} 2 & 1 \\ 4 & 2 \end{bmatrix}\begin{bmatrix*}[r] 3 & 4 \\ -1 & -2 \end{bmatrix*}\begin{bmatrix*}[r] -3 & 1 \\ 0 & -2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 \times 3 + 1 \times (-1) & 2 \times 4 + 1 \times (-2) \\ 4 \times 3 + 2 \times (-1) & 4 \times 4 + 2 \times (-2) \end{bmatrix*} \begin{bmatrix*}[r] -3 & 1 \\ 0 & -2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 6 - 1 & 8 - 2 \\ 12 - 2 & 16 - 4 \end{bmatrix*} \begin{bmatrix*}[r] -3 & 1 \\ 0 & -2 \end{bmatrix*} \\[1em] = \begin{bmatrix} 5 & 6 \\ 10 & 12 \end{bmatrix} \begin{bmatrix*}[r] -3 & 1 \\ 0 & -2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 5 \times (-3) + 6 \times 0 & 5 \times 1 + 6 \times (-2) \\ 10 \times (-3) + 12 \times 0 & 10 \times 1 + 12 \times (-2) \end{bmatrix*} \\[1em] = \begin{bmatrix} -15 + 0 & 5 - 12 \\ -30 + 0 & 10 + (-24) \end{bmatrix} \\[1em] = \begin{bmatrix} -15 & -7 \\ -30 & -14 \end{bmatrix} \\[1em] ABC = [ 2 4 1 2 ] [ 3 − 1 4 − 2 ] [ − 3 0 1 − 2 ] = [ 2 × 3 + 1 × ( − 1 ) 4 × 3 + 2 × ( − 1 ) 2 × 4 + 1 × ( − 2 ) 4 × 4 + 2 × ( − 2 ) ] [ − 3 0 1 − 2 ] = [ 6 − 1 12 − 2 8 − 2 16 − 4 ] [ − 3 0 1 − 2 ] = [ 5 10 6 12 ] [ − 3 0 1 − 2 ] = [ 5 × ( − 3 ) + 6 × 0 10 × ( − 3 ) + 12 × 0 5 × 1 + 6 × ( − 2 ) 10 × 1 + 12 × ( − 2 ) ] = [ − 15 + 0 − 30 + 0 5 − 12 10 + ( − 24 ) ] = [ − 15 − 30 − 7 − 14 ]
Hence, the matrix ABC = [ − 15 − 7 − 30 − 14 ] . \begin{bmatrix} -15 & -7 \\ -30 & -14 \end{bmatrix} . [ − 15 − 30 − 7 − 14 ] .
(ii)
ACB = [ 2 1 4 2 ] [ − 3 1 0 − 2 ] [ 3 4 − 1 − 2 ] = [ 2 × ( − 3 ) + 1 × 0 2 × 1 + 1 × ( − 2 ) 4 × ( − 3 ) + 2 × 0 4 × 1 + 2 × ( − 2 ) ] [ 3 4 − 1 − 2 ] = [ − 6 + 0 2 − 2 − 12 + 0 4 − 4 ] [ 3 4 − 1 − 2 ] = [ − 6 0 − 12 0 ] [ 3 4 − 1 − 2 ] = [ − 6 × 3 + 0 × ( − 1 ) − 6 × 4 + 0 × ( − 2 ) − 12 × 3 + 0 × ( − 1 ) − 12 × 4 + 0 × ( − 2 ) ] = [ − 18 + 0 − 24 + 0 − 36 + 0 − 48 + 0 ] = [ − 18 − 24 − 36 − 48 ] \text{ACB } = \begin{bmatrix*}[r] 2 & 1 \\ 4 & 2 \end{bmatrix*} \begin{bmatrix*}[r] -3 & 1 \\ 0 & -2 \end{bmatrix*} \begin{bmatrix*}[r] 3 & 4 \\ -1 & -2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 \times (-3) + 1 \times 0 & 2 \times 1 + 1 \times (-2) \\ 4 \times (-3) + 2 \times 0 & 4 \times 1 + 2 \times (-2) \end{bmatrix*} \begin{bmatrix*}[r] 3 & 4 \\ -1 & -2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -6 + 0 & 2 - 2 \\ -12 + 0 & 4 - 4 \end{bmatrix*} \begin{bmatrix*}[r] 3 & 4 \\ -1 & -2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -6 & 0 \\ -12 & 0 \end{bmatrix*} \begin{bmatrix*}[r] 3 & 4 \\ -1 & -2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -6 \times 3 + 0 \times (-1) & -6 \times 4 + 0 \times (-2) \\ -12 \times 3 + 0 \times (-1) & -12 \times 4 + 0 \times (-2) \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -18 + 0 & -24 + 0 \\ -36 + 0 & -48 + 0 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -18 & -24 \\ -36 & -48 \end{bmatrix*} \\[1em] ACB = [ 2 4 1 2 ] [ − 3 0 1 − 2 ] [ 3 − 1 4 − 2 ] = [ 2 × ( − 3 ) + 1 × 0 4 × ( − 3 ) + 2 × 0 2 × 1 + 1 × ( − 2 ) 4 × 1 + 2 × ( − 2 ) ] [ 3 − 1 4 − 2 ] = [ − 6 + 0 − 12 + 0 2 − 2 4 − 4 ] [ 3 − 1 4 − 2 ] = [ − 6 − 12 0 0 ] [ 3 − 1 4 − 2 ] = [ − 6 × 3 + 0 × ( − 1 ) − 12 × 3 + 0 × ( − 1 ) − 6 × 4 + 0 × ( − 2 ) − 12 × 4 + 0 × ( − 2 ) ] = [ − 18 + 0 − 36 + 0 − 24 + 0 − 48 + 0 ] = [ − 18 − 36 − 24 − 48 ]
Hence, the matrix ACB = [ − 18 − 24 − 36 − 48 ] , \begin{bmatrix} -18 & -24 \\ -36 & -48 \end{bmatrix} , [ − 18 − 36 − 24 − 48 ] ,
Evaluate : [ 4 sin 30° 2 cos 60° sin 90° 2 cos 0° ] [ 4 5 5 4 ] . \begin{bmatrix} \text{4 sin 30° } & \text {2 cos 60°} \\ \text{ sin 90° } & \text{ 2 cos 0°} \end{bmatrix} \begin{bmatrix} 4 & 5 \\ 5 & 4 \end{bmatrix}. [ 4 sin 30° sin 90° 2 cos 60° 2 cos 0° ] [ 4 5 5 4 ] .
Answer
[ 4 sin 30° 2 cos 60° sin 90° 2 cos 0° ] [ 4 5 5 4 ] Since, sin 30° = cos 60° = 1 2 , sin 90° = cos 0° = 1. ⇒ [ 4 × 1 2 2 × 1 2 1 2 × 1 ] [ 4 5 5 4 ] = [ 2 1 1 2 ] [ 4 5 5 4 ] = [ 2 × 4 + 1 × 5 2 × 5 + 1 × 4 1 × 4 + 2 × 5 1 × 5 + 2 × 4 ] = [ 8 + 5 10 + 4 4 + 10 5 + 8 ] = [ 13 14 14 13 ] . \begin{bmatrix} \text{4 sin 30° } & \text {2 cos 60°} \\ \text{ sin 90° } & \text{ 2 cos 0°} \end{bmatrix} \begin{bmatrix} 4 & 5 \\ 5 & 4 \end{bmatrix} \\[1em] \text{ Since, sin 30° = cos 60° } = \dfrac{1}{2}, \text { sin 90° = cos 0° = 1.} \\[1em] \Rightarrow \begin{bmatrix} 4 \times \dfrac{1}{2} & 2 \times \dfrac{1}{2} \\ 1 & 2 \times 1 \end{bmatrix} \begin{bmatrix} 4 & 5 \\ 5 & 4 \end{bmatrix} \\[1em] = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} 4 & 5 \\ 5 & 4 \end{bmatrix} \\[1em] = \begin{bmatrix} 2 \times 4 + 1 \times 5 & 2 \times 5 + 1 \times 4 \\ 1 \times 4 + 2 \times 5 & 1 \times 5 + 2 \times 4 \end{bmatrix} \\[1em] = \begin{bmatrix} 8 + 5 & 10 + 4 \\ 4 + 10 & 5 + 8 \end{bmatrix} \\[1em] = \begin{bmatrix} 13 & 14 \\ 14 & 13 \end{bmatrix} . [ 4 sin 30° sin 90° 2 cos 60° 2 cos 0° ] [ 4 5 5 4 ] Since, sin 30° = cos 60° = 2 1 , sin 90° = cos 0° = 1. ⇒ [ 4 × 2 1 1 2 × 2 1 2 × 1 ] [ 4 5 5 4 ] = [ 2 1 1 2 ] [ 4 5 5 4 ] = [ 2 × 4 + 1 × 5 1 × 4 + 2 × 5 2 × 5 + 1 × 4 1 × 5 + 2 × 4 ] = [ 8 + 5 4 + 10 10 + 4 5 + 8 ] = [ 13 14 14 13 ] .
Hence, the resultant matrix is [ 13 14 14 13 ] . \begin{bmatrix} 13 & 14 \\ 14 & 13 \end{bmatrix}. [ 13 14 14 13 ] .
If A = [ − 1 3 2 4 ] and B = [ 2 − 3 − 4 − 6 ] , \begin{bmatrix*}[r] -1 & 3 \\ 2 & 4 \end{bmatrix*} \text{ and B } = \begin{bmatrix*}[r] 2 & -3 \\ -4 & -6 \end{bmatrix*}, [ − 1 2 3 4 ] and B = [ 2 − 4 − 3 − 6 ] ,
Answer
AB = [ − 1 3 2 4 ] [ 2 − 3 − 4 − 6 ] = [ − 1 × 2 + 3 × − 4 − 1 × ( − 3 ) + 3 × ( − 6 ) 2 × 2 + 4 × ( − 4 ) 2 × ( − 3 ) + 4 × ( − 6 ) ] = [ − 2 − 12 3 − 18 4 − 16 − 6 − 24 ] = [ − 14 − 15 − 12 − 30 ] BA = [ 2 − 3 − 4 − 6 ] [ − 1 3 2 4 ] = [ 2 × ( − 1 ) + ( − 3 ) × 2 2 × 3 + ( − 3 ) × 4 ( − 4 ) × ( − 1 ) + ( − 6 ) × 2 ( − 4 ) × 3 + ( − 6 ) × 4 ] = [ − 2 − 6 6 − 12 4 − 12 − 12 − 24 ] = [ − 8 − 6 − 8 − 36 ] Given, AB + BA = [ − 14 − 15 − 12 − 30 ] + [ − 8 − 6 − 8 − 36 ] = [ − 14 + ( − 8 ) − 15 + ( − 6 ) − 12 + ( − 8 ) − 30 + ( − 36 ) ] = [ − 22 − 21 − 20 − 66 ] . \text{ AB } = \begin{bmatrix*}[r] -1 & 3 \\ 2 & 4 \end{bmatrix*} \begin{bmatrix*}[r] 2 & -3 \\ -4 & -6 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -1 \times 2 + 3 \times -4 & -1 \times (-3) + 3 \times (-6) \\ 2 \times 2 + 4 \times (-4) & 2 \times (-3) + 4 \times (-6) \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -2 - 12 & 3 - 18 \\ 4 - 16 & -6 - 24 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -14 & -15 \\ -12 & -30 \end{bmatrix*} \\[1em] \text{BA } = \begin{bmatrix*}[r] 2 & -3 \\ -4 & -6 \end{bmatrix*} \begin{bmatrix*}[r] -1 & 3 \\ 2 & 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 \times (-1) + (-3) \times 2 & 2 \times 3 + (-3) \times 4 \\ (-4) \times (-1) + (-6) \times 2 & (-4) \times 3 + (-6) \times 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -2 - 6 & 6 - 12 \\ 4 - 12 & -12 - 24 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -8 & -6 \\ -8 & -36 \end{bmatrix*} \\[1em] \text{Given, AB + BA } = \begin{bmatrix*}[r] -14 & -15 \\ -12 & -30 \end{bmatrix*} + \begin{bmatrix*}[r] -8 & -6 \\ -8 & -36 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -14 + (-8) & -15 + (-6) \\ -12 + (-8) & -30 + (-36) \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -22 & -21 \\ -20 & -66 \end{bmatrix*}. AB = [ − 1 2 3 4 ] [ 2 − 4 − 3 − 6 ] = [ − 1 × 2 + 3 × − 4 2 × 2 + 4 × ( − 4 ) − 1 × ( − 3 ) + 3 × ( − 6 ) 2 × ( − 3 ) + 4 × ( − 6 ) ] = [ − 2 − 12 4 − 16 3 − 18 − 6 − 24 ] = [ − 14 − 12 − 15 − 30 ] BA = [ 2 − 4 − 3 − 6 ] [ − 1 2 3 4 ] = [ 2 × ( − 1 ) + ( − 3 ) × 2 ( − 4 ) × ( − 1 ) + ( − 6 ) × 2 2 × 3 + ( − 3 ) × 4 ( − 4 ) × 3 + ( − 6 ) × 4 ] = [ − 2 − 6 4 − 12 6 − 12 − 12 − 24 ] = [ − 8 − 8 − 6 − 36 ] Given, AB + BA = [ − 14 − 12 − 15 − 30 ] + [ − 8 − 8 − 6 − 36 ] = [ − 14 + ( − 8 ) − 12 + ( − 8 ) − 15 + ( − 6 ) − 30 + ( − 36 ) ] = [ − 22 − 20 − 21 − 66 ] .
Hence, the matrix AB + BA = [ − 22 − 21 − 20 − 66 ] . \begin{bmatrix*}[r] -22 & -21 \\ -20 & -66 \end{bmatrix*}. [ − 22 − 20 − 21 − 66 ] .
If A = [ 1 − 2 2 − 1 ] and B = [ 3 2 − 2 1 ] , \begin{bmatrix*}[r] 1 & -2 \\ 2 & -1 \end{bmatrix*} \text{and B } = \begin{bmatrix*}[r] 3 & 2 \\ -2 & 1 \end{bmatrix*}, [ 1 2 − 2 − 1 ] and B = [ 3 − 2 2 1 ] , 2 .
Answer
2 B = 2 [ 3 2 − 2 1 ] = [ 6 4 − 4 2 ] A 2 = [ 1 − 2 2 − 1 ] [ 1 − 2 2 − 1 ] = [ 1 × 1 + ( − 2 ) × 2 1 × ( − 2 ) + ( − 2 ) × ( − 1 ) 2 × 1 + ( − 1 ) × 2 2 × ( − 2 ) + ( − 1 ) × ( − 1 ) ] = [ 1 − 4 − 2 + 2 2 − 2 − 4 + 1 ] = [ − 3 0 0 − 3 ] ∴ 2 B − A 2 = [ 6 4 − 4 2 ] − [ − 3 0 0 − 3 ] = [ 6 − ( − 3 ) 4 − 0 − 4 − 0 2 − ( − 3 ) ] = [ 9 4 − 4 5 ] 2B = 2\begin{bmatrix*}[r] 3 & 2 \\ -2 & 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 6 & 4 \\ -4 & 2 \end{bmatrix*} \\[1em] A^2 = \begin{bmatrix*}[r] 1 & -2 \\ 2 & -1 \end{bmatrix*} \begin{bmatrix*}[r] 1 & -2 \\ 2 & -1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 \times 1 + (-2) \times 2 & 1 \times (-2) + (-2) \times (-1) \\ 2 \times 1 + (-1) \times 2 & 2 \times (-2) + (-1)\times (-1) \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 - 4 & -2 + 2 \\ 2 - 2 & -4 + 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -3 & 0 \\ 0 & -3 \end{bmatrix*} \\[1em] \therefore 2B - A^2 = \begin{bmatrix*}[r] 6 & 4 \\ -4 & 2 \end{bmatrix*} - \begin{bmatrix*}[r] -3 & 0 \\ 0 & -3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 6 - (-3) & 4 - 0 \\ -4 - 0 & 2 - (-3) \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 9 & 4 \\ -4 & 5 \end{bmatrix*} 2 B = 2 [ 3 − 2 2 1 ] = [ 6 − 4 4 2 ] A 2 = [ 1 2 − 2 − 1 ] [ 1 2 − 2 − 1 ] = [ 1 × 1 + ( − 2 ) × 2 2 × 1 + ( − 1 ) × 2 1 × ( − 2 ) + ( − 2 ) × ( − 1 ) 2 × ( − 2 ) + ( − 1 ) × ( − 1 ) ] = [ 1 − 4 2 − 2 − 2 + 2 − 4 + 1 ] = [ − 3 0 0 − 3 ] ∴ 2 B − A 2 = [ 6 − 4 4 2 ] − [ − 3 0 0 − 3 ] = [ 6 − ( − 3 ) − 4 − 0 4 − 0 2 − ( − 3 ) ] = [ 9 − 4 4 5 ]
Hence, the matrix 2B - A2 = [ 9 4 − 4 5 ] . \begin{bmatrix*}[r] 9 & 4 \\ -4 & 5 \end{bmatrix*}. [ 9 − 4 4 5 ] .
If A = [ 1 2 3 4 ] , B = [ 2 1 4 2 ] and C = [ 5 1 7 4 ] , \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}, \text{ B } = \begin{bmatrix} 2 & 1 \\ 4 & 2 \end{bmatrix} \text{ and C } = \begin{bmatrix} 5 & 1 \\ 7 & 4 \end{bmatrix}, [ 1 3 2 4 ] , B = [ 2 4 1 2 ] and C = [ 5 7 1 4 ] ,
(i) A(B + C)
(ii) (B + C)A
Answer
(i) A(B + C)
B + C = [ 2 1 4 2 ] + [ 5 1 7 4 ] = [ 2 + 5 1 + 1 4 + 7 2 + 4 ] = [ 7 2 11 6 ] ∴ A(B + C) = [ 1 2 3 4 ] [ 7 2 11 6 ] = [ 1 × 7 + 2 × 11 1 × 2 + 2 × 6 3 × 7 + 4 × 11 3 × 2 + 4 × 6 ] = [ 7 + 22 2 + 12 21 + 44 6 + 24 ] = [ 29 14 65 30 ] . \text{B + C } = \begin{bmatrix} 2 & 1 \\ 4 & 2 \end{bmatrix} + \begin{bmatrix} 5 & 1 \\ 7 & 4 \end{bmatrix} \\[1em] = \begin{bmatrix} 2 + 5 & 1 + 1 \\ 4 + 7 & 2 + 4 \end{bmatrix} \\[1em] = \begin{bmatrix} 7 & 2 \\ 11 & 6 \end{bmatrix} \\[1em] \therefore \text{ A(B + C) } = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 7 & 2 \\ 11 & 6 \end{bmatrix} \\[1em] = \begin{bmatrix} 1 \times 7 + 2 \times 11 & 1 \times 2 + 2 \times 6 \\ 3 \times 7 + 4 \times 11 & 3 \times 2 + 4 \times 6 \end{bmatrix} \\[1em] = \begin{bmatrix} 7 + 22 & 2 + 12 \\ 21 + 44 & 6 + 24 \end{bmatrix} \\[1em] = \begin{bmatrix} 29 & 14 \\ 65 & 30 \end{bmatrix}. \\[1em] B + C = [ 2 4 1 2 ] + [ 5 7 1 4 ] = [ 2 + 5 4 + 7 1 + 1 2 + 4 ] = [ 7 11 2 6 ] ∴ A(B + C) = [ 1 3 2 4 ] [ 7 11 2 6 ] = [ 1 × 7 + 2 × 11 3 × 7 + 4 × 11 1 × 2 + 2 × 6 3 × 2 + 4 × 6 ] = [ 7 + 22 21 + 44 2 + 12 6 + 24 ] = [ 29 65 14 30 ] .
Hence, the matrix A(B + C) = [ 29 14 65 30 ] . \begin{bmatrix} 29 & 14 \\ 65 & 30 \end{bmatrix}. [ 29 65 14 30 ] .
(ii) (B + C)A
B + C = [ 2 1 4 2 ] + [ 5 1 7 4 ] = [ 2 + 5 1 + 1 4 + 7 2 + 4 ] = [ 7 2 11 6 ] ∴ (B + C)A = [ 7 2 11 6 ] [ 1 2 3 4 ] = [ 7 × 1 + 2 × 3 7 × 2 + 2 × 4 11 × 1 + 6 × 3 11 × 2 + 6 × 4 ] = [ 7 + 6 14 + 8 11 + 18 22 + 24 ] = [ 13 22 29 46 ] . \text{B + C } = \begin{bmatrix} 2 & 1 \\ 4 & 2 \end{bmatrix} + \begin{bmatrix} 5 & 1 \\ 7 & 4 \end{bmatrix} \\[1em] = \begin{bmatrix} 2 + 5 & 1 + 1 \\ 4 + 7 & 2 + 4 \end{bmatrix} \\[1em] = \begin{bmatrix} 7 & 2 \\ 11 & 6 \end{bmatrix} \\[1em] \therefore \text{(B + C)A } = \begin{bmatrix} 7 & 2 \\ 11 & 6 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \\[1em] = \begin{bmatrix} 7 \times 1 + 2 \times 3 & 7 \times 2 + 2 \times 4 \\ 11 \times 1 + 6 \times 3 & 11 \times 2 + 6 \times 4 \end{bmatrix} \\[1em] = \begin{bmatrix} 7 + 6 & 14 + 8 \\ 11 + 18 & 22 + 24 \end{bmatrix} \\[1em] = \begin{bmatrix} 13 & 22 \\ 29 & 46 \end{bmatrix}. B + C = [ 2 4 1 2 ] + [ 5 7 1 4 ] = [ 2 + 5 4 + 7 1 + 1 2 + 4 ] = [ 7 11 2 6 ] ∴ (B + C)A = [ 7 11 2 6 ] [ 1 3 2 4 ] = [ 7 × 1 + 2 × 3 11 × 1 + 6 × 3 7 × 2 + 2 × 4 11 × 2 + 6 × 4 ] = [ 7 + 6 11 + 18 14 + 8 22 + 24 ] = [ 13 29 22 46 ] .
Hence, the matrix (B + C)A = [ 13 22 29 46 ] . \begin{bmatrix} 13 & 22 \\ 29 & 46 \end{bmatrix}. [ 13 29 22 46 ] .
If A = [ 1 2 2 3 ] , B = [ 2 1 3 2 ] and C = [ 1 3 3 1 ] , \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix}, \text{ B } = \begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix} \text{ and C } = \begin{bmatrix} 1 & 3 \\ 3 & 1 \end{bmatrix}, [ 1 2 2 3 ] , B = [ 2 3 1 2 ] and C = [ 1 3 3 1 ] ,
Answer
B − A = [ 2 1 3 2 ] − [ 1 2 2 3 ] = [ 2 − 1 1 − 2 3 − 2 2 − 3 ] = [ 1 − 1 1 − 1 ] ∴ C ( B − A ) = [ 1 3 3 1 ] [ 1 − 1 1 − 1 ] = [ 1 × 1 + 3 × 1 1 × ( − 1 ) + 3 × ( − 1 ) 3 × 1 + 1 × 1 3 × ( − 1 ) + 1 × ( − 1 ) ] = [ 1 + 3 − 1 − 3 3 + 1 − 3 − 1 ] = [ 4 − 4 4 − 4 ] . B - A = \begin{bmatrix*}[r] 2 & 1 \\ 3 & 2 \end{bmatrix*} - \begin{bmatrix*}[r] 1 & 2 \\ 2 & 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 - 1 & 1 - 2 \\ 3 - 2 & 2 - 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 & -1 \\ 1 & -1 \end{bmatrix*} \\[1.5em] \therefore C(B - A) = \begin{bmatrix*}[r] 1 & 3 \\ 3 & 1 \end{bmatrix*} \begin{bmatrix*}[r] 1 & -1 \\ 1 & -1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 \times 1 + 3 \times 1 & 1 \times (-1) + 3 \times (-1) \\ 3 \times 1 + 1 \times 1 & 3 \times (-1) + 1 \times (-1) \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 + 3 & -1 - 3 \\ 3 + 1 & -3 - 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 4 & -4 \\ 4 & -4 \end{bmatrix*}. B − A = [ 2 3 1 2 ] − [ 1 2 2 3 ] = [ 2 − 1 3 − 2 1 − 2 2 − 3 ] = [ 1 1 − 1 − 1 ] ∴ C ( B − A ) = [ 1 3 3 1 ] [ 1 1 − 1 − 1 ] = [ 1 × 1 + 3 × 1 3 × 1 + 1 × 1 1 × ( − 1 ) + 3 × ( − 1 ) 3 × ( − 1 ) + 1 × ( − 1 ) ] = [ 1 + 3 3 + 1 − 1 − 3 − 3 − 1 ] = [ 4 4 − 4 − 4 ] .
Hence, the matrix C(B - A) = [ 4 − 4 4 − 4 ] \begin{bmatrix} 4 & -4 \\ 4 & -4 \end{bmatrix} [ 4 4 − 4 − 4 ]
Let A = [ 1 0 2 1 ] and B = [ 2 3 − 1 0 ] , \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} \text{ and B } = \begin{bmatrix*}[r] 2 & 3 \\ -1 & 0 \end{bmatrix*}, [ 1 2 0 1 ] and B = [ 2 − 1 3 0 ] , 2 + AB + B2 .
Answer
A 2 = [ 1 0 2 1 ] [ 1 0 2 1 ] = [ 1 × 1 + 0 × 2 1 × 0 + 0 × 1 2 × 1 + 1 × 2 2 × 0 + 1 × 1 ] = [ 1 + 0 0 + 0 2 + 2 0 + 1 ] = [ 1 0 4 1 ] A B = [ 1 0 2 1 ] [ 2 3 − 1 0 ] = [ 1 × 2 + 0 × ( − 1 ) 1 × 3 + 0 × 0 2 × 2 + 1 × ( − 1 ) 2 × 3 + 1 × 0 ] = [ 2 + 0 3 + 0 4 − 1 6 + 0 ] = [ 2 3 3 6 ] B 2 = [ 2 3 − 1 0 ] [ 2 3 − 1 0 ] = [ 2 × 2 + 3 × ( − 1 ) 2 × 3 + 3 × 0 ( − 1 ) × 2 + 0 × ( − 1 ) ( − 1 ) × 3 + 0 × 0 ] = [ 4 − 3 6 + 0 − 2 + 0 − 3 + 0 ] = [ 1 6 − 2 − 3 ] . ∴ A 2 + A B + B 2 = [ 1 0 4 1 ] + [ 2 3 3 6 ] + [ 1 6 − 2 − 3 ] = [ 1 + 2 + 1 0 + 3 + 6 4 + 3 + ( − 2 ) 1 + 6 + ( − 3 ) ] = [ 4 9 5 4 ] . A^2 = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} \\[1em] = \begin{bmatrix} 1 \times 1 + 0 \times 2 & 1 \times 0 + 0 \times 1 \\ 2 \times 1 + 1 \times 2 & 2 \times 0 + 1 \times 1 \end{bmatrix} \\[1em] = \begin{bmatrix} 1 + 0 & 0 + 0 \\ 2 + 2 & 0 + 1 \end{bmatrix} \\[1em] = \begin{bmatrix} 1 & 0 \\ 4 & 1 \end{bmatrix} \\[1em] AB = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} \begin{bmatrix*}[r] 2 & 3 \\ -1 & 0 \end{bmatrix*} \\[1em] = \begin{bmatrix} 1 \times 2 + 0 \times (-1) & 1 \times 3 + 0 \times 0 \\ 2 \times 2 + 1 \times (-1) & 2 \times 3 + 1 \times 0 \end{bmatrix} \\[1em] = \begin{bmatrix*}[r] 2 + 0 & 3 + 0 \\ 4 - 1 & 6 + 0 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 & 3 \\ 3 & 6 \end{bmatrix*} \\[1em] B^2 = \begin{bmatrix*}[r] 2 & 3 \\ -1 & 0 \end{bmatrix*} \begin{bmatrix*}[r] 2 & 3 \\ -1 & 0 \end{bmatrix*} \\[1em] = \begin{bmatrix} 2 \times 2 + 3 \times (-1) & 2 \times 3 + 3 \times 0 \\ (-1) \times 2 + 0 \times (-1) & (-1) \times 3 + 0 \times 0 \end{bmatrix} \\[1em] = \begin{bmatrix*}[r] 4 - 3 & 6 + 0 \\ -2 + 0 & -3 + 0 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 & 6 \\ -2 & -3 \end{bmatrix*}. \\[1em] \therefore A^2 + AB + B^2 = \begin{bmatrix} 1 & 0 \\ 4 & 1 \end{bmatrix} + \begin{bmatrix*}[r] 2 & 3 \\ 3 & 6 \end{bmatrix*} + \begin{bmatrix*}[r] 1 & 6 \\ -2 & -3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 + 2 + 1 & 0 + 3 + 6 \\ 4 + 3 + (-2) & 1 + 6 + (-3) \end{bmatrix*} \\[1em] = \begin{bmatrix} 4 & 9 \\ 5 & 4 \end{bmatrix}. A 2 = [ 1 2 0 1 ] [ 1 2 0 1 ] = [ 1 × 1 + 0 × 2 2 × 1 + 1 × 2 1 × 0 + 0 × 1 2 × 0 + 1 × 1 ] = [ 1 + 0 2 + 2 0 + 0 0 + 1 ] = [ 1 4 0 1 ] A B = [ 1 2 0 1 ] [ 2 − 1 3 0 ] = [ 1 × 2 + 0 × ( − 1 ) 2 × 2 + 1 × ( − 1 ) 1 × 3 + 0 × 0 2 × 3 + 1 × 0 ] = [ 2 + 0 4 − 1 3 + 0 6 + 0 ] = [ 2 3 3 6 ] B 2 = [ 2 − 1 3 0 ] [ 2 − 1 3 0 ] = [ 2 × 2 + 3 × ( − 1 ) ( − 1 ) × 2 + 0 × ( − 1 ) 2 × 3 + 3 × 0 ( − 1 ) × 3 + 0 × 0 ] = [ 4 − 3 − 2 + 0 6 + 0 − 3 + 0 ] = [ 1 − 2 6 − 3 ] . ∴ A 2 + A B + B 2 = [ 1 4 0 1 ] + [ 2 3 3 6 ] + [ 1 − 2 6 − 3 ] = [ 1 + 2 + 1 4 + 3 + ( − 2 ) 0 + 3 + 6 1 + 6 + ( − 3 ) ] = [ 4 5 9 4 ] .
Hence, the matrix A2 + AB + B2 = [ 4 9 5 4 ] \begin{bmatrix} 4 & 9 \\ 5 & 4 \end{bmatrix} [ 4 5 9 4 ]
If A = [ 3 0 5 1 ] and B [ − 4 2 1 0 ] \begin{bmatrix*}[r] 3 & 0 \\ 5 & 1 \end{bmatrix*} \text{ and B} \begin{bmatrix*}[r] -4 & 2 \\ 1 & 0 \end{bmatrix*} [ 3 5 0 1 ] and B [ − 4 1 2 0 ] 2 - 2AB + B2 .
Answer
Given, A = [ 3 0 5 1 ] and B = [ − 4 2 1 0 ] \begin{bmatrix*}[r] 3 & 0 \\ 5 & 1 \end{bmatrix*} \text{ and B} = \begin{bmatrix*}[r] -4 & 2 \\ 1 & 0 \end{bmatrix*} [ 3 5 0 1 ] and B = [ − 4 1 2 0 ]
⇒ A 2 = [ 3 0 5 1 ] × [ 3 0 5 1 ] = [ 3 × 3 + 0 × 5 3 × 0 + 0 × 1 5 × 3 + 1 × 5 5 × 0 + 1 × 1 ] = [ 9 + 0 0 + 0 15 + 5 0 + 1 ] = [ 9 0 20 1 ] ⇒ 2AB = 2 × [ 3 0 5 1 ] × [ − 4 2 1 0 ] = [ 6 0 10 2 ] × [ − 4 2 1 0 ] = [ 6 × ( − 4 ) + 0 × 1 6 × 2 + 0 × 0 10 × ( − 4 ) + 2 × 1 10 × 2 + 2 × 0 ] = [ − 24 + 0 12 + 0 − 40 + 2 20 + 0 ] = [ − 24 12 − 38 20 ] ⇒ B 2 = [ − 4 2 1 0 ] × [ − 4 2 1 0 ] = [ − 4 × ( − 4 ) + 2 × 1 − 4 × 2 + 2 × 0 1 × ( − 4 ) + 0 × 1 1 × 2 + 0 × 0 ] = [ 16 + 2 − 8 + 0 − 4 + 0 2 + 0 ] = [ 18 − 8 − 4 2 ] \Rightarrow A^2 =\begin{bmatrix*}[r] 3 & 0 \\ 5 & 1 \end{bmatrix*} \times \begin{bmatrix*}[r] 3 & 0 \\ 5 & 1 \end{bmatrix*}\\[1em] = \begin{bmatrix*}[r] 3 \times 3 + 0 \times 5 & 3 \times 0 + 0 \times 1 \\ 5 \times 3 + 1 \times 5 & 5 \times 0 + 1 \times 1 \end{bmatrix*}\\[1em] = \begin{bmatrix*}[r] 9 + 0 & 0 + 0 \\ 15 + 5 & 0 + 1 \end{bmatrix*}\\[1em] = \begin{bmatrix*}[r] 9 & 0 \\ 20 & 1 \end{bmatrix*}\\[1em] \Rightarrow \text{2AB} = 2 \times \begin{bmatrix*}[r] 3 & 0 \\ 5 & 1 \end{bmatrix*} \times \begin{bmatrix*}[r] -4 & 2 \\ 1 & 0 \end{bmatrix*}\\[1em] = \begin{bmatrix*}[r] 6 & 0 \\ 10 & 2 \end{bmatrix*} \times \begin{bmatrix*}[r] -4 & 2 \\ 1 & 0 \end{bmatrix*}\\[1em] = \begin{bmatrix*}[r] 6 \times (-4) + 0 \times 1 & 6 \times 2 + 0 \times 0 \\ 10 \times (-4) + 2 \times 1 & 10 \times 2 + 2 \times 0 \end{bmatrix*}\\[1em] = \begin{bmatrix*}[r] -24 + 0 & 12 + 0 \\ -40 + 2 & 20 + 0 \end{bmatrix*}\\[1em] = \begin{bmatrix*}[r] -24 & 12 \\ -38 & 20 \end{bmatrix*}\\[1em] \Rightarrow \text{B}^2 = \begin{bmatrix*}[r] -4 & 2 \\ 1 & 0 \end{bmatrix*} \times \begin{bmatrix*}[r] -4 & 2 \\ 1 & 0 \end{bmatrix*}\\[1em] = \begin{bmatrix*}[r] -4 \times (-4) + 2 \times 1 & -4 \times 2 + 2 \times 0 \\ 1 \times (-4) + 0 \times 1 & 1 \times 2 + 0 \times 0 \end{bmatrix*}\\[1em] = \begin{bmatrix*}[r] 16 + 2 & -8 + 0 \\ -4 + 0 & 2 + 0 \end{bmatrix*}\\[1em] = \begin{bmatrix*}[r] 18 & -8 \\ -4 & 2 \end{bmatrix*} ⇒ A 2 = [ 3 5 0 1 ] × [ 3 5 0 1 ] = [ 3 × 3 + 0 × 5 5 × 3 + 1 × 5 3 × 0 + 0 × 1 5 × 0 + 1 × 1 ] = [ 9 + 0 15 + 5 0 + 0 0 + 1 ] = [ 9 20 0 1 ] ⇒ 2AB = 2 × [ 3 5 0 1 ] × [ − 4 1 2 0 ] = [ 6 10 0 2 ] × [ − 4 1 2 0 ] = [ 6 × ( − 4 ) + 0 × 1 10 × ( − 4 ) + 2 × 1 6 × 2 + 0 × 0 10 × 2 + 2 × 0 ] = [ − 24 + 0 − 40 + 2 12 + 0 20 + 0 ] = [ − 24 − 38 12 20 ] ⇒ B 2 = [ − 4 1 2 0 ] × [ − 4 1 2 0 ] = [ − 4 × ( − 4 ) + 2 × 1 1 × ( − 4 ) + 0 × 1 − 4 × 2 + 2 × 0 1 × 2 + 0 × 0 ] = [ 16 + 2 − 4 + 0 − 8 + 0 2 + 0 ] = [ 18 − 4 − 8 2 ]
Substituting values in A2 - 2AB + B2 , we get :
A 2 − 2 A B + B 2 = [ 9 0 20 1 ] − [ − 24 12 − 38 20 ] + [ 18 − 8 − 4 2 ] = [ 9 − ( − 24 ) + 18 0 − 12 + ( − 8 ) 20 − ( − 38 ) + ( − 4 ) 1 − 20 + 2 ] = [ 51 − 20 54 − 17 ] A^2 - 2AB + B^2 = \begin{bmatrix*}[r] 9 & 0 \\ 20 & 1 \end{bmatrix*} - \begin{bmatrix*}[r] -24 & 12 \\ -38 & 20 \end{bmatrix*} + \begin{bmatrix*}[r] 18 & -8 \\ -4 & 2 \end{bmatrix*}\\[1em] = \begin{bmatrix*}[r] 9 - (-24) + 18 & 0 - 12 + (-8) \\ 20 - (-38) + (-4) & 1 - 20 + 2 \end{bmatrix*}\\[1em] = \begin{bmatrix*}[r] 51 & -20 \\ 54 & -17 \end{bmatrix*} A 2 − 2 A B + B 2 = [ 9 20 0 1 ] − [ − 24 − 38 12 20 ] + [ 18 − 4 − 8 2 ] = [ 9 − ( − 24 ) + 18 20 − ( − 38 ) + ( − 4 ) 0 − 12 + ( − 8 ) 1 − 20 + 2 ] = [ 51 54 − 20 − 17 ]
Hence, the value of A2 - 2AB + B2 = [ 51 − 20 54 − 17 ] \begin{bmatrix*}[r] 51 & -20 \\ 54 & -17 \end{bmatrix*} [ 51 54 − 20 − 17 ]
Let A = [ 2 1 0 − 2 ] , B = [ 4 1 − 3 − 2 ] and C = [ − 3 2 − 1 4 ] , \begin{bmatrix*}[r] 2 & 1 \\ 0 & -2 \end{bmatrix*}, \text{ B } = \begin{bmatrix*}[r] 4 & 1 \\ -3 & -2 \end{bmatrix*} \text{ and C } = \begin{bmatrix*}[r] -3 & 2 \\ -1 & 4 \end{bmatrix*}, [ 2 0 1 − 2 ] , B = [ 4 − 3 1 − 2 ] and C = [ − 3 − 1 2 4 ] , 2 + AC - 5B.
Answer
A 2 = [ 2 1 0 − 2 ] [ 2 1 0 − 2 ] = [ 2 × 2 + 1 × 0 2 × 1 + 1 × ( − 2 ) 0 × 2 + ( − 2 ) × 0 0 × 1 + ( − 2 ) × ( − 2 ) ] = [ 4 + 0 2 − 2 0 + 0 0 + 4 ] = [ 4 0 0 4 ] . A C = [ 2 1 0 − 2 ] [ − 3 2 − 1 4 ] = [ 2 × ( − 3 ) + 1 × ( − 1 ) 2 × 2 + 1 × 4 0 × ( − 3 ) + ( − 2 ) × ( − 1 ) 0 × 2 + ( − 2 ) × 4 ] = [ − 6 + ( − 1 ) 4 + 4 0 + 2 0 + ( − 8 ) ] = [ − 7 8 2 − 8 ] . 5 B = 5 [ 4 1 − 3 − 2 ] = [ 20 5 − 15 − 10 ] ∴ A 2 + A C − 5 B = [ 4 0 0 4 ] + [ − 7 8 2 − 8 ] − [ 20 5 − 15 − 10 ] = [ 4 + ( − 7 ) − 20 0 + 8 − 5 0 + 2 − ( − 15 ) 4 + ( − 8 ) − ( − 10 ) ] = [ − 23 3 17 6 ] . A^2 = \begin{bmatrix*}[r] 2 & 1 \\ 0 & -2 \end{bmatrix*} \begin{bmatrix*}[r] 2 & 1 \\ 0 & -2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 \times 2 + 1 \times 0 & 2 \times 1 + 1 \times (-2) \\ 0 \times 2 + (-2) \times 0 & 0 \times 1 + (-2) \times (-2) \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 4 + 0 & 2 - 2 \\ 0 + 0 & 0 + 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 4 & 0 \\ 0 & 4 \end{bmatrix*}. \\[1.5em] AC = \begin{bmatrix*}[r] 2 & 1 \\ 0 & -2 \end{bmatrix*} \begin{bmatrix*}[r] -3 & 2 \\ -1 & 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 \times (-3) + 1 \times (-1) & 2 \times 2 + 1 \times 4 \\ 0 \times (-3) + (-2) \times (-1) & 0 \times 2 + (-2) \times 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -6 + (-1) & 4 + 4 \\ 0 + 2 & 0 + (-8) \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -7 & 8 \\ 2 & -8 \end{bmatrix*}. \\[1.5em] 5B = 5 \begin{bmatrix*}[r] 4 & 1 \\ -3 & -2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 20 & 5 \\ -15 & -10 \end{bmatrix*} \\[1.5em] \therefore A^2 + AC - 5B = \begin{bmatrix*}[r] 4 & 0 \\ 0 & 4 \end{bmatrix*} + \begin{bmatrix*}[r] -7 & 8 \\ 2 & -8 \end{bmatrix*} - \begin{bmatrix*}[r] 20 & 5 \\ -15 & -10 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 4 + (-7) - 20 & 0 + 8 - 5 \\ 0 + 2 - (-15) & 4 + (-8) - (-10) \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -23 & 3 \\ 17 & 6 \end{bmatrix*} . A 2 = [ 2 0 1 − 2 ] [ 2 0 1 − 2 ] = [ 2 × 2 + 1 × 0 0 × 2 + ( − 2 ) × 0 2 × 1 + 1 × ( − 2 ) 0 × 1 + ( − 2 ) × ( − 2 ) ] = [ 4 + 0 0 + 0 2 − 2 0 + 4 ] = [ 4 0 0 4 ] . A C = [ 2 0 1 − 2 ] [ − 3 − 1 2 4 ] = [ 2 × ( − 3 ) + 1 × ( − 1 ) 0 × ( − 3 ) + ( − 2 ) × ( − 1 ) 2 × 2 + 1 × 4 0 × 2 + ( − 2 ) × 4 ] = [ − 6 + ( − 1 ) 0 + 2 4 + 4 0 + ( − 8 ) ] = [ − 7 2 8 − 8 ] . 5 B = 5 [ 4 − 3 1 − 2 ] = [ 20 − 15 5 − 10 ] ∴ A 2 + A C − 5 B = [ 4 0 0 4 ] + [ − 7 2 8 − 8 ] − [ 20 − 15 5 − 10 ] = [ 4 + ( − 7 ) − 20 0 + 2 − ( − 15 ) 0 + 8 − 5 4 + ( − 8 ) − ( − 10 ) ] = [ − 23 17 3 6 ] .
Hence, the matrix A2 + AC - 5B = [ − 23 3 17 6 ] . \begin{bmatrix*}[r] -23 & 3 \\ 17 & 6 \end{bmatrix*} . [ − 23 17 3 6 ] .
If A = [ 2 3 5 7 ] , B = [ 0 4 − 1 7 ] and C = [ 1 0 − 1 4 ] , \begin{bmatrix*}[r] 2 & 3 \\ 5 & 7 \end{bmatrix*}, \text{ B } = \begin{bmatrix*}[r] 0 & 4 \\ -1 & 7 \end{bmatrix*} \text{ and C } = \begin{bmatrix*}[r] 1 & 0 \\ -1 & 4 \end{bmatrix*}, [ 2 5 3 7 ] , B = [ 0 − 1 4 7 ] and C = [ 1 − 1 0 4 ] , 2 - 10C.
Answer
A C = [ 2 3 5 7 ] [ 1 0 − 1 4 ] = [ 2 × 1 + 3 × ( − 1 ) 2 × 0 + 3 × 4 5 × 1 + 7 × ( − 1 ) 5 × 0 + 7 × 4 ] = [ 2 − 3 0 + 12 5 − 7 0 + 28 ] = [ − 1 12 − 2 28 ] . AC = \begin{bmatrix*}[r] 2 & 3 \\ 5 & 7 \end{bmatrix*} \begin{bmatrix*}[r] 1 & 0 \\ -1 & 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 \times 1 + 3 \times (-1) & 2 \times 0 + 3 \times 4 \\ 5 \times 1 + 7 \times (-1) & 5 \times 0 + 7 \times 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 - 3 & 0 + 12 \\ 5 - 7 & 0 + 28 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -1 & 12 \\ -2 & 28 \end{bmatrix*}. A C = [ 2 5 3 7 ] [ 1 − 1 0 4 ] = [ 2 × 1 + 3 × ( − 1 ) 5 × 1 + 7 × ( − 1 ) 2 × 0 + 3 × 4 5 × 0 + 7 × 4 ] = [ 2 − 3 5 − 7 0 + 12 0 + 28 ] = [ − 1 − 2 12 28 ] .
B 2 = [ 0 4 − 1 7 ] [ 0 4 − 1 7 ] = [ 0 × 0 + 4 × ( − 1 ) 0 × 4 + 4 × 7 ( − 1 ) × 0 + 7 × ( − 1 ) ( − 1 ) × 4 + 7 × 7 ] = [ 0 − 4 0 + 28 0 − 7 − 4 + 49 ] = [ − 4 28 − 7 45 ] . B^2 = \begin{bmatrix*}[r] 0 & 4 \\ -1 & 7 \end{bmatrix*} \begin{bmatrix*}[r] 0 & 4 \\ -1 & 7 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 0 \times 0 + 4 \times (-1) & 0 \times 4 + 4 \times 7 \\ (-1) \times 0 + 7 \times (-1) & (-1) \times 4 + 7 \times 7 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 0 - 4 & 0 + 28 \\ 0 - 7 & -4 + 49 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -4 & 28 \\ -7 & 45 \end{bmatrix*}. B 2 = [ 0 − 1 4 7 ] [ 0 − 1 4 7 ] = [ 0 × 0 + 4 × ( − 1 ) ( − 1 ) × 0 + 7 × ( − 1 ) 0 × 4 + 4 × 7 ( − 1 ) × 4 + 7 × 7 ] = [ 0 − 4 0 − 7 0 + 28 − 4 + 49 ] = [ − 4 − 7 28 45 ] .
10 C = 10 [ 1 0 − 1 4 ] = [ 10 0 − 10 40 ] . ∴ A C + B 2 − 10 C = [ − 1 12 − 2 28 ] + [ − 4 28 − 7 45 ] − [ 10 0 − 10 40 ] = [ − 1 + ( − 4 ) − 10 12 + 28 − 0 − 2 + ( − 7 ) − ( − 10 ) 28 + 45 − 40 ] = [ − 15 40 1 33 ] 10C = 10 \begin{bmatrix*}[r] 1 & 0 \\ -1 & 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 10 & 0 \\ -10 & 40 \end{bmatrix*}. \\[1em] \therefore AC + B^2 - 10C = \begin{bmatrix*}[r] -1 & 12 \\ -2 & 28 \end{bmatrix*} + \begin{bmatrix*}[r] -4 & 28 \\ -7 & 45 \end{bmatrix*} - \begin{bmatrix*}[r] 10 & 0 \\ -10 & 40 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -1 + (-4) - 10 & 12 + 28 - 0 \\ -2 + (-7) - (-10) & 28 + 45 - 40 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -15 & 40 \\ 1 & 33 \end{bmatrix*} 10 C = 10 [ 1 − 1 0 4 ] = [ 10 − 10 0 40 ] . ∴ A C + B 2 − 10 C = [ − 1 − 2 12 28 ] + [ − 4 − 7 28 45 ] − [ 10 − 10 0 40 ] = [ − 1 + ( − 4 ) − 10 − 2 + ( − 7 ) − ( − 10 ) 12 + 28 − 0 28 + 45 − 40 ] = [ − 15 1 40 33 ]
Hence, the matrix AC + B2 - 10C = [ − 15 40 1 33 ] \begin{bmatrix*}[r] -15 & 40 \\ 1 & 33 \end{bmatrix*} [ − 15 1 40 33 ]
If A = [ 1 0 0 − 1 ] , \begin{bmatrix*}[r] 1 & 0 \\ 0 & -1 \end{bmatrix*}, [ 1 0 0 − 1 ] , 2 and A3 . Also state which of these is equal to A.
Answer
A 2 = [ 1 0 0 − 1 ] [ 1 0 0 − 1 ] = [ 1 × 1 + 0 × 0 1 × 0 + 0 × ( − 1 ) 0 × 1 + ( − 1 ) × 0 0 × 0 + ( − 1 ) × ( − 1 ) ] = [ 1 0 0 1 ] A^2 = \begin{bmatrix*}[r] 1 & 0 \\ 0 & -1 \end{bmatrix*} \begin{bmatrix*}[r] 1 & 0 \\ 0 & -1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 \times 1 + 0 \times 0 & 1 \times 0 + 0 \times (-1) \\ 0 \times 1 + (-1) \times 0 & 0 \times 0 + (-1) \times (-1) \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} A 2 = [ 1 0 0 − 1 ] [ 1 0 0 − 1 ] = [ 1 × 1 + 0 × 0 0 × 1 + ( − 1 ) × 0 1 × 0 + 0 × ( − 1 ) 0 × 0 + ( − 1 ) × ( − 1 ) ] = [ 1 0 0 1 ]
A 3 = A 2 × A = [ 1 0 0 1 ] [ 1 0 0 − 1 ] = [ 1 × 1 + 0 × 0 1 × 0 + 0 × ( − 1 ) 0 × 1 + 1 × 0 0 × 0 + 1 × ( − 1 ) ] [ 1 + 0 0 + 0 0 + 0 0 − 1 ] = [ 1 0 0 − 1 ] A^3 = A^2 \times A \\[1em] = \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \begin{bmatrix*}[r] 1 & 0 \\ 0 & -1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 \times 1 + 0 \times 0 & 1 \times 0 + 0 \times (-1) \\ 0 \times 1 + 1 \times 0 & 0 \times 0 + 1 \times (-1) \end{bmatrix*} \\[1em] \begin{bmatrix*}[r] 1 + 0 & 0 + 0 \\ 0 + 0 & 0 -1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 & 0 \\ 0 & -1 \end{bmatrix*} A 3 = A 2 × A = [ 1 0 0 1 ] [ 1 0 0 − 1 ] = [ 1 × 1 + 0 × 0 0 × 1 + 1 × 0 1 × 0 + 0 × ( − 1 ) 0 × 0 + 1 × ( − 1 ) ] [ 1 + 0 0 + 0 0 + 0 0 − 1 ] = [ 1 0 0 − 1 ]
Hence, the matrix A 2 = [ 1 0 0 1 ] and A 3 = [ 1 0 0 − 1 ] . A^2 = \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \text{ and } A^3 = \begin{bmatrix*}[r] 1 & 0 \\ 0 & -1 \end{bmatrix*}. A 2 = [ 1 0 0 1 ] and A 3 = [ 1 0 0 − 1 ] . 3 = A.
If X = [ 4 1 − 1 2 ] , \begin{bmatrix*}[r] 4 & 1 \\ -1 & 2 \end{bmatrix*}, [ 4 − 1 1 2 ] , 2 = 9I where I is the unit matrix.
Answer
We have to prove 6X - X2 = 9I,
L.H.S. = 6 X − X 2 6 X − X 2 = 6 [ 4 1 − 1 2 ] − [ 4 1 − 1 2 ] [ 4 1 − 1 2 ] = [ 24 6 − 6 12 ] − [ 4 × 4 + 1 × ( − 1 ) 4 × 1 + 1 × 2 ( − 1 ) × 4 + 2 × ( − 1 ) ( − 1 ) × 1 + 2 × 2 ] = [ 24 6 − 6 12 ] − [ 16 − 1 4 + 2 − 4 − 2 − 1 + 4 ] = [ 24 6 − 6 12 ] − [ 15 6 − 6 3 ] = [ 24 − 15 6 − 6 − 6 + 6 12 − 3 ] = [ 9 0 0 9 ] \text{L.H.S. } = 6X - X^2 \\[1em] 6X - X^2 = 6\begin{bmatrix*}[r] 4 & 1 \\ -1 & 2 \end{bmatrix*} - \begin{bmatrix*}[r] 4 & 1 \\ -1 & 2 \end{bmatrix*}\begin{bmatrix*}[r] 4 & 1 \\ -1 & 2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 24 & 6 \\ -6 & 12 \end{bmatrix*} - \begin{bmatrix*}[r] 4 \times 4 + 1 \times (-1) & 4 \times 1 + 1 \times 2 \\ (-1) \times 4 + 2 \times (-1) & (-1) \times 1 + 2 \times 2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 24 & 6 \\ -6 & 12 \end{bmatrix*} - \begin{bmatrix*}[r] 16 - 1 & 4 + 2 \\ -4 - 2 & -1 + 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 24 & 6 \\ -6 & 12 \end{bmatrix*} - \begin{bmatrix*}[r] 15 & 6 \\ -6 & 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 24 - 15 & 6 - 6 \\ -6 + 6 & 12 - 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 9 & 0 \\ 0 & 9 \end{bmatrix*} L.H.S. = 6 X − X 2 6 X − X 2 = 6 [ 4 − 1 1 2 ] − [ 4 − 1 1 2 ] [ 4 − 1 1 2 ] = [ 24 − 6 6 12 ] − [ 4 × 4 + 1 × ( − 1 ) ( − 1 ) × 4 + 2 × ( − 1 ) 4 × 1 + 1 × 2 ( − 1 ) × 1 + 2 × 2 ] = [ 24 − 6 6 12 ] − [ 16 − 1 − 4 − 2 4 + 2 − 1 + 4 ] = [ 24 − 6 6 12 ] − [ 15 − 6 6 3 ] = [ 24 − 15 − 6 + 6 6 − 6 12 − 3 ] = [ 9 0 0 9 ]
R.H.S. = 9 I 9 I = 9 [ 1 0 0 1 ] = [ 9 0 0 9 ] . \text{R.H.S. } = 9I \\[1em] 9I = 9\begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 9 & 0 \\ 0 & 9 \end{bmatrix*}. R.H.S. = 9 I 9 I = 9 [ 1 0 0 1 ] = [ 9 0 0 9 ] .
Since, L.H.S. = [ 9 0 0 9 ] \begin{bmatrix*}[r] 9 & 0 \\ 0 & 9 \end{bmatrix*} [ 9 0 0 9 ] 2 = 9I.
Show that [ 1 2 2 1 ] \begin{bmatrix*}[r] 1 & 2 \\ 2 & 1 \end{bmatrix*} [ 1 2 2 1 ] 2 - 2X - 3I = 0 where I is the unit matrix of order 2.
Answer
I = [ 1 0 0 1 ] , X = [ 1 2 2 1 ] I = \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*}, X = \begin{bmatrix*}[r] 1 & 2 \\ 2 & 1 \end{bmatrix*} \\[1em] I = [ 1 0 0 1 ] , X = [ 1 2 2 1 ]
Given,
X2 - 2X - 3I = 0
Putting value of X and I in above equation we get,
L.H.S. = [ 1 2 2 1 ] [ 1 2 2 1 ] − 2 [ 1 2 2 1 ] − 3 [ 1 0 0 1 ] = [ 1 × 1 + 2 × 2 1 × 2 + 2 × 1 2 × 1 + 1 × 2 2 × 2 + 1 × 1 ] − [ 2 4 4 2 ] − [ 3 0 0 3 ] [ 1 + 4 2 + 2 2 + 2 4 + 1 ] − [ 2 4 4 2 ] − [ 3 0 0 3 ] [ 5 4 4 5 ] − [ 2 4 4 2 ] − [ 3 0 0 3 ] = [ 5 − 2 − 3 4 − 4 − 0 4 − 4 − 0 5 − 2 − 3 ] = [ 0 0 0 0 ] \text{L.H.S. } = \begin{bmatrix*}[r] 1 & 2 \\ 2 & 1 \end{bmatrix*} \begin{bmatrix*}[r] 1 & 2 \\ 2 & 1 \end{bmatrix*} - 2 \begin{bmatrix*}[r] 1 & 2 \\ 2 & 1 \end{bmatrix*} - 3 \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 \times 1 + 2 \times 2 & 1 \times 2 + 2 \times 1 \\ 2 \times 1 + 1 \times 2 & 2 \times 2 + 1 \times 1 \end{bmatrix*} - \begin{bmatrix*}[r] 2 & 4 \\ 4 & 2 \end{bmatrix*} - \begin{bmatrix*}[r] 3 & 0 \\ 0 & 3 \end{bmatrix*} \\[1em] \begin{bmatrix*}[r] 1 + 4 & 2 + 2 \\ 2 + 2 & 4 + 1 \end{bmatrix*} - \begin{bmatrix*}[r] 2 & 4 \\ 4 & 2 \end{bmatrix*} - \begin{bmatrix*}[r] 3 & 0 \\ 0 & 3 \end{bmatrix*} \\[1em] \begin{bmatrix*}[r] 5 & 4 \\ 4 & 5 \end{bmatrix*} - \begin{bmatrix*}[r] 2 & 4 \\ 4 & 2 \end{bmatrix*} - \begin{bmatrix*}[r] 3 & 0 \\ 0 & 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 5 - 2 - 3 & 4 - 4 - 0 \\ 4 - 4 - 0 & 5 - 2 - 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*} L.H.S. = [ 1 2 2 1 ] [ 1 2 2 1 ] − 2 [ 1 2 2 1 ] − 3 [ 1 0 0 1 ] = [ 1 × 1 + 2 × 2 2 × 1 + 1 × 2 1 × 2 + 2 × 1 2 × 2 + 1 × 1 ] − [ 2 4 4 2 ] − [ 3 0 0 3 ] [ 1 + 4 2 + 2 2 + 2 4 + 1 ] − [ 2 4 4 2 ] − [ 3 0 0 3 ] [ 5 4 4 5 ] − [ 2 4 4 2 ] − [ 3 0 0 3 ] = [ 5 − 2 − 3 4 − 4 − 0 4 − 4 − 0 5 − 2 − 3 ] = [ 0 0 0 0 ]
Since, L.H.S. = [ 0 0 0 0 ] \begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*} [ 0 0 0 0 ] 2 - 2X -3I = 0.
∴[ 1 2 2 1 ] \begin{bmatrix*}[r] 1 & 2 \\ 2 & 1 \end{bmatrix*} [ 1 2 2 1 ] is a solution of the matrix equation X2 - 2X - 3I = 0
Find the matrix X of order 2 x 2 which satisfies the equation :
[ 3 7 2 4 ] [ 0 2 5 3 ] + 2 X = [ 1 − 5 − 4 6 ] . \begin{bmatrix*}[r] 3 & 7 \\ 2 & 4 \end{bmatrix*} \begin{bmatrix*}[r] 0 & 2 \\ 5 & 3 \end{bmatrix*} + 2X = \begin{bmatrix*}[r] 1 & -5 \\ -4 & 6 \end{bmatrix*}. [ 3 2 7 4 ] [ 0 5 2 3 ] + 2 X = [ 1 − 4 − 5 6 ] .
Answer
Given,
[ 3 7 2 4 ] [ 0 2 5 3 ] + 2 X = [ 1 − 5 − 4 6 ] , ⇒ [ 3 × 0 + 7 × 5 3 × 2 + 7 × 3 2 × 0 + 4 × 5 2 × 2 + 4 × 3 ] + 2 X = [ 1 − 5 − 4 6 ] ⇒ [ 0 + 35 6 + 21 0 + 20 4 + 12 ] + 2 X = [ 1 − 5 − 4 6 ] ⇒ [ 35 27 20 16 ] + 2 X = [ 1 − 5 − 4 6 ] ⇒ 2 X = [ 1 − 5 − 4 6 ] − [ 35 27 20 16 ] ⇒ 2 X = [ 1 − 35 − 5 − 27 − 4 − 20 6 − 16 ] ⇒ 2 X = [ − 34 − 32 − 24 − 10 ] ⇒ X = 1 2 [ − 34 − 32 − 24 − 10 ] X = [ − 17 − 16 − 12 − 5 ] . \begin{bmatrix*}[r] 3 & 7 \\ 2 & 4 \end{bmatrix*} \begin{bmatrix*}[r] 0 & 2 \\ 5 & 3 \end{bmatrix*} + 2X = \begin{bmatrix*}[r] 1 & -5 \\ -4 & 6 \end{bmatrix*}, \\[1em] \Rightarrow \begin{bmatrix*}[r] 3 \times 0 + 7 \times 5 & 3 \times 2 + 7 \times 3 \\ 2 \times 0 + 4 \times 5 & 2 \times 2 + 4 \times 3 \end{bmatrix*} + 2X = \begin{bmatrix*}[r] 1 & -5 \\ -4 & 6 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 0 + 35 & 6 + 21 \\ 0 + 20 & 4 + 12 \end{bmatrix*} + 2X = \begin{bmatrix*}[r] 1 & -5 \\ -4 & 6 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 35 & 27 \\ 20 & 16 \end{bmatrix*} + 2X = \begin{bmatrix*}[r] 1 & -5 \\ -4 & 6 \end{bmatrix*} \\[1em] \Rightarrow 2X = \begin{bmatrix*}[r] 1 & -5 \\ -4 & 6 \end{bmatrix*} - \begin{bmatrix*}[r] 35 & 27 \\ 20 & 16 \end{bmatrix*} \\[1em] \Rightarrow 2X = \begin{bmatrix*}[r] 1 - 35 & -5 - 27 \\ -4 - 20 & 6 - 16 \end{bmatrix*} \\[1em] \Rightarrow 2X = \begin{bmatrix*}[r] -34 & -32 \\ -24 & -10 \end{bmatrix*} \\[1em] \Rightarrow X = \dfrac{1}{2} \begin{bmatrix*}[r] -34 & -32 \\ -24 & -10 \end{bmatrix*} \\[1em] X = \begin{bmatrix*}[r] -17 & -16 \\ -12 & -5 \end{bmatrix*}. [ 3 2 7 4 ] [ 0 5 2 3 ] + 2 X = [ 1 − 4 − 5 6 ] , ⇒ [ 3 × 0 + 7 × 5 2 × 0 + 4 × 5 3 × 2 + 7 × 3 2 × 2 + 4 × 3 ] + 2 X = [ 1 − 4 − 5 6 ] ⇒ [ 0 + 35 0 + 20 6 + 21 4 + 12 ] + 2 X = [ 1 − 4 − 5 6 ] ⇒ [ 35 20 27 16 ] + 2 X = [ 1 − 4 − 5 6 ] ⇒ 2 X = [ 1 − 4 − 5 6 ] − [ 35 20 27 16 ] ⇒ 2 X = [ 1 − 35 − 4 − 20 − 5 − 27 6 − 16 ] ⇒ 2 X = [ − 34 − 24 − 32 − 10 ] ⇒ X = 2 1 [ − 34 − 24 − 32 − 10 ] X = [ − 17 − 12 − 16 − 5 ] .
Hence, the matrix X = [ − 17 − 16 − 12 − 5 ] \begin{bmatrix*}[r] -17 & -16 \\ -12 & -5 \end{bmatrix*} [ − 17 − 12 − 16 − 5 ]
If A = [ 1 1 x x ] , \begin{bmatrix*}[r] 1 & 1 \\ x & x \end{bmatrix*}, [ 1 x 1 x ] , 2 = O.
Answer
Given, A2 = O.
or , [ 1 1 x x ] [ 1 1 x x ] = [ 0 0 0 0 ] L.H.S. = [ 1 1 x x ] [ 1 1 x x ] = [ 1 × 1 + 1 × x 1 × 1 + 1 × x x × 1 + x × x x × 1 + x × x ] = [ 1 + x 1 + x x + x 2 x + x 2 ] \text{or}, \begin{bmatrix*}[r] 1 & 1 \\ x & x \end{bmatrix*} \begin{bmatrix*}[r] 1 & 1 \\ x & x \end{bmatrix*} = \begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*} \\[1em] \text{L.H.S.} = \begin{bmatrix*}[r] 1 & 1 \\ x & x \end{bmatrix*} \begin{bmatrix*}[r] 1 & 1 \\ x & x \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 \times 1 + 1 \times x & 1 \times 1 + 1 \times x \\ x \times 1 + x \times x & x \times 1 + x \times x \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 + x & 1 + x \\ x + x^2 & x + x^2 \end{bmatrix*} or , [ 1 x 1 x ] [ 1 x 1 x ] = [ 0 0 0 0 ] L.H.S. = [ 1 x 1 x ] [ 1 x 1 x ] = [ 1 × 1 + 1 × x x × 1 + x × x 1 × 1 + 1 × x x × 1 + x × x ] = [ 1 + x x + x 2 1 + x x + x 2 ]
Comparing with R.H.S. we get,
⇒ 1 + x = 0 or x = -1 (...Eq 1)
⇒ x + x2 = 0 (...Eq 2)
Putting the value x = -1 in equation 2,
⇒ -1 + (-1)2 = -1 + 1 = 0.
Since, x = -1 satisfies the equation,
Hence, the required value of x = -1.
Find x and y, if [ − 3 2 0 − 5 ] [ x 2 ] = [ − 5 y ] . \begin{bmatrix*}[r] -3 & 2 \\ 0 & -5 \end{bmatrix*} \begin{bmatrix*}[r] x \\ 2 \end{bmatrix*} = \begin{bmatrix*}[r] -5 \\ y \end{bmatrix*}. [ − 3 0 2 − 5 ] [ x 2 ] = [ − 5 y ] .
Answer
Given,
[ − 3 2 0 − 5 ] [ x 2 ] = [ − 5 y ] L.H.S. = [ − 3 2 0 − 5 ] [ x 2 ] = [ − 3 × x + 2 × 2 0 × x + ( − 5 ) × 2 ] = [ − 3 x + 4 − 10 ] \begin{bmatrix*}[r] -3 & 2 \\ 0 & -5 \end{bmatrix*} \begin{bmatrix*}[r] x \\ 2 \end{bmatrix*} = \begin{bmatrix*}[r] -5 \\ y \end{bmatrix*} \\[1em] \text{L.H.S.} = \begin{bmatrix*}[r] -3 & 2 \\ 0 & -5 \end{bmatrix*} \begin{bmatrix*}[r] x \\ 2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -3 \times x + 2 \times 2 \\ 0 \times x + (-5) \times 2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -3x + 4 \\ -10 \end{bmatrix*} \\[1em] [ − 3 0 2 − 5 ] [ x 2 ] = [ − 5 y ] L.H.S. = [ − 3 0 2 − 5 ] [ x 2 ] = [ − 3 × x + 2 × 2 0 × x + ( − 5 ) × 2 ] = [ − 3 x + 4 − 10 ]
Comparing L.H.S. with R.H.S.,
⇒ [ − 3 x + 4 − 10 ] = [ − 5 y ] \Rightarrow \begin{bmatrix*}[r] -3x + 4 \\ -10 \end{bmatrix*} = \begin{bmatrix*}[r] -5 \\ y \end{bmatrix*} \\[1em] ⇒ [ − 3 x + 4 − 10 ] = [ − 5 y ]
By equality of matrices,
⇒ -3x + 4 = -5 and -10 = y
Hence, the values are x = 3 and y = -10.
Find x and y, if [ 2 x x y 3 y ] [ 3 2 ] = [ 16 9 ] . \begin{bmatrix*}[r] 2x & x \\ y & 3y \end{bmatrix*} \begin{bmatrix*}[r] 3 \\ 2 \end{bmatrix*} = \begin{bmatrix*}[r] 16 \\ 9 \end{bmatrix*}. [ 2 x y x 3 y ] [ 3 2 ] = [ 16 9 ] .
Answer
Given,
[ 2 x x y 3 y ] [ 3 2 ] = [ 16 9 ] L.H.S. = [ 2 x x y 3 y ] [ 3 2 ] = [ 2 x × 3 + x × 2 y × 3 + 3 y × 2 ] = [ 6 x + 2 x 3 y + 6 y ] = [ 8 x 9 y ] \begin{bmatrix*}[r] 2x & x \\ y & 3y \end{bmatrix*} \begin{bmatrix*}[r] 3 \\ 2 \end{bmatrix*} = \begin{bmatrix*}[r] 16 \\ 9 \end{bmatrix*} \\[1em] \text{L.H.S.} = \begin{bmatrix*}[r] 2x & x \\ y & 3y \end{bmatrix*} \begin{bmatrix*}[r] 3 \\ 2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2x \times 3 + x \times 2 \\ y \times 3 + 3y \times 2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 6x + 2x \\ 3y + 6y \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 8x \\ 9y \end{bmatrix*} [ 2 x y x 3 y ] [ 3 2 ] = [ 16 9 ] L.H.S. = [ 2 x y x 3 y ] [ 3 2 ] = [ 2 x × 3 + x × 2 y × 3 + 3 y × 2 ] = [ 6 x + 2 x 3 y + 6 y ] = [ 8 x 9 y ]
Comparing L.H.S. with R.H.S. we get,
⇒ [ 8 x 9 y ] = [ 16 9 ] ⇒ 8 x = 16 and 9 y = 9 ∴ x = 2 and y = 1. \Rightarrow \begin{bmatrix*}[r] 8x \\ 9y \end{bmatrix*} = \begin{bmatrix*}[r] 16 \\ 9 \end{bmatrix*} \\[1em] \Rightarrow 8x = 16 \text{ and } 9y = 9 \\[1em] \therefore x = 2 \text{ and } y = 1. ⇒ [ 8 x 9 y ] = [ 16 9 ] ⇒ 8 x = 16 and 9 y = 9 ∴ x = 2 and y = 1.
Hence, the value of x = 2 and y = 1.
Find the values of x and y if [ x + y y 2 x x − y ] [ 2 − 1 ] = [ 3 2 ] . \begin{bmatrix*}[r] x + y & y \\ 2x & x - y \end{bmatrix*} \begin{bmatrix*}[r] 2 \\ -1 \end{bmatrix*} = \begin{bmatrix*}[r] 3 \\ 2 \end{bmatrix*}. [ x + y 2 x y x − y ] [ 2 − 1 ] = [ 3 2 ] .
Answer
Given,
[ x + y y 2 x x − y ] [ 2 − 1 ] = [ 3 2 ] ⇒ [ ( x + y ) × 2 + y × ( − 1 ) 2 x × 2 + ( x − y ) × ( − 1 ) ] = [ 3 2 ] ⇒ [ 2 x + 2 y − y 4 x − x + y ] = [ 3 2 ] ⇒ [ 2 x + y 3 x + y ] = [ 3 2 ] \begin{bmatrix*}[r] x + y & y \\ 2x & x - y \end{bmatrix*} \begin{bmatrix*}[r] 2 \\ -1 \end{bmatrix*} = \begin{bmatrix*}[r] 3 \\ 2 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] (x + y) \times 2 + y \times (-1) \\ 2x \times 2 + (x - y) \times (-1) \end{bmatrix*} = \begin{bmatrix*}[r] 3 \\ 2 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2x + 2y - y \\ 4x - x + y \end{bmatrix*} = \begin{bmatrix*}[r] 3 \\ 2 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2x + y \\ 3x + y \end{bmatrix*} = \begin{bmatrix*}[r] 3 \\ 2 \end{bmatrix*} \\[1em] [ x + y 2 x y x − y ] [ 2 − 1 ] = [ 3 2 ] ⇒ [ ( x + y ) × 2 + y × ( − 1 ) 2 x × 2 + ( x − y ) × ( − 1 ) ] = [ 3 2 ] ⇒ [ 2 x + 2 y − y 4 x − x + y ] = [ 3 2 ] ⇒ [ 2 x + y 3 x + y ] = [ 3 2 ]
By definition of equality of matrices we have,
⇒ 2x + y = 3 or y = 3 - 2x (...Eq 1)
Putting value of y from equation 1 in equation 2,
⇒ 3x + 3 - 2x = 2
Now finding value of y,
⇒ y = 3 - 2x
Hence, the value of x = -1 and y = 5.
If [ 1 2 3 3 ] [ x 0 0 y ] = [ x 0 9 0 ] , \begin{bmatrix*}[r] 1 & 2 \\ 3 & 3 \end{bmatrix*} \begin{bmatrix*}[r] x & 0 \\ 0 & y \end{bmatrix*} = \begin{bmatrix*}[r] x & 0 \\ 9 & 0 \end{bmatrix*}, [ 1 3 2 3 ] [ x 0 0 y ] = [ x 9 0 0 ] ,
Answer
Given,
[ 1 2 3 3 ] [ x 0 0 y ] = [ x 0 9 0 ] ⇒ [ 1 × x + 2 × 0 1 × 0 + 2 × y 3 × x + 3 × 0 3 × 0 + 3 × y ] = [ x 0 9 0 ] ⇒ [ x + 0 0 + 2 y 3 x + 0 0 + 3 y ] = [ x 0 9 0 ] ⇒ [ x 2 y 3 x 3 y ] = [ x 0 9 0 ] \begin{bmatrix*}[r] 1 & 2 \\ 3 & 3 \end{bmatrix*} \begin{bmatrix*}[r] x & 0 \\ 0 & y \end{bmatrix*} = \begin{bmatrix*}[r] x & 0 \\ 9 & 0 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 1 \times x + 2 \times 0 & 1 \times 0 + 2 \times y \\ 3 \times x + 3 \times 0 & 3 \times 0 + 3 \times y \end{bmatrix*} = \begin{bmatrix*}[r] x & 0 \\ 9 & 0 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] x + 0 & 0 + 2y \\ 3x + 0 & 0 + 3y \end{bmatrix*} = \begin{bmatrix*}[r] x & 0 \\ 9 & 0 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] x & 2y \\ 3x & 3y \end{bmatrix*} = \begin{bmatrix*}[r] x & 0 \\ 9 & 0 \end{bmatrix*} \\[1em] [ 1 3 2 3 ] [ x 0 0 y ] = [ x 9 0 0 ] ⇒ [ 1 × x + 2 × 0 3 × x + 3 × 0 1 × 0 + 2 × y 3 × 0 + 3 × y ] = [ x 9 0 0 ] ⇒ [ x + 0 3 x + 0 0 + 2 y 0 + 3 y ] = [ x 9 0 0 ] ⇒ [ x 3 x 2 y 3 y ] = [ x 9 0 0 ]
By definition of equality of matrices,
⇒ 2y = 0, 3x = 9 and 3y = 0
Hence, the values are x = 3 and y = 0.
If [ 3 4 2 5 ] = [ a b c d ] [ 1 0 0 1 ] , \begin{bmatrix*}[r] 3 & 4 \\ 2 & 5 \end{bmatrix*} = \begin{bmatrix*}[r] a & b \\ c & d \end{bmatrix*} \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*}, [ 3 2 4 5 ] = [ a c b d ] [ 1 0 0 1 ] ,
Answer
Given,
[ 3 4 2 5 ] = [ a b c d ] [ 1 0 0 1 ] ⇒ [ 3 4 2 5 ] = [ a × 1 + b × 0 a × 0 + b × 1 c × 1 + d × 0 c × 0 + d × 1 ] ⇒ [ 3 4 2 5 ] = [ a b c d ] \begin{bmatrix*}[r] 3 & 4 \\ 2 & 5 \end{bmatrix*} = \begin{bmatrix*}[r] a & b \\ c & d \end{bmatrix*} \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 3 & 4 \\ 2 & 5 \end{bmatrix*} = \begin{bmatrix*}[r] a \times 1 + b \times 0 & a \times 0 + b \times 1 \\ c \times 1 + d \times 0 & c \times 0 + d \times 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 3 & 4 \\ 2 & 5 \end{bmatrix*} = \begin{bmatrix*}[r] a & b \\ c & d \end{bmatrix*} [ 3 2 4 5 ] = [ a c b d ] [ 1 0 0 1 ] ⇒ [ 3 2 4 5 ] = [ a × 1 + b × 0 c × 1 + d × 0 a × 0 + b × 1 c × 0 + d × 1 ] ⇒ [ 3 2 4 5 ] = [ a c b d ]
By definition of equality of matrices,
⇒ a = 3, b = 4, c = 2 and d = 5
Hence, the value of a = 3, b = 4, c = 2 and d = 5.
Find the value of x given that A2 = B where,
A = [ 2 12 0 1 ] and B = [ 4 x 0 1 ] . A = \begin{bmatrix*}[r] 2 & 12 \\ 0 & 1 \end{bmatrix*} \text{ and } B = \begin{bmatrix*}[r] 4 & x \\ 0 & 1 \end{bmatrix*}. A = [ 2 0 12 1 ] and B = [ 4 0 x 1 ] .
Answer
Given A 2 = B , ⇒ [ 2 12 0 1 ] [ 2 12 0 1 ] = [ 4 x 0 1 ] ⇒ [ 2 × 2 + 12 × 0 2 × 12 + 12 × 1 0 × 2 + 1 × 0 0 × 12 + 1 × 1 ] = [ 4 x 0 1 ] ⇒ [ 4 + 0 24 + 12 0 + 0 0 + 1 ] = [ 4 x 0 1 ] ⇒ [ 4 36 0 1 ] = [ 4 x 0 1 ] \text{Given } A^2 = B, \\[1em] \Rightarrow \begin{bmatrix*}[r] 2 & 12 \\ 0 & 1 \end{bmatrix*} \begin{bmatrix*}[r] 2 & 12 \\ 0 & 1 \end{bmatrix*} = \begin{bmatrix*}[r] 4 & x \\ 0 & 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2 \times 2 + 12 \times 0 & 2 \times 12 + 12 \times 1 \\ 0 \times 2 + 1 \times 0 & 0 \times 12 + 1 \times 1 \end{bmatrix*} = \begin{bmatrix*}[r] 4 & x \\ 0 & 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 + 0 & 24 + 12 \\ 0 + 0 & 0 + 1 \end{bmatrix*} = \begin{bmatrix*}[r] 4 & x \\ 0 & 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 & 36 \\ 0 & 1 \end{bmatrix*} = \begin{bmatrix*}[r] 4 & x \\ 0 & 1 \end{bmatrix*} \\[1em] Given A 2 = B , ⇒ [ 2 0 12 1 ] [ 2 0 12 1 ] = [ 4 0 x 1 ] ⇒ [ 2 × 2 + 12 × 0 0 × 2 + 1 × 0 2 × 12 + 12 × 1 0 × 12 + 1 × 1 ] = [ 4 0 x 1 ] ⇒ [ 4 + 0 0 + 0 24 + 12 0 + 1 ] = [ 4 0 x 1 ] ⇒ [ 4 0 36 1 ] = [ 4 0 x 1 ]
By definition of equality of matrices,
⇒ x = 36.
Hence, the value of x = 36.
If A = [ 2 x 0 1 ] and B = [ 4 36 0 1 ] , \begin{bmatrix*}[r] 2 & x \\ 0 & 1 \end{bmatrix*} \text{and B } = \begin{bmatrix*}[r] 4 & 36 \\ 0 & 1 \end{bmatrix*}, [ 2 0 x 1 ] and B = [ 4 0 36 1 ] , 2 = B.
Answer
Given,
A2 = B
⇒ [ 2 x 0 1 ] [ 2 x 0 1 ] = [ 4 36 0 1 ] ⇒ [ 2 × 2 + x × 0 2 × x + x × 1 0 × 2 + 1 × 0 0 × x + 1 × 1 ] = [ 4 36 0 1 ] ⇒ [ 4 + 0 2 x + x 0 + 0 0 + 1 ] = [ 4 36 0 1 ] ⇒ [ 4 3 x 0 1 ] = [ 4 36 0 1 ] \Rightarrow \begin{bmatrix*}[r] 2 & x \\ 0 & 1 \end{bmatrix*} \begin{bmatrix*}[r] 2 & x \\ 0 & 1 \end{bmatrix*} = \begin{bmatrix*}[r] 4 & 36 \\ 0 & 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2 \times 2 + x \times 0 & 2 \times x + x \times 1 \\ 0 \times 2 + 1 \times 0 & 0 \times x + 1 \times 1 \end{bmatrix*} = \begin{bmatrix*}[r] 4 & 36 \\ 0 & 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 + 0 & 2x + x \\ 0 + 0 & 0 + 1 \end{bmatrix*} = \begin{bmatrix*}[r] 4 & 36 \\ 0 & 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 & 3x \\ 0 & 1 \end{bmatrix*} = \begin{bmatrix*}[r] 4 & 36 \\ 0 & 1 \end{bmatrix*} \\[1em] ⇒ [ 2 0 x 1 ] [ 2 0 x 1 ] = [ 4 0 36 1 ] ⇒ [ 2 × 2 + x × 0 0 × 2 + 1 × 0 2 × x + x × 1 0 × x + 1 × 1 ] = [ 4 0 36 1 ] ⇒ [ 4 + 0 0 + 0 2 x + x 0 + 1 ] = [ 4 0 36 1 ] ⇒ [ 4 0 3 x 1 ] = [ 4 0 36 1 ]
By definition of equality of matrices we get,
⇒ 3x = 36
Hence, the value of x = 12.
If A = [ 3 x 0 1 ] and B = [ 9 16 0 − y ] , \begin{bmatrix*}[r] 3 & x \\ 0 & 1 \end{bmatrix*} \text{ and B} = \begin{bmatrix*}[r] 9 & 16 \\ 0 & -y \end{bmatrix*}, [ 3 0 x 1 ] and B = [ 9 0 16 − y ] , 2 = B.
Answer
Given,
A2 = B
⇒ [ 3 x 0 1 ] [ 3 x 0 1 ] = [ 9 16 0 − y ] ⇒ [ 3 × 3 + x × 0 3 × x + x × 1 0 × 3 + 1 × 0 0 × x + 1 × 1 ] = [ 9 16 0 − y ] ⇒ [ 9 + 0 3 x + x 0 + 0 0 + 1 ] = [ 9 16 0 − y ] ⇒ [ 9 4 x 0 1 ] = [ 9 16 0 − y ] \Rightarrow \begin{bmatrix*}[r] 3 & x \\ 0 & 1 \end{bmatrix*} \begin{bmatrix*}[r] 3 & x \\ 0 & 1 \end{bmatrix*} = \begin{bmatrix*}[r] 9 & 16 \\ 0 & -y \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 3 \times 3 + x \times 0 & 3 \times x + x \times 1 \\ 0 \times 3 + 1 \times 0 & 0 \times x + 1 \times 1 \end{bmatrix*} = \begin{bmatrix*}[r] 9 & 16 \\ 0 & -y \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 9 + 0 & 3x + x \\ 0 + 0 & 0 + 1 \end{bmatrix*} = \begin{bmatrix*}[r] 9 & 16 \\ 0 & -y \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 9 & 4x \\ 0 & 1 \end{bmatrix*} = \begin{bmatrix*}[r] 9 & 16 \\ 0 & -y \end{bmatrix*} \\[1em] ⇒ [ 3 0 x 1 ] [ 3 0 x 1 ] = [ 9 0 16 − y ] ⇒ [ 3 × 3 + x × 0 0 × 3 + 1 × 0 3 × x + x × 1 0 × x + 1 × 1 ] = [ 9 0 16 − y ] ⇒ [ 9 + 0 0 + 0 3 x + x 0 + 1 ] = [ 9 0 16 − y ] ⇒ [ 9 0 4 x 1 ] = [ 9 0 16 − y ]
By definition of equality of matrices we get,
⇒ 4x = 16 and -y = 1
Hence, the values are x = 4 and y = -1.
Find x, y if [ − 2 0 3 1 ] [ − 1 2 x ] + 3 [ − 2 1 ] = 2 [ y 3 ] . \begin{bmatrix*}[r] -2 & 0 \\ 3 & 1 \end{bmatrix*} \begin{bmatrix*}[r] -1 \\ 2x \end{bmatrix*} + 3\begin{bmatrix*}[r] -2 \\ 1 \end{bmatrix*} = 2\begin{bmatrix*}[r] y \\ 3 \end{bmatrix*}. [ − 2 3 0 1 ] [ − 1 2 x ] + 3 [ − 2 1 ] = 2 [ y 3 ] .
Answer
Given,
[ − 2 0 3 1 ] [ − 1 2 x ] + 3 [ − 2 1 ] = 2 [ y 3 ] ⇒ [ ( − 2 ) × ( − 1 ) + 0 × 2 x 3 × ( − 1 ) + 1 × 2 x ] + [ − 6 3 ] = [ 2 y 6 ] ⇒ [ 2 − 3 + 2 x ] + [ − 6 3 ] = [ 2 y 6 ] ⇒ [ 2 + ( − 6 ) − 3 + 2 x + 3 ] = [ 2 y 6 ] ⇒ [ − 4 2 x ] = [ 2 y 6 ] \begin{bmatrix*}[r] -2 & 0 \\ 3 & 1 \end{bmatrix*} \begin{bmatrix*}[r] -1 \\ 2x \end{bmatrix*} + 3\begin{bmatrix*}[r] -2 \\ 1 \end{bmatrix*} = 2\begin{bmatrix*}[r] y \\ 3 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] (-2) \times (-1) + 0 \times 2x \\ 3 \times (-1) + 1 \times 2x \end{bmatrix*} + \begin{bmatrix*}[r] -6 \\ 3 \end{bmatrix*} = \begin{bmatrix*}[r] 2y \\ 6 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2 \\ -3 + 2x \end{bmatrix*} + \begin{bmatrix*}[r] -6 \\ 3 \end{bmatrix*} = \begin{bmatrix*}[r] 2y \\ 6 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2 + (-6) \\ -3 + 2x + 3 \end{bmatrix*} = \begin{bmatrix*}[r] 2y \\ 6 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] -4 \\ 2x \end{bmatrix*} = \begin{bmatrix*}[r] 2y \\ 6 \end{bmatrix*} \\[1em] [ − 2 3 0 1 ] [ − 1 2 x ] + 3 [ − 2 1 ] = 2 [ y 3 ] ⇒ [ ( − 2 ) × ( − 1 ) + 0 × 2 x 3 × ( − 1 ) + 1 × 2 x ] + [ − 6 3 ] = [ 2 y 6 ] ⇒ [ 2 − 3 + 2 x ] + [ − 6 3 ] = [ 2 y 6 ] ⇒ [ 2 + ( − 6 ) − 3 + 2 x + 3 ] = [ 2 y 6 ] ⇒ [ − 4 2 x ] = [ 2 y 6 ]
By definition of equality of matrices we get,
2y = -4 and 2x = 6
Hence, the values are x = 3 and y = -2.
If [ a 1 1 0 ] [ 4 3 − 3 2 ] = [ b 11 4 c ] , \begin{bmatrix*}[r] a & 1 \\ 1 & 0 \end{bmatrix*} \begin{bmatrix*}[r] 4 & 3 \\ -3 & 2 \end{bmatrix*} = \begin{bmatrix*}[r] b & 11 \\ 4 & c \end{bmatrix*}, [ a 1 1 0 ] [ 4 − 3 3 2 ] = [ b 4 11 c ] ,
Answer
Given,
[ a 1 1 0 ] [ 4 3 − 3 2 ] = [ b 11 4 c ] ⇒ [ a × 4 + 1 × ( − 3 ) a × 3 + 1 × 2 1 × 4 + 0 × ( − 3 ) 1 × 3 + 0 × 2 ] = [ b 11 4 c ] ⇒ [ 4 a − 3 3 a + 2 4 + 0 3 + 0 ] = [ b 11 4 c ] ⇒ [ 4 a − 3 3 a + 2 4 3 ] = [ b 11 4 c ] \begin{bmatrix*}[r] a & 1 \\ 1 & 0 \end{bmatrix*} \begin{bmatrix*}[r] 4 & 3 \\ -3 & 2 \end{bmatrix*} = \begin{bmatrix*}[r] b & 11 \\ 4 & c \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] a \times 4 + 1 \times (-3) & a \times 3 + 1 \times 2 \\ 1 \times 4 + 0 \times (-3) & 1 \times 3 + 0 \times 2 \end{bmatrix*} = \begin{bmatrix*}[r] b & 11 \\ 4 & c \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 4a - 3 & 3a + 2 \\ 4 + 0 & 3 + 0 \end{bmatrix*} = \begin{bmatrix*}[r] b & 11 \\ 4 & c \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 4a - 3 & 3a + 2 \\ 4 & 3 \end{bmatrix*} = \begin{bmatrix*}[r] b & 11 \\ 4 & c \end{bmatrix*} \\[1em] [ a 1 1 0 ] [ 4 − 3 3 2 ] = [ b 4 11 c ] ⇒ [ a × 4 + 1 × ( − 3 ) 1 × 4 + 0 × ( − 3 ) a × 3 + 1 × 2 1 × 3 + 0 × 2 ] = [ b 4 11 c ] ⇒ [ 4 a − 3 4 + 0 3 a + 2 3 + 0 ] = [ b 4 11 c ] ⇒ [ 4 a − 3 4 3 a + 2 3 ] = [ b 4 11 c ]
By definition of equality of matrices we get,
4a - 3 = b (...Eq 1)
Solving (Eq 2) first,
⇒ 3a + 2 = 11
Putting value of a in Eq 1,
⇒ 4a - 3 = b
∴ a = 3, b = 9 and c = 3.
Hence, the value of a = 3, b = 9 and c = 3.
If A = [ 2 3 1 2 ] , \begin{bmatrix*}[r] 2 & 3 \\ 1 & 2 \end{bmatrix*}, [ 2 1 3 2 ] , 2 = xA + yI.
Answer
Given,
A = [ 2 3 1 2 ] , I = [ 1 0 0 1 ] A 2 = x A + y I ⇒ [ 2 3 1 2 ] [ 2 3 1 2 ] = x [ 2 3 1 2 ] + y [ 1 0 0 1 ] ⇒ [ 2 × 2 + 3 × 1 2 × 3 + 3 × 2 1 × 2 + 2 × 1 1 × 3 + 2 × 2 ] = [ 2 x 3 x x 2 x ] + [ y 0 0 y ] ⇒ [ 4 + 3 6 + 6 2 + 2 3 + 4 ] = [ 2 x + y 3 x + 0 x + 0 2 x + y ] ⇒ [ 7 12 4 7 ] = [ 2 x + y 3 x + 0 x + 0 2 x + y ] ⇒ [ 7 12 4 7 ] = [ 2 x + y 3 x x 2 x + y ] \text{A } = \begin{bmatrix*}[r] 2 & 3 \\ 1 & 2 \end{bmatrix*}, \text{ I } = \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em] \text{A}^2 = x\text{A} + y\text{I } \\[1em] \Rightarrow \begin{bmatrix*}[r] 2 & 3 \\ 1 & 2 \end{bmatrix*} \begin{bmatrix*}[r] 2 & 3 \\ 1 & 2 \end{bmatrix*} = x \begin{bmatrix*}[r] 2 & 3 \\ 1 & 2 \end{bmatrix*} + y \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2 \times 2 + 3 \times 1 & 2 \times 3 + 3 \times 2 \\ 1 \times 2 + 2 \times 1 & 1 \times 3 + 2 \times 2 \end{bmatrix*} = \begin{bmatrix*}[r] 2x & 3x \\ x & 2x \end{bmatrix*} + \begin{bmatrix*}[r] y & 0 \\ 0 & y \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 + 3 & 6 + 6 \\ 2 + 2 & 3 + 4 \end{bmatrix*} = \begin{bmatrix*}[r] 2x + y & 3x + 0 \\ x + 0 & 2x + y \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 7 & 12 \\ 4 & 7 \end{bmatrix*} = \begin{bmatrix*}[r] 2x + y & 3x + 0 \\ x + 0 & 2x + y \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 7 & 12 \\ 4 & 7 \end{bmatrix*} = \begin{bmatrix*}[r] 2x + y & 3x \\ x & 2x + y \end{bmatrix*} \\[1em] A = [ 2 1 3 2 ] , I = [ 1 0 0 1 ] A 2 = x A + y I ⇒ [ 2 1 3 2 ] [ 2 1 3 2 ] = x [ 2 1 3 2 ] + y [ 1 0 0 1 ] ⇒ [ 2 × 2 + 3 × 1 1 × 2 + 2 × 1 2 × 3 + 3 × 2 1 × 3 + 2 × 2 ] = [ 2 x x 3 x 2 x ] + [ y 0 0 y ] ⇒ [ 4 + 3 2 + 2 6 + 6 3 + 4 ] = [ 2 x + y x + 0 3 x + 0 2 x + y ] ⇒ [ 7 4 12 7 ] = [ 2 x + y x + 0 3 x + 0 2 x + y ] ⇒ [ 7 4 12 7 ] = [ 2 x + y x 3 x 2 x + y ]
By definition of equality of matrices we get,
3x = 12 or x = 4 (...Eq 1)
2x + y = 7 (...Eq 2)
Putting value of x from Eq1 in Eq 2,
⇒ 2(4) + y = 7
Hence, the value of x = 4 and y = -1.
If P = [ 2 6 3 9 ] and Q = [ 3 x y 2 ] , \begin{bmatrix*}[r] 2 & 6 \\ 3 & 9 \end{bmatrix*} \text{ and Q } = \begin{bmatrix*}[r] 3 & x \\ y & 2 \end{bmatrix*}, [ 2 3 6 9 ] and Q = [ 3 y x 2 ] ,
Answer
Given,
PQ = 0
⇒ [ 2 6 3 9 ] [ 3 x y 2 ] = [ 0 0 0 0 ] ⇒ [ 2 × 3 + 6 × y 2 × x + 6 × 2 3 × 3 + 9 × y 3 × x + 9 × 2 ] = [ 0 0 0 0 ] ⇒ [ 6 + 6 y 2 x + 12 9 + 9 y 3 x + 18 ] = [ 0 0 0 0 ] \Rightarrow \begin{bmatrix*}[r] 2 & 6 \\ 3 & 9 \end{bmatrix*} \begin{bmatrix*}[r] 3 & x \\ y & 2 \end{bmatrix*} = \begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2 \times 3 + 6 \times y & 2 \times x + 6 \times 2 \\ 3 \times 3 + 9 \times y & 3 \times x + 9 \times 2 \end{bmatrix*} = \begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 6 + 6y & 2x + 12 \\ 9 + 9y & 3x + 18 \end{bmatrix*} = \begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*} \\[1em] ⇒ [ 2 3 6 9 ] [ 3 y x 2 ] = [ 0 0 0 0 ] ⇒ [ 2 × 3 + 6 × y 3 × 3 + 9 × y 2 × x + 6 × 2 3 × x + 9 × 2 ] = [ 0 0 0 0 ] ⇒ [ 6 + 6 y 9 + 9 y 2 x + 12 3 x + 18 ] = [ 0 0 0 0 ]
By definition of equality of matrices we get,
⇒ 6 + 6y = 0 and 2x + 12 = 0
Checking whether x = -6 and y = -1, satisfies other equations 9 + 9y = 0 and 3x + 18 = 0,
⇒ 9 + 9y = 0
⇒ 3x + 18 = 0
∴ x = -6 and y = -1.
Hence, the value of x = -6 and y = -1.
Let M × [ 1 1 0 2 ] = [ 1 2 ] \times \begin{bmatrix*}[r] 1 & 1 \\ 0 & 2 \end{bmatrix*} = \begin{bmatrix*}[r] 1 & 2 \end{bmatrix*} × [ 1 0 1 2 ] = [ 1 2 ]
(i) State the order of the matrix M.
(ii) Find the matrix M.
Answer
(i)
Since,
M × [ 1 1 0 2 ] = [ 1 2 ] ⇒ M × [ 1 1 0 2 ] is a 1 × 2 matrix, but [ 1 1 0 2 ] is a 2 × 2 matrix . ⇒ M is a 1 × 2 matrix . \text{M} \times \begin{bmatrix*}[r] 1 & 1 \\ 0 & 2 \end{bmatrix*} = \begin{bmatrix*}[r] 1 & 2 \end{bmatrix*} \\[1em] \Rightarrow \text{M} \times \begin{bmatrix*}[r] 1 & 1 \\ 0 & 2 \end{bmatrix*} \text{is a } 1 \times 2 \text{ matrix, but} \begin{bmatrix*}[r] 1 & 1 \\ 0 & 2 \end{bmatrix*} \text{ is a } 2 \times 2 \text{ matrix}. \\[1em] \Rightarrow \text{M is a } 1 \times 2 \text{ matrix}. M × [ 1 0 1 2 ] = [ 1 2 ] ⇒ M × [ 1 0 1 2 ] is a 1 × 2 matrix, but [ 1 0 1 2 ] is a 2 × 2 matrix . ⇒ M is a 1 × 2 matrix .
The order of matrix M is 1 × 2.
(ii)
Let M = [ x y ] Given, M × [ 1 1 0 2 ] = [ 1 2 ] ⇒ [ x y ] × [ 1 1 0 2 ] = [ 1 2 ] ⇒ [ x × 1 + y × 0 x × 1 + y × 2 ] = [ 1 2 ] ⇒ [ x x + 2 y ] = [ 1 2 ] \text{Let M =} \begin{bmatrix*}[r] x & y \end{bmatrix*} \\[1em] \text{Given, } \text{M} \times \begin{bmatrix*}[r] 1 & 1 \\ 0 & 2 \end{bmatrix*} = \begin{bmatrix*}[r] 1 & 2 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] x & y \end{bmatrix*} \times \begin{bmatrix*}[r] 1 & 1 \\ 0 & 2 \end{bmatrix*} = \begin{bmatrix*}[r] 1 & 2 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] x \times 1 + y \times 0 & x \times 1 + y \times 2 \\ \end{bmatrix*} = \begin{bmatrix*}[r] 1 & 2 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] x & x + 2y \\ \end{bmatrix*} = \begin{bmatrix*}[r] 1 & 2 \end{bmatrix*} \\[1em] Let M = [ x y ] Given, M × [ 1 0 1 2 ] = [ 1 2 ] ⇒ [ x y ] × [ 1 0 1 2 ] = [ 1 2 ] ⇒ [ x × 1 + y × 0 x × 1 + y × 2 ] = [ 1 2 ] ⇒ [ x x + 2 y ] = [ 1 2 ]
By definition of equality of matrices we get,
x = 1 (...Eq 1)
Putting value of x from Eq 1 in Eq 2,
⇒ x + 2y = 21 2 \dfrac{1}{2} 2 1
Since, M = [ x y ] ∴ M = [ 1 1 2 ] . \text{Since, M }= \begin{bmatrix*}[r] x & y \\ \end{bmatrix*} \\[1em] \therefore \text{M} = \begin{bmatrix*}[r] 1 & \dfrac{1}{2} \\ \end{bmatrix*}. Since, M = [ x y ] ∴ M = [ 1 2 1 ] .
Hence, the matrix M = [ 1 1 2 ] \begin{bmatrix*}[r] 1 & \dfrac{1}{2} \end{bmatrix*} [ 1 2 1 ]
Given [ 2 1 − 3 4 ] X = [ 7 6 ] , \begin{bmatrix*}[r] 2 & 1 \\ -3 & 4 \end{bmatrix*}X = \begin{bmatrix*}[r] 7 \\ 6 \end{bmatrix*}, [ 2 − 3 1 4 ] X = [ 7 6 ] ,
(i) the order of the matrix X
(ii) the matrix X.
Answer
(i) Since,
[ 2 1 − 3 4 ] X = [ 7 6 ] ⇒ [ 2 1 − 3 4 ] X is a 2 × 1 matrix, but [ 2 1 − 3 4 ] is a 2 × 2 matrix . ⇒ X is a 2 × 1 matrix . \begin{bmatrix*}[r] 2 & 1 \\ -3 & 4 \end{bmatrix*}X = \begin{bmatrix*}[r] 7 \\ 6 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2 & 1 \\ -3 & 4 \end{bmatrix*}X \text{ is a } 2 \times 1 \text{ matrix, but} \begin{bmatrix*}[r] 2 & 1 \\ -3 & 4 \end{bmatrix*} \text{ is a } 2 \times 2 \text{ matrix}. \\[1em] \Rightarrow \text{X is a } 2 \times 1 \text{ matrix}. [ 2 − 3 1 4 ] X = [ 7 6 ] ⇒ [ 2 − 3 1 4 ] X is a 2 × 1 matrix, but [ 2 − 3 1 4 ] is a 2 × 2 matrix . ⇒ X is a 2 × 1 matrix .
The order of the matrix is 2 × 1.
(ii) Let X = [ x y ] \text{X} = \begin{bmatrix*}[r] x \\ y \end{bmatrix*} X = [ x y ]
Given,
[ 2 1 − 3 4 ] X = [ 7 6 ] ⇒ [ 2 1 − 3 4 ] [ x y ] = [ 7 6 ] ⇒ [ 2 × x + 1 × y − 3 × x + 4 × y ] = [ 7 6 ] ⇒ [ 2 x + y − 3 x + 4 y ] = [ 7 6 ] \begin{bmatrix*}[r] 2 & 1 \\ -3 & 4 \end{bmatrix*}X = \begin{bmatrix*}[r] 7 \\ 6 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2 & 1 \\ -3 & 4 \end{bmatrix*} \begin{bmatrix*}[r] x \\ y \end{bmatrix*} = \begin{bmatrix*}[r] 7 \\ 6 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2 \times x + 1 \times y \\ -3 \times x + 4 \times y \end{bmatrix*} = \begin{bmatrix*}[r] 7 \\ 6 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2x + y \\ -3x + 4y \end{bmatrix*} = \begin{bmatrix*}[r] 7 \\ 6 \end{bmatrix*} \\[1em] [ 2 − 3 1 4 ] X = [ 7 6 ] ⇒ [ 2 − 3 1 4 ] [ x y ] = [ 7 6 ] ⇒ [ 2 × x + 1 × y − 3 × x + 4 × y ] = [ 7 6 ] ⇒ [ 2 x + y − 3 x + 4 y ] = [ 7 6 ]
By definition of equality of matrices we get,
2x + y = 7 or y = 7 - 2x (...Eq 1)
-3x + 4y = 6 (...Eq 2)
Putting value of y from Eq 1 in Eq 2
⇒ -3x + 4y = 6
∴ x = 2 and y = 7 - 2x = 7 - 2(2) = 7 - 4 = 3.
Since, X = [ x y ] ∴ X = [ 2 3 ] \text{X} = \begin{bmatrix*}[r] x \\ y \end{bmatrix*} \\[1em] \therefore \text{X} = \begin{bmatrix*}[r] 2 \\ 3 \end{bmatrix*} X = [ x y ] ∴ X = [ 2 3 ]
Hence, the matrix X = [ 2 3 ] . \text{X} = \begin{bmatrix*}[r] 2 \\ 3 \end{bmatrix*} . X = [ 2 3 ] .
Solve the matrix equation [ 4 1 ] X = [ − 4 8 − 1 2 ] . \begin{bmatrix*}[r] 4 \\ 1 \end{bmatrix*} X = \begin{bmatrix*}[r] -4 & 8 \\ -1 & 2 \end{bmatrix*} . [ 4 1 ] X = [ − 4 − 1 8 2 ] .
Answer
Since,
[ 4 1 ] X = [ − 4 8 − 1 2 ] ⇒ [ 4 1 ] X is a 2 × 2 matrix, but [ 4 1 ] is a 2 × 1 matrix . ⇒ X is a 1 × 2 matrix . \begin{bmatrix*}[r] 4 \\ 1 \end{bmatrix*} X = \begin{bmatrix*}[r] -4 & 8 \\ -1 & 2 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 \\ 1 \end{bmatrix*} X \text{ is a } 2 \times 2 \text{ matrix, but} \begin{bmatrix*}[r] 4 \\ 1 \end{bmatrix*} \text{ is a } 2 \times 1 \text{ matrix}. \\[1em] \Rightarrow \text{X is a } 1 \times 2 \text{ matrix}. [ 4 1 ] X = [ − 4 − 1 8 2 ] ⇒ [ 4 1 ] X is a 2 × 2 matrix, but [ 4 1 ] is a 2 × 1 matrix . ⇒ X is a 1 × 2 matrix .
We know that X matrix will be of order 1 × 2. So, let matrix X be [ x y ] . \begin{bmatrix*}[r] x & y \end{bmatrix*}. [ x y ] .
Given,
[ 4 1 ] X = [ − 4 8 − 1 2 ] ⇒ [ 4 1 ] [ x y ] = [ − 4 8 − 1 2 ] ⇒ [ 4 x 4 y x y ] = [ − 4 8 − 1 2 ] \begin{bmatrix*}[r] 4 \\ 1 \end{bmatrix*} X = \begin{bmatrix*}[r] -4 & 8 \\ -1 & 2 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 \\ 1 \end{bmatrix*} \begin{bmatrix*}[r] x & y \end{bmatrix*} = \begin{bmatrix*}[r] -4 & 8 \\ -1 & 2 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 4x & 4y \\ x & y \end{bmatrix*} = \begin{bmatrix*}[r] -4 & 8 \\ -1 & 2 \end{bmatrix*} \\[1em] [ 4 1 ] X = [ − 4 − 1 8 2 ] ⇒ [ 4 1 ] [ x y ] = [ − 4 − 1 8 2 ] ⇒ [ 4 x x 4 y y ] = [ − 4 − 1 8 2 ]
From definition of equality of matrices we get,
⇒ x = -1 and y = 2.
Since, X = [ x y ] ∴ X = [ − 1 2 ] \text{Since, X }= \begin{bmatrix*}[r] x & y \\ \end{bmatrix*} \\[1em] \therefore \text{X } = \begin{bmatrix*}[r] -1 & 2 \\ \end{bmatrix*} Since, X = [ x y ] ∴ X = [ − 1 2 ]
Hence, the matrix X = [ − 1 2 ] . \text{X} = \begin{bmatrix*}[r] -1 & 2 \end{bmatrix*}. X = [ − 1 2 ] .
If A = [ 2 − 1 − 4 5 ] and B = [ − 3 2 ] , \begin{bmatrix*}[r] 2 & -1 \\ -4 & 5 \end{bmatrix*} \text{and B } = \begin{bmatrix*}[r] -3 \\ 2 \end{bmatrix*}, [ 2 − 4 − 1 5 ] and B = [ − 3 2 ] ,
Answer
Given,
AC = B
⇒ [ 2 − 1 − 4 5 ] C = [ − 3 2 ] ⇒ [ 2 − 1 − 4 5 ] C is a 2 × 1 matrix, but [ 2 − 1 − 4 5 ] is a 2 × 2 matrix . ∴ C is a 1 × 2 matrix . Let matrix C = [ x y ] ⇒ [ 2 − 1 − 4 5 ] [ x y ] = [ − 3 2 ] ⇒ [ 2 × x + ( − 1 ) × y − 4 × x + 5 × y ] = [ − 3 2 ] ⇒ [ 2 x − y − 4 x + 5 y ] = [ − 3 2 ] \Rightarrow \begin{bmatrix*}[r] 2 & -1 \\ -4 & 5 \end{bmatrix*} C = \begin{bmatrix*}[r] -3 \\ 2 \end{bmatrix*} \\[0.5em] \Rightarrow \begin{bmatrix*}[r] 2 & -1 \\ -4 & 5 \end{bmatrix*} C \text{ is a } 2 \times 1 \text{ matrix, but} \begin{bmatrix*}[r] 2 & -1 \\ -4 & 5 \end{bmatrix*} \text{ is a } 2 \times 2 \text{ matrix}. \\[0.5em] \therefore \text{C is a } 1 \times 2 \text{ matrix}. \\[0.5em] \text{Let matrix C } = \begin{bmatrix*}[r] x \\ y \end{bmatrix*} \\[0.5em] \Rightarrow \begin{bmatrix*}[r] 2 & -1 \\ -4 & 5 \end{bmatrix*} \begin{bmatrix*}[r] x \\ y \end{bmatrix*} = \begin{bmatrix*}[r] -3 \\ 2 \end{bmatrix*} \\[0.5em] \Rightarrow \begin{bmatrix*}[r] 2 \times x + (-1) \times y \\ -4 \times x + 5 \times y \end{bmatrix*} = \begin{bmatrix*}[r] -3 \\ 2 \end{bmatrix*} \\[0.5em] \Rightarrow \begin{bmatrix*}[r] 2x - y \\ -4x + 5y \end{bmatrix*} = \begin{bmatrix*}[r] -3 \\ 2 \end{bmatrix*} \\[0.5em] ⇒ [ 2 − 4 − 1 5 ] C = [ − 3 2 ] ⇒ [ 2 − 4 − 1 5 ] C is a 2 × 1 matrix, but [ 2 − 4 − 1 5 ] is a 2 × 2 matrix . ∴ C is a 1 × 2 matrix . Let matrix C = [ x y ] ⇒ [ 2 − 4 − 1 5 ] [ x y ] = [ − 3 2 ] ⇒ [ 2 × x + ( − 1 ) × y − 4 × x + 5 × y ] = [ − 3 2 ] ⇒ [ 2 x − y − 4 x + 5 y ] = [ − 3 2 ]
By definition of equality of matrices we get,
⇒ 2x - y = -3 or y = 2x + 3 (...Eq 1)
⇒ -4x + 5y = 2 (...Eq 2)
Putting value of y from Eq 1 in Eq 2,
⇒ -4x + 5y = 2− 13 6 -\dfrac{13}{6} − 6 13
Now finding value of y,
y = 2 x + 3 = 2 ( − 13 6 ) + 3 = − 26 6 + 3 = − 26 + 18 6 = − 8 6 = − 4 3 . y = 2x + 3 \\[1em] = 2\Big(-\dfrac{13}{6}\Big) + 3 \\[1em] = -\dfrac{26}{6} + 3 \\[1em] = \dfrac{-26 + 18}{6} \\[1em] = -\dfrac{8}{6} \\[1em] = -\dfrac{4}{3}. y = 2 x + 3 = 2 ( − 6 13 ) + 3 = − 6 26 + 3 = 6 − 26 + 18 = − 6 8 = − 3 4 .
∴ x = − 13 6 and y = − 4 3 . -\dfrac{13}{6} \text{ and y } = -\dfrac{4}{3}. − 6 13 and y = − 3 4 .
Since,
C = [ x y ] ∴ C = [ − 13 6 − 4 3 ] \text{C }= \begin{bmatrix*}[r] x \\ y \end{bmatrix*} \\[1em] \therefore C = \begin{bmatrix*}[r] -\dfrac{13}{6} \\ -\dfrac{4}{3} \end{bmatrix*} C = [ x y ] ∴ C = − 6 13 − 3 4
Hence, the matrix C = [ − 13 6 − 4 3 ] . \begin{bmatrix*}[r] -\dfrac{13}{6} \\ -\dfrac{4}{3} \end{bmatrix*}. − 6 13 − 3 4 .
If A = [ 2 − 1 − 4 5 ] and B = [ 0 − 3 ] , \begin{bmatrix*}[r] 2 & -1 \\ -4 & 5 \end{bmatrix*} \text{and B } = \begin{bmatrix*}[r] 0 & -3 \\ \end{bmatrix*}, [ 2 − 4 − 1 5 ] and B = [ 0 − 3 ] ,
Answer
Given,
CA = B.
C [ 2 − 1 − 4 5 ] = [ 0 − 3 ] ⇒ C [ 2 − 1 − 4 5 ] is a 1 × 2 matrix, but [ 2 − 1 − 4 5 ] is a 2 × 2 matrix . ∴ C is a 1 × 2 matrix . Let matrix C = [ x y ] ⇒ [ x y ] [ 2 − 1 − 4 5 ] = [ 0 − 3 ] ⇒ [ x × 2 + y × ( − 4 ) x × ( − 1 ) + y × 5 ] = [ 0 − 3 ] ⇒ [ 2 x − 4 y − x + 5 y ] = [ 0 − 3 ] C\begin{bmatrix*}[r] 2 & -1 \\ -4 & 5 \end{bmatrix*} = \begin{bmatrix*}[r] 0 & -3 \\ \end{bmatrix*} \\[1em] \Rightarrow C\begin{bmatrix*}[r] 2 & -1 \\ -4 & 5 \end{bmatrix*} \text{ is a } 1 \times 2 \text{ matrix, but} \begin{bmatrix*}[r] 2 & -1 \\ -4 & 5 \end{bmatrix*} \text{ is a } 2 \times 2 \text{ matrix}. \\[1em] \therefore \text{C is a } 1 \times 2 \text{ matrix}. \\[1em] \text{Let matrix C } = \begin{bmatrix*}[r] x & y \\ \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] x & y \\ \end{bmatrix*} \begin{bmatrix*}[r] 2 & -1 \\ -4 & 5 \end{bmatrix*} = \begin{bmatrix*}[r] 0 & -3 \\ \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] x \times 2 + y \times (-4) & x \times (-1) + y \times 5 \end{bmatrix*} = \begin{bmatrix*}[r] 0 & -3 \\ \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2x - 4y & -x + 5y \\ \end{bmatrix*} = \begin{bmatrix*}[r] 0 & -3 \\ \end{bmatrix*} C [ 2 − 4 − 1 5 ] = [ 0 − 3 ] ⇒ C [ 2 − 4 − 1 5 ] is a 1 × 2 matrix, but [ 2 − 4 − 1 5 ] is a 2 × 2 matrix . ∴ C is a 1 × 2 matrix . Let matrix C = [ x y ] ⇒ [ x y ] [ 2 − 4 − 1 5 ] = [ 0 − 3 ] ⇒ [ x × 2 + y × ( − 4 ) x × ( − 1 ) + y × 5 ] = [ 0 − 3 ] ⇒ [ 2 x − 4 y − x + 5 y ] = [ 0 − 3 ]
By definition of equality of matrices we get,
2x - 4y = 0 (...Eq 1)
-x + 5y = -3 or x = 5y + 3 (...Eq 2)
Putting value of x from Eq 2 in Eq 1,
⇒ 2x - 4y = 0
Now finding value of x,
⇒ x = 5y + 3 = 5(-1) + 3 = -5 + 3 = -2.
∴ x = -2, y = -1.
Since, C = [ x y ] ∴ C = [ − 2 − 1 ] \text{Since, C }= \begin{bmatrix*}[r] x & y \\ \end{bmatrix*} \\[0.5em] \therefore C = \begin{bmatrix*}[r] -2 & -1 \\ \end{bmatrix*} Since, C = [ x y ] ∴ C = [ − 2 − 1 ]
Hence, the matrix C = [ − 2 − 1 ] \begin{bmatrix*}[r] -2 & -1 \\ \end{bmatrix*} [ − 2 − 1 ]
If A = [ 3 − 4 − 1 2 ] , \begin{bmatrix*}[r] 3 & -4 \\ -1 & 2 \end{bmatrix*}, [ 3 − 1 − 4 2 ] ,
Answer
Given,
BA = I
B [ 3 − 4 − 1 2 ] = [ 1 0 0 1 ] ⇒ B [ 3 − 4 − 1 2 ] is a 2 × 2 matrix, and [ 3 − 4 − 1 2 ] is a 2 × 2 matrix . ∴ B is a 2 × 2 matrix . I = [ 1 0 0 1 ] \text{B}\begin{bmatrix*}[r] 3 & -4 \\ -1 & 2 \end{bmatrix*} = \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em] \Rightarrow \text{B}\begin{bmatrix*}[r] 3 & -4 \\ -1 & 2 \end{bmatrix*} \text{ is a } 2 \times 2 \text{ matrix, and} \begin{bmatrix*}[r] 3 & -4 \\ -1 & 2 \end{bmatrix*} \text{ is a } 2 \times 2 \text{ matrix}. \\[1em] \therefore \text{B is a } 2 \times 2 \text{ matrix}. \\[1em] \text{I } = \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} B [ 3 − 1 − 4 2 ] = [ 1 0 0 1 ] ⇒ B [ 3 − 1 − 4 2 ] is a 2 × 2 matrix, and [ 3 − 1 − 4 2 ] is a 2 × 2 matrix . ∴ B is a 2 × 2 matrix . I = [ 1 0 0 1 ]
We know that B will be of order 2 × 2. So, let
B = [ a b c d ] ⇒ [ a b c d ] [ 3 − 4 − 1 2 ] = [ 1 0 0 1 ] ⇒ [ a × 3 + b × ( − 1 ) a × ( − 4 ) + b × 2 c × 3 + d × ( − 1 ) c × ( − 4 ) + d × 2 ] = [ 1 0 0 1 ] ⇒ [ 3 a − b − 4 a + 2 b 3 c − d − 4 c + 2 d ] = [ 1 0 0 1 ] \text{B} = \begin{bmatrix*}[r] a & b \\ c & d \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] a & b \\ c & d \end{bmatrix*} \begin{bmatrix*}[r] 3 & -4 \\ -1 & 2 \end{bmatrix*} = \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] a \times 3 + b \times (-1) & a \times (-4) + b \times 2 \\ c \times 3 + d \times (-1) & c \times (-4) + d \times 2 \end{bmatrix*} = \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 3a - b & -4a + 2b \\ 3c - d & -4c + 2d \end{bmatrix*} = \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em] B = [ a c b d ] ⇒ [ a c b d ] [ 3 − 1 − 4 2 ] = [ 1 0 0 1 ] ⇒ [ a × 3 + b × ( − 1 ) c × 3 + d × ( − 1 ) a × ( − 4 ) + b × 2 c × ( − 4 ) + d × 2 ] = [ 1 0 0 1 ] ⇒ [ 3 a − b 3 c − d − 4 a + 2 b − 4 c + 2 d ] = [ 1 0 0 1 ]
By definition of equality of matrices we get,
3a - b = 1 (...Eq 1)
-4a + 2b = 0
3c - d = 0
-4c + 2d = 1 (...Eq 4)
Putting value of b from Eq 2 in Eq 1
⇒ 3a - b = 1
∴ a = 1, b = 2a = 2.
Putting value of d from Eq 3 in Eq 4
⇒ -4c + 2d = 11 2 \dfrac{1}{2} 2 1
∴ c = 1 2 \dfrac{1}{2} 2 1 3 2 \dfrac{3}{2} 2 3
Since, B = [ a b c d ] ∴ B = [ 1 2 1 2 3 2 ] \text{Since, B }= \begin{bmatrix*}[r] a & b \\ c & d \end{bmatrix*} \\[1em] \therefore \text{B} = \begin{bmatrix*}[r] 1 & 2 \\ \dfrac{1}{2} & \dfrac{3}{2} \end{bmatrix*} Since, B = [ a c b d ] ∴ B = [ 1 2 1 2 2 3 ]
Hence, the matrix B = [ 1 2 1 2 3 2 ] \begin{bmatrix*}[r] 1 & 2 \\ \dfrac{1}{2} & \dfrac{3}{2} \end{bmatrix*} [ 1 2 1 2 2 3 ]
Given [ 4 2 − 1 1 ] \begin{bmatrix*}[r] 4 & 2 \\ -1 & 1 \end{bmatrix*} [ 4 − 1 2 1 ]
(i) State the order of matrix M.
(ii) Find the matrix M.
Answer
(i) Given,
[ 4 2 − 1 1 ] M = 6 I ⇒ [ 4 2 − 1 1 ] M is a 2 × 2 matrix, and [ 4 2 − 1 1 ] is a 2 × 2 matrix . ∴ M is a 2 × 2 matrix . \begin{bmatrix*}[r] 4 & 2 \\ -1 & 1 \end{bmatrix*} M = 6I \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 & 2 \\ -1 & 1 \end{bmatrix*} M \text{ is a } 2 \times 2 \text{ matrix, and} \begin{bmatrix*}[r] 4 & 2 \\ -1 & 1 \end{bmatrix*} \text{ is a } 2 \times 2 \text{ matrix}. \\[1em] \therefore \text{M is a } 2 \times 2 \text{ matrix}. \\[1em] [ 4 − 1 2 1 ] M = 6 I ⇒ [ 4 − 1 2 1 ] M is a 2 × 2 matrix, and [ 4 − 1 2 1 ] is a 2 × 2 matrix . ∴ M is a 2 × 2 matrix .
Hence, the matrix M is of order 2 × 2.
(ii) Let matrix M be [ a b c d ] . \begin{bmatrix*}[r] a & b \\ c & d \end{bmatrix*}. [ a c b d ] .
Given,
[ 4 2 − 1 1 ] M = 6 I ⇒ [ 4 2 − 1 1 ] [ a b c d ] = 6 [ 1 0 0 1 ] ⇒ [ 4 × a + 2 × c 4 × b + 2 × d ( − 1 ) × a + 1 × c ( − 1 ) × b + 1 × d ] = [ 6 0 0 6 ] ⇒ [ 4 a + 2 c 4 b + 2 d − a + c − b + d ] = [ 6 0 0 6 ] \begin{bmatrix*}[r] 4 & 2 \\ -1 & 1 \end{bmatrix*} M = 6I \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 & 2 \\ -1 & 1 \end{bmatrix*} \begin{bmatrix*}[r] a & b \\ c & d \end{bmatrix*} = 6\begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 \times a + 2 \times c & 4 \times b + 2 \times d \\ (-1) \times a + 1 \times c & (-1) \times b + 1 \times d \end{bmatrix*} = \begin{bmatrix*}[r] 6 & 0 \\ 0 & 6 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 4a + 2c & 4b + 2d \\ -a + c & -b + d \end{bmatrix*} = \begin{bmatrix*}[r] 6 & 0 \\ 0 & 6 \end{bmatrix*} \\[1em] [ 4 − 1 2 1 ] M = 6 I ⇒ [ 4 − 1 2 1 ] [ a c b d ] = 6 [ 1 0 0 1 ] ⇒ [ 4 × a + 2 × c ( − 1 ) × a + 1 × c 4 × b + 2 × d ( − 1 ) × b + 1 × d ] = [ 6 0 0 6 ] ⇒ [ 4 a + 2 c − a + c 4 b + 2 d − b + d ] = [ 6 0 0 6 ]
By definition of equality of matrices we get,
4a + 2c = 6 (...Eq 1)
4b + 2d = 0
-a + c = 0
-b + d = 6 (...Eq 4)
Putting value of a from Eq 3 in Eq 1
⇒ 4a + 2c = 6
∴ c = 1 and a = c = 1.
Putting value of d from Eq 2 in Eq 4
⇒ -b + d = 6
∴ b = -2 and d = -2b = 4.
Since,
M = [ a b c d ] ∴ M = [ 1 − 2 1 4 ] \text{M} = \begin{bmatrix*}[r] a & b \\ c & d \end{bmatrix*} \\[1em] \therefore \text{M} = \begin{bmatrix*}[r] 1 & -2 \\ 1 & 4 \end{bmatrix*} M = [ a c b d ] ∴ M = [ 1 1 − 2 4 ]
Hence, the matrix M = [ 1 − 2 1 4 ] \text{M} = \begin{bmatrix*}[r] 1 & -2 \\ 1 & 4 \end{bmatrix*} M = [ 1 1 − 2 4 ]
If B = [ − 4 2 5 − 1 ] and C = [ 17 − 1 47 − 13 ] , \begin{bmatrix*}[r] -4 & 2 \\ 5 & -1 \end{bmatrix*} \text{ and C} = \begin{bmatrix*}[r] 17 & -1 \\ 47 & -13 \end{bmatrix*}, [ − 4 5 2 − 1 ] and C = [ 17 47 − 1 − 13 ] ,
Answer
Given,
AB = C
⇒ A [ − 4 2 5 − 1 ] is a 2 × 2 matrix, and [ 4 2 5 − 1 ] is a 2 × 2 matrix . ∴ A is a 2 × 2 matrix . \Rightarrow A\begin{bmatrix*}[r] -4 & 2 \\ 5 & -1 \end{bmatrix*} \text{ is a } 2 \times 2 \text{ matrix, and} \begin{bmatrix*}[r] 4 & 2 \\ 5 & -1 \end{bmatrix*} \text{ is a } 2 \times 2 \text{ matrix}. \\[1em] \therefore \text{A is a } 2 \times 2 \text{ matrix}. ⇒ A [ − 4 5 2 − 1 ] is a 2 × 2 matrix, and [ 4 5 2 − 1 ] is a 2 × 2 matrix . ∴ A is a 2 × 2 matrix .
We know that matrix A will be of order 2 × 2. Let matrix be
A = [ a b c d ] ⇒ [ a b c d ] [ − 4 2 5 − 1 ] = [ 17 − 1 47 − 13 ] ⇒ [ a × ( − 4 ) + b × 5 a × 2 + b × ( − 1 ) c × ( − 4 ) + d × 5 c × 2 + d × ( − 1 ) ] = [ 17 − 1 47 − 13 ] ⇒ [ − 4 a + 5 b 2 a − b − 4 c + 5 d 2 c − d ] = [ 17 − 1 47 − 13 ] \text{A} = \begin{bmatrix*}[r] a & b \\ c & d \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] a & b \\ c & d \end{bmatrix*} \begin{bmatrix*}[r] -4 & 2 \\ 5 & -1 \end{bmatrix*} = \begin{bmatrix*}[r] 17 & -1 \\ 47 & -13 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] a \times (-4) + b \times 5 & a \times 2 + b \times (-1) \\ c \times (-4) + d \times 5 & c \times 2 + d \times (-1) \end{bmatrix*} = \begin{bmatrix*}[r] 17 & -1 \\ 47 & -13 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] -4a + 5b & 2a - b \\ -4c + 5d & 2c - d \end{bmatrix*} = \begin{bmatrix*}[r] 17 & -1 \\ 47 & -13 \end{bmatrix*} \\[1em] A = [ a c b d ] ⇒ [ a c b d ] [ − 4 5 2 − 1 ] = [ 17 47 − 1 − 13 ] ⇒ [ a × ( − 4 ) + b × 5 c × ( − 4 ) + d × 5 a × 2 + b × ( − 1 ) c × 2 + d × ( − 1 ) ] = [ 17 47 − 1 − 13 ] ⇒ [ − 4 a + 5 b − 4 c + 5 d 2 a − b 2 c − d ] = [ 17 47 − 1 − 13 ]
By definition of equality of matrices we get,
-4a + 5b = 17 (...Eq 1)
2a - b = -1
-4c + 5d = 47 (...Eq 3)
2c - d = -13
Putting value of b from Eq 2 in Eq 1
⇒ -4a + 5b = 17
∴ a = 2, b = 2a + 1 = 2(2) + 1 = 5.
Putting value of d from Eq 4 in Eq 3
⇒ -4c + 5d = 47
∴ c = -3, d = 2c + 13 = 2(-3) + 13 = 7.
Since,
A = [ a b c d ] ∴ A = [ 2 5 − 3 7 ] \text{A} = \begin{bmatrix*}[r] a & b \\ c & d \end{bmatrix*} \\[1em] \therefore \text{A} = \begin{bmatrix*}[r] 2 & 5 \\ -3 & 7 \end{bmatrix*} A = [ a c b d ] ∴ A = [ 2 − 3 5 7 ]
Hence, the matrix A = [ 2 5 − 3 7 ] \begin{bmatrix*}[r] 2 & 5 \\ -3 & 7 \end{bmatrix*} [ 2 − 3 5 7 ]
If A = [ 4 − 4 − 4 4 ] \begin{bmatrix*}[r] 4 & -4 \\ -4 & 4 \end{bmatrix*} [ 4 − 4 − 4 4 ] 2 . If A2 = pA, then find the value of p.
Answer
Given,
⇒ A2 = pA
⇒ [ 4 − 4 − 4 4 ] [ 4 − 4 − 4 4 ] = p [ 4 − 4 − 4 4 ] ⇒ [ 4 × 4 + ( − 4 ) × ( − 4 ) 4 × ( − 4 ) + ( − 4 ) × 4 − 4 × 4 + 4 × ( − 4 ) ( − 4 ) × ( − 4 ) + 4 × 4 ] = p [ 4 − 4 − 4 4 ] ⇒ [ 16 + 16 − 16 − 16 − 16 − 16 16 + 16 ] = p [ 4 − 4 − 4 4 ] ⇒ [ 32 − 32 − 32 32 ] = p [ 4 − 4 − 4 4 ] ⇒ 32 [ 1 − 1 − 1 1 ] = 4 p [ 1 − 1 − 1 1 ] ⇒ 32 = 4 p ⇒ p = 32 4 = 8. \Rightarrow \begin{bmatrix*}[r] 4 & -4 \\ -4 & 4 \end{bmatrix*}\begin{bmatrix*}[r] 4 & -4 \\ -4 & 4 \end{bmatrix*} = p\begin{bmatrix*}[r] 4 & -4 \\ -4 & 4 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 \times 4 + (-4) \times (-4) & 4 \times (-4) + (-4) \times 4 \\ -4 \times 4 + 4 \times (-4) & (-4) \times (-4) + 4 \times 4 \end{bmatrix*} = p\begin{bmatrix*}[r] 4 & -4 \\ -4 & 4 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 16 + 16 & -16 - 16 \\ -16 - 16 & 16 + 16 \end{bmatrix*} = p\begin{bmatrix*}[r] 4 & -4 \\ -4 & 4 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 32 & -32 \\ -32 & 32 \end{bmatrix*} = p\begin{bmatrix*}[r] 4 & -4 \\ -4 & 4 \end{bmatrix*} \\[1em] \Rightarrow 32\begin{bmatrix*}[r] 1 & -1 \\ -1 & 1 \end{bmatrix*} = 4p\begin{bmatrix*}[r] 1 & -1 \\ -1 & 1 \end{bmatrix*} \\[1em] \Rightarrow 32 = 4p \\[1em] \Rightarrow p = \dfrac{32}{4} = 8. ⇒ [ 4 − 4 − 4 4 ] [ 4 − 4 − 4 4 ] = p [ 4 − 4 − 4 4 ] ⇒ [ 4 × 4 + ( − 4 ) × ( − 4 ) − 4 × 4 + 4 × ( − 4 ) 4 × ( − 4 ) + ( − 4 ) × 4 ( − 4 ) × ( − 4 ) + 4 × 4 ] = p [ 4 − 4 − 4 4 ] ⇒ [ 16 + 16 − 16 − 16 − 16 − 16 16 + 16 ] = p [ 4 − 4 − 4 4 ] ⇒ [ 32 − 32 − 32 32 ] = p [ 4 − 4 − 4 4 ] ⇒ 32 [ 1 − 1 − 1 1 ] = 4 p [ 1 − 1 − 1 1 ] ⇒ 32 = 4 p ⇒ p = 4 32 = 8.
Hence, p = 8.
Multiple Choice Questions
If [ x + 3 4 y − 4 x + y ] = [ 5 4 3 9 ] , \begin{bmatrix*}[r] x + 3 & 4 \\ y - 4 & x + y \end{bmatrix*} = \begin{bmatrix*}[r] 5 & 4 \\ 3 & 9 \end{bmatrix*}, [ x + 3 y − 4 4 x + y ] = [ 5 3 4 9 ] ,
x = 2, y = 7 x = 7, y = 2 x = 3, y = 6 x = -2, y = 7 Answer
Given,
[ x + 3 4 y − 4 x + y ] = [ 5 4 3 9 ] \begin{bmatrix*}[r] x + 3 & 4 \\ y - 4 & x + y \end{bmatrix*} = \begin{bmatrix*}[r] 5 & 4 \\ 3 & 9 \end{bmatrix*} \\[0.5em] [ x + 3 y − 4 4 x + y ] = [ 5 3 4 9 ]
By definition of equality of matrices we get,
⇒ x + 3 = 5 or x = 5 - 3 = 2.
⇒ y - 4 = 3 or y = 3 + 4 = 7.
⇒ x + y = 9.
Since, x = 2 and y = 7 satisfies the equation x + y = 9, hence, value of x = 2 and y = 7.
∴ Option 1 is the correct option.
If [ x + 2 y − y 3 x 4 ] = [ − 4 3 6 4 ] , \begin{bmatrix*}[r] x + 2y & -y \\ 3x & 4 \end{bmatrix*} = \begin{bmatrix*}[r] -4 & 3 \\ 6 & 4 \end{bmatrix*}, [ x + 2 y 3 x − y 4 ] = [ − 4 6 3 4 ] ,
x = 2, y = 3 x = 2, y = -3 x = -2, y = 3 x = 3, y = 2 Answer
Given,
[ x + 2 y − y 3 x 4 ] = [ − 4 3 6 4 ] \begin{bmatrix*}[r] x + 2y & -y \\ 3x & 4 \end{bmatrix*} = \begin{bmatrix*}[r] -4 & 3 \\ 6 & 4 \end{bmatrix*} \\[0.5em] [ x + 2 y 3 x − y 4 ] = [ − 4 6 3 4 ]
By definition of equality of matrices we get,
⇒ x + 2y = -4 (...Eq 1)
⇒ -y = 3 or y = -3
⇒ 3x = 6 or x = 2
Putting, x = 2 and y = -3 in Eq 1,
⇒ x + 2y = -4
Since, x = 2 and y = -3 satisfies the equation x + 2y = -4,
∴ x = 2 and y = -3.
∴ Option 2 is the correct option.
If [ x − 2 y 5 3 y ] = [ 6 5 3 − 2 ] , \begin{bmatrix*}[r] x - 2y & 5 \\ 3 & y \end{bmatrix*} = \begin{bmatrix*}[r] 6 & 5 \\ 3 & -2 \end{bmatrix*}, [ x − 2 y 3 5 y ] = [ 6 3 5 − 2 ] ,
-2 0 1 2 Answer
Given,
[ x − 2 y 5 3 y ] = [ 6 5 3 − 2 ] \begin{bmatrix*}[r] x - 2y & 5 \\ 3 & y \end{bmatrix*} = \begin{bmatrix*}[r] 6 & 5 \\ 3 & -2 \end{bmatrix*} \\[0.5em] [ x − 2 y 3 5 y ] = [ 6 3 5 − 2 ]
By definition of equality of matrices,
⇒ x - 2y = 6 (...Eq 1)
⇒ y = -2.
Putting value of y in Eq 1 we get,
⇒ x - 2y = 6
∴ x = 2.
∴ Option 4 is the correct option .
If [ x + 2 y 3 y 4 x 2 ] = [ 0 − 3 8 2 ] , \begin{bmatrix*}[r] x + 2y & 3y \\ 4x & 2 \end{bmatrix*} = \begin{bmatrix*}[r] 0 & -3 \\ 8 & 2 \end{bmatrix*}, [ x + 2 y 4 x 3 y 2 ] = [ 0 8 − 3 2 ] ,
-3 1 3 5 Answer
Given,
[ x + 2 y 3 y 4 x 2 ] = [ 0 − 3 8 2 ] \begin{bmatrix*}[r] x + 2y & 3y \\ 4x & 2 \end{bmatrix*} = \begin{bmatrix*}[r] 0 & -3 \\ 8 & 2 \end{bmatrix*} \\[0.5em] [ x + 2 y 4 x 3 y 2 ] = [ 0 8 − 3 2 ]
By definition of equality of matrices we get,
⇒ x + 2y = 0 (...Eq 1)
⇒ 3y = -3 or y = -1
⇒ 4x = 8 or x = 2
Putting the value of x = 2 and y = -1 in Eq 1
⇒ x + 2y = 0
Since, x = 2 and y = -1 satisfies Eq 1
∴ x = 2, y = -1 and x - y = 2 - (-1) = 2 + 1 = 3.
∴ Option 3 is the correct option.
If x [ 2 3 ] + y [ − 1 0 ] = [ 10 6 ] , x\begin{bmatrix*}[r] 2 \\ 3 \end{bmatrix*} + y\begin{bmatrix*}[r] -1 \\ 0 \end{bmatrix*} = \begin{bmatrix*}[r] 10 \\ 6 \end{bmatrix*}, x [ 2 3 ] + y [ − 1 0 ] = [ 10 6 ] ,
x = 2, y = 6 x = 2, y = -6 x = 3, y = -4 x = 3, y = -6 Answer
Given,
x [ 2 3 ] + y [ − 1 0 ] = [ 10 6 ] ⇒ [ 2 x 3 x ] + [ − y 0 ] = [ 10 6 ] ⇒ [ 2 x − y 3 x ] = [ 10 6 ] x\begin{bmatrix*}[r] 2 \\ 3 \end{bmatrix*} + y\begin{bmatrix*}[r] -1 \\ 0 \end{bmatrix*} = \begin{bmatrix*}[r] 10 \\ 6 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2x \\ 3x \end{bmatrix*} + \begin{bmatrix*}[r] -y \\ 0 \end{bmatrix*} = \begin{bmatrix*}[r] 10 \\ 6 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2x - y \\ 3x \end{bmatrix*} = \begin{bmatrix*}[r] 10 \\ 6 \end{bmatrix*} \\[1em] x [ 2 3 ] + y [ − 1 0 ] = [ 10 6 ] ⇒ [ 2 x 3 x ] + [ − y 0 ] = [ 10 6 ] ⇒ [ 2 x − y 3 x ] = [ 10 6 ]
By definition of equality of matrices we get,
⇒ 2x - y = 10 (...Eq 1)
⇒ 3x = 6 or x = 2.
Putting value of x in Eq 1
⇒ 2x - y = 10
∴ x = 2 and y = -6.
∴ Option 2 is the correct option.
If A = [ 0 1 1 0 ] , \text{If A }= \begin{bmatrix*}[r] 0 & 1 \\ 1 & 0 \end{bmatrix*}, If A = [ 0 1 1 0 ] , 2 =
[ 1 1 0 0 ] \begin{bmatrix*}[r] 1 & 1 \\ 0 & 0 \end{bmatrix*} [ 1 0 1 0 ]
[ 0 0 1 1 ] \begin{bmatrix*}[r] 0 & 0 \\ 1 & 1 \end{bmatrix*} [ 0 1 0 1 ]
[ 0 1 1 0 ] \begin{bmatrix*}[r] 0 & 1 \\ 1 & 0 \end{bmatrix*} [ 0 1 1 0 ]
[ 1 0 0 1 ] \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} [ 1 0 0 1 ]
Answer
Given,
A = [ 0 1 1 0 ] ⇒ A 2 = [ 0 1 1 0 ] [ 0 1 1 0 ] = [ 0 × 0 + 1 × 1 0 × 1 + 1 × 0 1 × 0 + 0 × 1 1 × 1 + 0 × 0 ] = [ 1 0 0 1 ] ∴ A 2 = [ 1 0 0 1 ] . \text{A} = \begin{bmatrix*}[r] 0 & 1 \\ 1 & 0 \end{bmatrix*} \\[1em] \Rightarrow \text{A}^2 = \begin{bmatrix*}[r] 0 & 1 \\ 1 & 0 \end{bmatrix*}\begin{bmatrix*}[r] 0 & 1 \\ 1 & 0 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 0 \times 0 + 1 \times 1 & 0 \times 1 + 1 \times 0 \\ 1 \times 0 + 0 \times 1 & 1 \times 1 + 0 \times 0 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em] \therefore \text{A}^2 = \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*}. A = [ 0 1 1 0 ] ⇒ A 2 = [ 0 1 1 0 ] [ 0 1 1 0 ] = [ 0 × 0 + 1 × 1 1 × 0 + 0 × 1 0 × 1 + 1 × 0 1 × 1 + 0 × 0 ] = [ 1 0 0 1 ] ∴ A 2 = [ 1 0 0 1 ] .
∴ Option 4 is the correct option.
If A = [ 0 0 1 0 ] , \text{If A }= \begin{bmatrix*}[r] 0 & 0 \\ 1 & 0 \end{bmatrix*}, If A = [ 0 1 0 0 ] , 2 =
A O I 2A Answer
Given,
A = [ 0 0 1 0 ] ⇒ A 2 = [ 0 0 1 0 ] [ 0 0 1 0 ] = [ 0 × 0 + 0 × 1 0 × 0 + 0 × 0 1 × 0 + 0 × 1 1 × 0 + 0 × 0 ] = [ 0 0 0 0 ] ∴ A 2 = O . \text{A} = \begin{bmatrix*}[r] 0 & 0 \\ 1 & 0 \end{bmatrix*} \\[1em] \Rightarrow \text{A}^2 = \begin{bmatrix*}[r] 0 & 0 \\ 1 & 0 \end{bmatrix*}\begin{bmatrix*}[r] 0 & 0 \\ 1 & 0 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 0 \times 0 + 0 \times 1 & 0 \times 0 + 0 \times 0 \\ 1 \times 0 + 0 \times 1 & 1 \times 0 + 0 \times 0 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*} \\[1em] \therefore \text{A}^2 = O. A = [ 0 1 0 0 ] ⇒ A 2 = [ 0 1 0 0 ] [ 0 1 0 0 ] = [ 0 × 0 + 0 × 1 1 × 0 + 0 × 1 0 × 0 + 0 × 0 1 × 0 + 0 × 0 ] = [ 0 0 0 0 ] ∴ A 2 = O .
∴ Option 2 is the correct option.
If A = [ 1 0 1 1 ] , \text{If A }= \begin{bmatrix*}[r] 1 & 0 \\ 1 & 1 \end{bmatrix*}, If A = [ 1 1 0 1 ] , 2 =
[ 2 0 1 1 ] \begin{bmatrix*}[r] 2 & 0 \\ 1 & 1 \end{bmatrix*} [ 2 1 0 1 ]
[ 1 0 1 2 ] \begin{bmatrix*}[r] 1 & 0 \\ 1 & 2 \end{bmatrix*} [ 1 1 0 2 ]
[ 1 0 2 1 ] \begin{bmatrix*}[r] 1 & 0 \\ 2 & 1 \end{bmatrix*} [ 1 2 0 1 ]
none of these
Answer
Given,
A = [ 1 0 1 1 ] ⇒ A 2 = [ 1 0 1 1 ] [ 1 0 1 1 ] = [ 1 × 1 + 0 × 1 1 × 0 + 0 × 1 1 × 1 + 1 × 1 1 × 0 + 1 × 1 ] = [ 1 0 2 1 ] ∴ A 2 = [ 1 0 2 1 ] . \text{A} = \begin{bmatrix*}[r] 1 & 0 \\ 1 & 1 \end{bmatrix*} \\[0.5em] \Rightarrow \text{A}^2 = \begin{bmatrix*}[r] 1 & 0 \\ 1 & 1 \end{bmatrix*}\begin{bmatrix*}[r] 1 & 0 \\ 1 & 1 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 1 \times 1 + 0 \times 1 & 1 \times 0 + 0 \times 1 \\ 1 \times 1 + 1 \times 1 & 1 \times 0 + 1 \times 1 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 1 & 0 \\ 2 & 1 \end{bmatrix*} \\[0.5em] \therefore \text{A}^2 = \begin{bmatrix*}[r] 1 & 0 \\ 2 & 1 \end{bmatrix*}. A = [ 1 1 0 1 ] ⇒ A 2 = [ 1 1 0 1 ] [ 1 1 0 1 ] = [ 1 × 1 + 0 × 1 1 × 1 + 1 × 1 1 × 0 + 0 × 1 1 × 0 + 1 × 1 ] = [ 1 2 0 1 ] ∴ A 2 = [ 1 2 0 1 ] .
∴ Option 3 is the correct option.
If A = [ 3 1 − 1 2 ] , \text{If A }= \begin{bmatrix*}[r] 3 & 1 \\ -1 & 2 \end{bmatrix*}, If A = [ 3 − 1 1 2 ] , 2 =
[ 8 5 − 5 3 ] \begin{bmatrix*}[r] 8 & 5 \\ -5 & 3 \end{bmatrix*} [ 8 − 5 5 3 ]
[ 8 − 5 5 3 ] \begin{bmatrix*}[r] 8 & -5 \\ 5 & 3 \end{bmatrix*} [ 8 5 − 5 3 ]
[ 8 − 5 − 5 − 3 ] \begin{bmatrix*}[r] 8 & -5 \\ -5 & -3 \end{bmatrix*} [ 8 − 5 − 5 − 3 ]
[ 8 − 5 − 5 3 ] \begin{bmatrix*}[r] 8 & -5 \\ -5 & 3 \end{bmatrix*} [ 8 − 5 − 5 3 ]
Answer
Given,
A = [ 3 1 − 1 2 ] ⇒ A 2 = [ 3 1 − 1 2 ] [ 3 1 − 1 2 ] = [ 3 × 3 + 1 × ( − 1 ) 3 × 1 + 1 × 2 ( − 1 ) × 3 + 2 × ( − 1 ) ( − 1 ) × 1 + 2 × 2 ] = [ 9 − 1 3 + 2 − 3 − 2 − 1 + 4 ] = [ 8 5 − 5 3 ] ∴ A 2 = [ 8 5 − 5 3 ] . \text{A} = \begin{bmatrix*}[r] 3 & 1 \\ -1 & 2 \end{bmatrix*} \\[0.5em] \Rightarrow \text{A}^2 = \begin{bmatrix*}[r] 3 & 1 \\ -1 & 2 \end{bmatrix*}\begin{bmatrix*}[r] 3 & 1 \\ -1 & 2 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 3 \times 3 + 1 \times (-1) & 3 \times 1 + 1 \times 2 \\ (-1) \times 3 + 2 \times (-1) & (-1) \times 1 + 2 \times 2 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 9 - 1 & 3 + 2 \\ -3 - 2 & -1 + 4 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 8 & 5 \\ -5 & 3 \end{bmatrix*} \\[0.5em] \therefore \text{A}^2 = \begin{bmatrix*}[r] 8 & 5 \\ -5 & 3 \end{bmatrix*}. A = [ 3 − 1 1 2 ] ⇒ A 2 = [ 3 − 1 1 2 ] [ 3 − 1 1 2 ] = [ 3 × 3 + 1 × ( − 1 ) ( − 1 ) × 3 + 2 × ( − 1 ) 3 × 1 + 1 × 2 ( − 1 ) × 1 + 2 × 2 ] = [ 9 − 1 − 3 − 2 3 + 2 − 1 + 4 ] = [ 8 − 5 5 3 ] ∴ A 2 = [ 8 − 5 5 3 ] .
∴ Option 1 is the correct option.
If matrix 𝐴 = [ 2 2 0 2 ] and A 2 = [ 4 x 0 4 ] \begin{bmatrix*}[r] 2 & 2 \\ 0 & 2 \end{bmatrix*}\text{ and } A^2 = \begin{bmatrix*}[r] 4 & x \\ 0 & 4 \end{bmatrix*} [ 2 0 2 2 ] and A 2 = [ 4 0 x 4 ]
2
4
8
10
Answer
⇒ A 2 = [ 2 2 0 2 ] [ 2 2 0 2 ] ⇒ A 2 = [ 2 × 2 + 2 × 0 2 × 2 + 2 × 2 0 × 2 + 2 × 0 0 × 2 + 2 × 2 ] ⇒ [ 4 x 0 4 ] = [ 4 + 0 4 + 4 0 + 0 0 + 4 ] ⇒ [ 4 x 0 4 ] = [ 4 8 0 4 ] ⇒ x = 8. \phantom{\Rightarrow} A^2 = \begin{bmatrix*}[r] 2 & 2 \\ 0 & 2 \end{bmatrix*}\begin{bmatrix*}[r] 2 & 2 \\ 0 & 2 \end{bmatrix*} \\[1em] \Rightarrow A^2 = \begin{bmatrix*}[r] 2 \times 2 + 2\times 0 & 2 \times 2 + 2 \times 2 \\ 0 \times 2 + 2 \times 0 & 0 \times 2 + 2 \times 2 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 & x \\ 0 & 4 \end{bmatrix*} = \begin{bmatrix*}[r] 4 + 0 & 4 + 4 \\ 0 + 0 & 0 + 4 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 & x \\ 0 & 4 \end{bmatrix*} = \begin{bmatrix*}[r] 4 & 8 \\ 0 & 4 \end{bmatrix*} \\[1em] \Rightarrow x = 8. ⇒ A 2 = [ 2 0 2 2 ] [ 2 0 2 2 ] ⇒ A 2 = [ 2 × 2 + 2 × 0 0 × 2 + 2 × 0 2 × 2 + 2 × 2 0 × 2 + 2 × 2 ] ⇒ [ 4 0 x 4 ] = [ 4 + 0 0 + 0 4 + 4 0 + 4 ] ⇒ [ 4 0 x 4 ] = [ 4 0 8 4 ] ⇒ x = 8.
Hence, Option 3 is the correct option.
If A = [ 2 − 2 − 2 2 ] , \begin{bmatrix*}[r] 2 & -2 \\ -2 & 2 \end{bmatrix*}, [ 2 − 2 − 2 2 ] , 2 = pA, then the value of p is
2
4
-2
-4
Answer
Given,
A 2 = p A ⇒ [ 2 − 2 − 2 2 ] [ 2 − 2 − 2 2 ] = p [ 2 − 2 − 2 2 ] ⇒ [ 2 × 2 + ( − 2 ) × ( − 2 ) 2 × ( − 2 ) + ( − 2 ) × 2 − 2 × 2 + 2 × ( − 2 ) ( − 2 ) × ( − 2 ) + 2 × 2 ] = p [ 2 − 2 − 2 2 ] ⇒ [ 4 + 4 − 4 − 4 − 4 − 4 4 + 4 ] = p [ 2 − 2 − 2 2 ] ⇒ [ 8 − 8 − 8 8 ] = [ 2 p − 2 p − 2 p 2 p ] \text{A}^2 = pA \\[0.5em] \Rightarrow \begin{bmatrix*}[r] 2 & -2 \\ -2 & 2 \end{bmatrix*} \begin{bmatrix*}[r] 2 & -2 \\ -2 & 2 \end{bmatrix*} = p\begin{bmatrix*}[r] 2 & -2 \\ -2 & 2 \end{bmatrix*} \\[0.5em] \Rightarrow \begin{bmatrix*}[r] 2 \times 2 + (-2) \times (-2) & 2 \times (-2) + (-2) \times 2 \\ -2 \times 2 + 2 \times (-2) & (-2) \times (-2) + 2 \times 2 \end{bmatrix*} = p\begin{bmatrix*}[r] 2 & -2 \\ -2 & 2 \end{bmatrix*} \\[0.5em] \Rightarrow \begin{bmatrix*}[r] 4 + 4 & -4 - 4 \\ -4 - 4 & 4 + 4 \end{bmatrix*} = p\begin{bmatrix*}[r] 2 & -2 \\ -2 & 2 \end{bmatrix*} \\[0.5em] \Rightarrow \begin{bmatrix*}[r] 8 & -8 \\ -8 & 8 \end{bmatrix*} = \begin{bmatrix*}[r] 2p & -2p \\ -2p & 2p \end{bmatrix*} \\[0.5em] A 2 = p A ⇒ [ 2 − 2 − 2 2 ] [ 2 − 2 − 2 2 ] = p [ 2 − 2 − 2 2 ] ⇒ [ 2 × 2 + ( − 2 ) × ( − 2 ) − 2 × 2 + 2 × ( − 2 ) 2 × ( − 2 ) + ( − 2 ) × 2 ( − 2 ) × ( − 2 ) + 2 × 2 ] = p [ 2 − 2 − 2 2 ] ⇒ [ 4 + 4 − 4 − 4 − 4 − 4 4 + 4 ] = p [ 2 − 2 − 2 2 ] ⇒ [ 8 − 8 − 8 8 ] = [ 2 p − 2 p − 2 p 2 p ]
By definition of equality of matrices we get,
⇒ 2p = 8
∴ p = 4.
∴ Option 2 is the correct option.
The product AB of two matrices A and B is possible if
A and B have same number of rows.
A and B have same number of columns.
The number of columns of A is equal to the number of rows of B.
The number of rows of A is equal to the number of columns of B.
Answer
Matrix multiplication is possible if the number of columns of the first matrix is the same as the number of rows as the second matrix.
So, AB is possible if the number of columns of A is equal to the number of rows of B.
Hence, option 3 is the correct option.
If A = [ − 1 2 ] and B = [ 1 − 2 0 3 ] \begin{bmatrix*}[r] -1 & 2 \end{bmatrix*} \text{ and } B = \begin{bmatrix*}[r] 1 & -2 \\ 0 & 3 \end{bmatrix*} [ − 1 2 ] and B = [ 1 0 − 2 3 ]
A - B
A + B
AB
BA
Answer
Order of A : 1 × 2 and order of B : 2 × 2
Since, no. of columns in A is same as the no. of rows in B.
∴ AB is possible.
Hence, Option 3 is the correct option.
Assertion-Reason Type Questions
A, B and C are square matrices of order 2 such that AB = C.
Assertion (A): BA = C
Reason (R): Matrix multiplication is not always commutative.
Assertion (A) is true, but Reason (R) is false.
Assertion (A) is false, but Reason (R) is true.
Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).
Answer
Given, A, B and C are square matrices of order 2 such that AB = C.
Matrix multiplication is not always commutative.
So, reason (R) is true.
⇒ AB ≠ BA
So, BA is not necessarily equal to C.
So, assertion (A) is false.
Thus, Assertion (A) is false, but Reason (R) is true.
Hence, option 2 is the correct option.
A, B and C are square matrices of order 2 such that AB = C.
Assertion (A): Product BA need not be equal to C.
Reason (R): Matrix multiplication is not associative.
Assertion (A) is true, but Reason (R) is false.
Assertion (A) is false, but Reason (R) is true.
Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).
Answer
Given, A, B and C are square matrices of order 2 such that AB = C.
Matrix multiplication is not always commutative.
∴ AB is not necessarily equal to BA.
So, BA is not necessarily equal to C.
So, assertion (A) is true.
Matrix multiplication is associative.
So, reason (R) is false.
Assertion (A) is true, but Reason (R) is false.
Hence, option 1 is the correct option.
If A = [ 3 − 2 ] and B = [ − 1 4 2 0 ] \begin{bmatrix*}[r] 3 & -2 \end{bmatrix*} \text{ and B} = \begin{bmatrix*}[r] -1 & 4 \\ 2 & 0 \end{bmatrix*} [ 3 − 2 ] and B = [ − 1 2 4 0 ]
Assertion (A): Product AB of the two matrices A and B is possible.
Reason (R): Number of columns of matrix A is equal to number of rows in matrix B.
Assertion (A) is true, but Reason (R) is false.
Assertion (A) is false, but Reason (R) is true.
Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).
Answer
Given, A = [ 3 − 2 ] and B = [ − 1 4 2 0 ] \begin{bmatrix*}[r] 3 & -2 \end{bmatrix*} \text{ and B} = \begin{bmatrix*}[r] -1 & 4 \\ 2 & 0 \end{bmatrix*} [ 3 − 2 ] and B = [ − 1 2 4 0 ]
To multiply two matrices, the number of columns of the first matrix must equal the number of rows of the second matrix.
Matrix A has 1 row and 2 columns.
Matrix B has 2 rows and 2 columns.
So, reason (R) is true.
The number of columns in A is 2 and number of rows in B is also 2.
So, product AB is defined.
So, both A and R are true and R is the correct explanation of assertion A.
Thus, both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
Hence, option 3 is the correct option.
Find the values of a and b if,
[ a + 3 b 2 + 2 0 − 6 ] = [ 2 a + 1 3 b 0 b 2 − 5 b ] . \begin{bmatrix*}[r] a + 3 & b^2 + 2 \\ 0 & -6 \end{bmatrix*} = \begin{bmatrix*}[r] 2a + 1 & 3b \\ 0 & b^2 - 5b \end{bmatrix*}. [ a + 3 0 b 2 + 2 − 6 ] = [ 2 a + 1 0 3 b b 2 − 5 b ] .
Answer
By definition of equality of matrices we get,
⇒ a + 3 = 2a + 1 or a = 2
⇒ b2 + 2 = 3b (...Eq 1)
⇒ b2 - 5b = -6 (...Eq 2)
Solving Eq 1 for b,
⇒ b2 + 2 = 3b2 - 3b + 2 = 02 - 2b - b + 2 = 0
Checking whether the value of b = 1 satisfies Eq 2
⇒ b2 - 5b = -6
L.H.S. = b2 - 5b = (1)2 - 5(1) = -4.
L.H.S. ≠ \neq =
Checking whether the value of b = 2 satisfies Eq 2
⇒ b2 - 5b = -6
L.H.S. = b2 - 5b2 - 5(2)
∴ a = 2 and b = 2.
Hence, the values are a = 2 and b = 2.
Find a, b, c and d if 3 [ a b c d ] = [ 4 a + b c + d 3 ] + [ a 6 − 1 2 d ] 3\begin{bmatrix*}[r] a & b \\ c & d \end{bmatrix*} = \begin{bmatrix*}[r] 4 & a + b \\ c + d & 3 \end{bmatrix*} + \begin{bmatrix*}[r] a & 6 \\ -1 & 2d \end{bmatrix*} 3 [ a c b d ] = [ 4 c + d a + b 3 ] + [ a − 1 6 2 d ]
Answer
Given,
3 [ a b c d ] = [ 4 a + b c + d 3 ] + [ a 6 − 1 2 d ] ⇒ [ 3 a 3 b 3 c 3 d ] = [ 4 + a a + b + 6 c + d − 1 3 + 2 d ] 3\begin{bmatrix*}[r] a & b \\ c & d \end{bmatrix*} = \begin{bmatrix*}[r] 4 & a + b \\ c + d & 3 \end{bmatrix*} + \begin{bmatrix*}[r] a & 6 \\ -1 & 2d \end{bmatrix*} \\[0.5em] \Rightarrow \begin{bmatrix*}[r] 3a & 3b \\ 3c & 3d \end{bmatrix*} = \begin{bmatrix*}[r] 4 + a & a + b + 6 \\ c + d - 1 & 3 + 2d \end{bmatrix*} \\[0.5em] 3 [ a c b d ] = [ 4 c + d a + b 3 ] + [ a − 1 6 2 d ] ⇒ [ 3 a 3 c 3 b 3 d ] = [ 4 + a c + d − 1 a + b + 6 3 + 2 d ]
By definition of equality of matrices
⇒ 3a = 4 + a (...Eq 1)
⇒ 3b = a + b + 6 (...Eq 2)
⇒ 3c = c + d - 1 (...Eq 3)
⇒ 3d = 3 + 2d (...Eq 4)
Solving Eq 1 we get,
⇒ 3a = 4 + a
Solving Eq 2 by putting value of a from Eq 1 we get,
⇒ 3b = a + b + 6
Solving Eq 4 we get,
⇒ 3d = 3 + 2d
Solving Eq 3 by putting value of d from Eq 4 we get,
⇒ 3c = c + d - 1
∴ a = 2, b = 4, c = 1 and d = 3.
Hence, the value of a = 2, b = 4, c = 1 and d = 3.
Determine the matrices A and B when
A + 2B = [ 1 2 6 − 3 ] and 2A - B = [ 2 − 1 2 − 1 ] . \begin{bmatrix*}[r] 1 & 2 \\ 6 & -3 \end{bmatrix*} \text{ and 2A - B} = \begin{bmatrix*}[r] 2 & -1 \\ 2 & -1 \end{bmatrix*}. [ 1 6 2 − 3 ] and 2A - B = [ 2 2 − 1 − 1 ] .
Answer
Given,
A + 2B = [ 1 2 6 − 3 ] ....[Eq 1] 2A - B = [ 2 − 1 2 − 1 ] ....[Eq 2] \text{A + 2B} = \begin{bmatrix*}[r] 1 & 2 \\ 6 & -3 \end{bmatrix*} \qquad \text{....[Eq 1] } \\[1em] \text{2A - B} = \begin{bmatrix*}[r] 2 & -1 \\ 2 & -1 \end{bmatrix*} \qquad \text{....[Eq 2] } \\[1em] A + 2B = [ 1 6 2 − 3 ] ....[Eq 1] 2A - B = [ 2 2 − 1 − 1 ] ....[Eq 2]
Multiplying Eq 1 by 2,
⇒ 2 A + 4 B = [ 2 4 12 − 6 ] \Rightarrow 2A + 4B = \begin{bmatrix*}[r] 2 & 4 \\ 12 & -6 \end{bmatrix*} \\[1em] ⇒ 2 A + 4 B = [ 2 12 4 − 6 ]
Subtracting Eq 2 from above equation we get,
⇒ 2 A + 4 B − ( 2 A − B ) = [ 2 4 12 − 6 ] − [ 2 − 1 2 − 1 ] ⇒ 2 A − 2 A + 4 B − ( − B ) = [ 2 − 2 4 − ( − 1 ) 12 − 2 − 6 − ( − 1 ) ] ⇒ 5 B = [ 0 5 10 − 5 ] ⇒ B = 1 5 [ 0 5 10 − 5 ] ⇒ B = [ 0 1 2 − 1 ] . \Rightarrow 2A + 4B - (2A - B) = \begin{bmatrix*}[r] 2 & 4 \\ 12 & -6 \end{bmatrix*} - \begin{bmatrix*}[r] 2 & -1 \\ 2 & -1 \end{bmatrix*} \\[1em] \Rightarrow 2A - 2A + 4B - (-B) = \begin{bmatrix*}[r] 2 - 2 & 4 - (-1) \\ 12 - 2 & -6 - (-1) \end{bmatrix*} \\[1em] \Rightarrow 5B = \begin{bmatrix*}[r] 0 & 5 \\ 10 & -5 \end{bmatrix*} \\[1em] \Rightarrow B = \dfrac{1}{5}\begin{bmatrix*}[r] 0 & 5 \\ 10 & -5 \end{bmatrix*} \\[1em] \Rightarrow B = \begin{bmatrix*}[r] 0 & 1 \\ 2 & -1 \end{bmatrix*}. \\[1em] ⇒ 2 A + 4 B − ( 2 A − B ) = [ 2 12 4 − 6 ] − [ 2 2 − 1 − 1 ] ⇒ 2 A − 2 A + 4 B − ( − B ) = [ 2 − 2 12 − 2 4 − ( − 1 ) − 6 − ( − 1 ) ] ⇒ 5 B = [ 0 10 5 − 5 ] ⇒ B = 5 1 [ 0 10 5 − 5 ] ⇒ B = [ 0 2 1 − 1 ] .
Putting value of matrix B in Eq 1,
⇒ A + 2 [ 0 1 2 − 1 ] = [ 1 2 6 − 3 ] ⇒ A + [ 0 2 4 − 2 ] = [ 1 2 6 − 3 ] ⇒ A = [ 1 2 6 − 3 ] − [ 0 2 4 − 2 ] ⇒ A = [ 1 − 0 2 − 2 6 − 4 − 3 − ( − 2 ) ] ⇒ A = [ 1 0 2 − 1 ] . ∴ A = [ 1 0 2 − 1 ] and B = [ 0 1 2 − 1 ] . \Rightarrow A + 2 \begin{bmatrix*}[r] 0 & 1 \\ 2 & -1 \end{bmatrix*} = \begin{bmatrix*}[r] 1 & 2 \\ 6 & -3 \end{bmatrix*} \\[1em] \Rightarrow A + \begin{bmatrix*}[r] 0 & 2 \\ 4 & -2 \end{bmatrix*} = \begin{bmatrix*}[r] 1 & 2 \\ 6 & -3 \end{bmatrix*} \\[1em] \Rightarrow A = \begin{bmatrix*}[r] 1 & 2 \\ 6 & -3 \end{bmatrix*} - \begin{bmatrix*}[r] 0 & 2 \\ 4 & -2 \end{bmatrix*} \\[1em] \Rightarrow A = \begin{bmatrix*}[r] 1 - 0 & 2 - 2 \\ 6 - 4 & -3 - (-2) \end{bmatrix*} \\[1em] \Rightarrow A = \begin{bmatrix*}[r] 1 & 0 \\ 2 & -1 \end{bmatrix*}. \\[1em] \therefore A = \begin{bmatrix*}[r] 1 & 0 \\ 2 & -1 \end{bmatrix*} \text{ and } B = \begin{bmatrix*}[r] 0 & 1 \\ 2 & -1 \end{bmatrix*}. ⇒ A + 2 [ 0 2 1 − 1 ] = [ 1 6 2 − 3 ] ⇒ A + [ 0 4 2 − 2 ] = [ 1 6 2 − 3 ] ⇒ A = [ 1 6 2 − 3 ] − [ 0 4 2 − 2 ] ⇒ A = [ 1 − 0 6 − 4 2 − 2 − 3 − ( − 2 ) ] ⇒ A = [ 1 2 0 − 1 ] . ∴ A = [ 1 2 0 − 1 ] and B = [ 0 2 1 − 1 ] .
Hence, the value of A = [ 1 0 2 − 1 ] and B = [ 0 1 2 − 1 ] . \text{A} = \begin{bmatrix*}[r] 1 & 0 \\ 2 & -1 \end{bmatrix*} \text{and B} = \begin{bmatrix*}[r] 0 & 1 \\ 2 & -1 \end{bmatrix*}. A = [ 1 2 0 − 1 ] and B = [ 0 2 1 − 1 ] .
Find the matrix B if A = [ 4 1 2 3 ] \begin{bmatrix*}[r] 4 & 1 \\ 2 & 3 \end{bmatrix*} [ 4 2 1 3 ] 2 = A + 2B.
Answer
Given, A2 = A + 2B.
⇒ [ 4 1 2 3 ] [ 4 1 2 3 ] = [ 4 1 2 3 ] + 2 B ⇒ [ 4 × 4 + 1 × 2 4 × 1 + 1 × 3 2 × 4 + 3 × 2 2 × 1 + 3 × 3 ] = [ 4 1 2 3 ] + 2 B ⇒ [ 16 + 2 4 + 3 8 + 6 2 + 9 ] = [ 4 1 2 3 ] + 2 B ⇒ [ 18 7 14 11 ] = [ 4 1 2 3 ] + 2 B ⇒ 2 B = [ 18 7 14 11 ] − [ 4 1 2 3 ] ⇒ 2 B = [ 18 − 4 7 − 1 14 − 2 11 − 3 ] ⇒ 2 B = [ 14 6 12 8 ] ∴ B = [ 7 3 6 4 ] . \Rightarrow \begin{bmatrix*}[r] 4 & 1 \\ 2 & 3 \end{bmatrix*} \begin{bmatrix*}[r] 4 & 1 \\ 2 & 3 \end{bmatrix*} = \begin{bmatrix*}[r] 4 & 1 \\ 2 & 3 \end{bmatrix*} + 2B \\[0.5em] \Rightarrow \begin{bmatrix*}[r] 4 \times 4 + 1 \times 2 & 4 \times 1 + 1 \times 3 \\ 2 \times 4 + 3 \times 2 & 2 \times 1 + 3 \times 3 \end{bmatrix*} = \begin{bmatrix*}[r] 4 & 1 \\ 2 & 3 \end{bmatrix*} + 2B \\[0.5em] \Rightarrow \begin{bmatrix*}[r] 16 + 2 & 4 + 3 \\ 8 + 6 & 2 + 9 \end{bmatrix*} = \begin{bmatrix*}[r] 4 & 1 \\ 2 & 3 \end{bmatrix*} + 2B \\[0.5em] \Rightarrow \begin{bmatrix*}[r] 18 & 7 \\ 14 & 11 \end{bmatrix*} = \begin{bmatrix*}[r] 4 & 1 \\ 2 & 3 \end{bmatrix*} + 2B \\[0.5em] \Rightarrow 2B = \begin{bmatrix*}[r] 18 & 7 \\ 14 & 11 \end{bmatrix*} - \begin{bmatrix*}[r] 4 & 1 \\ 2 & 3 \end{bmatrix*} \\[0.5em] \Rightarrow 2B = \begin{bmatrix*}[r] 18 - 4 & 7 - 1 \\ 14 - 2 & 11 - 3 \end{bmatrix*} \\[0.5em] \Rightarrow 2B = \begin{bmatrix*}[r] 14 & 6 \\ 12 & 8 \end{bmatrix*} \\[0.5em] \therefore B = \begin{bmatrix*}[r] 7 & 3 \\ 6 & 4 \end{bmatrix*}. ⇒ [ 4 2 1 3 ] [ 4 2 1 3 ] = [ 4 2 1 3 ] + 2 B ⇒ [ 4 × 4 + 1 × 2 2 × 4 + 3 × 2 4 × 1 + 1 × 3 2 × 1 + 3 × 3 ] = [ 4 2 1 3 ] + 2 B ⇒ [ 16 + 2 8 + 6 4 + 3 2 + 9 ] = [ 4 2 1 3 ] + 2 B ⇒ [ 18 14 7 11 ] = [ 4 2 1 3 ] + 2 B ⇒ 2 B = [ 18 14 7 11 ] − [ 4 2 1 3 ] ⇒ 2 B = [ 18 − 4 14 − 2 7 − 1 11 − 3 ] ⇒ 2 B = [ 14 12 6 8 ] ∴ B = [ 7 6 3 4 ] .
Hence, the matrix B = [ 7 3 6 4 ] . \begin{bmatrix*}[r] 7 & 3 \\ 6 & 4 \end{bmatrix*}. [ 7 6 3 4 ] .
If A = [ 1 2 − 3 4 ] , B = [ 0 1 − 2 5 ] and C = [ − 2 0 − 1 1 ] , A = \begin{bmatrix*}[r] 1 & 2 \\ -3 & 4 \end{bmatrix*}, B = \begin{bmatrix*}[r] 0 & 1 \\ -2 & 5 \end{bmatrix*} \text{ and C } = \begin{bmatrix*}[r] -2 & 0 \\ -1 & 1 \end{bmatrix*}, A = [ 1 − 3 2 4 ] , B = [ 0 − 2 1 5 ] and C = [ − 2 − 1 0 1 ] ,
Answer
We need to find the value of A(4B - 3C).
A(4B - 3C) = [ 1 2 − 3 4 ] ( 4 [ 0 1 − 2 5 ] − 3 [ − 2 0 − 1 1 ] ) = [ 1 2 − 3 4 ] ( [ 0 4 − 8 20 ] − [ − 6 0 − 3 3 ] ) = [ 1 2 − 3 4 ] [ 0 − ( − 6 ) 4 − 0 − 8 − ( − 3 ) 20 − 3 ] = [ 1 2 − 3 4 ] [ 6 4 − 5 17 ] = [ 1 × 6 + 2 × ( − 5 ) 1 × 4 + 2 × 17 − 3 × 6 + 4 × ( − 5 ) − 3 × 4 + 4 × 17 ] = [ 6 − 10 4 + 34 − 18 − 20 − 12 + 68 ] = [ − 4 38 − 38 56 ] . \text{A(4B - 3C)} = \begin{bmatrix*}[r] 1 & 2 \\ -3 & 4 \end{bmatrix*} \Big(4\begin{bmatrix*}[r] 0 & 1 \\ -2 & 5 \end{bmatrix*} - 3\begin{bmatrix*}[r] -2 & 0 \\ -1 & 1 \end{bmatrix*}\Big) \\[0.5em] = \begin{bmatrix*}[r] 1 & 2 \\ -3 & 4 \end{bmatrix*} \Big(\begin{bmatrix*}[r] 0 & 4 \\ -8 & 20 \end{bmatrix*} - \begin{bmatrix*}[r] -6 & 0 \\ -3 & 3 \end{bmatrix*}\Big) \\[0.5em] = \begin{bmatrix*}[r] 1 & 2 \\ -3 & 4 \end{bmatrix*} \begin{bmatrix*}[r] 0 - (-6) & 4 - 0 \\ -8 - (-3) & 20 - 3 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 1 & 2 \\ -3 & 4 \end{bmatrix*} \begin{bmatrix*}[r] 6 & 4 \\ -5 & 17 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 1 \times 6 + 2 \times (-5) & 1 \times 4 + 2 \times 17 \\ -3 \times 6 + 4 \times (-5) & -3 \times 4 + 4 \times 17 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 6 - 10 & 4 + 34 \\ -18 - 20 & -12 + 68 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] -4 & 38 \\ -38 & 56 \end{bmatrix*}. A(4B - 3C) = [ 1 − 3 2 4 ] ( 4 [ 0 − 2 1 5 ] − 3 [ − 2 − 1 0 1 ] ) = [ 1 − 3 2 4 ] ( [ 0 − 8 4 20 ] − [ − 6 − 3 0 3 ] ) = [ 1 − 3 2 4 ] [ 0 − ( − 6 ) − 8 − ( − 3 ) 4 − 0 20 − 3 ] = [ 1 − 3 2 4 ] [ 6 − 5 4 17 ] = [ 1 × 6 + 2 × ( − 5 ) − 3 × 6 + 4 × ( − 5 ) 1 × 4 + 2 × 17 − 3 × 4 + 4 × 17 ] = [ 6 − 10 − 18 − 20 4 + 34 − 12 + 68 ] = [ − 4 − 38 38 56 ] .
Hence, the value of A(4B - 3C) = [ − 4 38 − 38 56 ] . \begin{bmatrix*}[r] -4 & 38 \\ -38 & 56 \end{bmatrix*}. [ − 4 − 38 38 56 ] .
If A = [ 1 3 2 4 ] , B = [ 1 2 2 4 ] , C = [ 4 1 1 5 ] and I = [ 1 0 0 1 ] A = \begin{bmatrix*}[r] 1 & 3 \\ 2 & 4 \end{bmatrix*}, B = \begin{bmatrix*}[r] 1 & 2 \\ 2 & 4 \end{bmatrix*}, C = \begin{bmatrix*}[r] 4 & 1 \\ 1 & 5 \end{bmatrix*} \text{ and } I = \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} A = [ 1 2 3 4 ] , B = [ 1 2 2 4 ] , C = [ 4 1 1 5 ] and I = [ 1 0 0 1 ]
Answer
B + C = [ 1 2 2 4 ] + [ 4 1 1 5 ] = [ 5 3 3 9 ] \begin{bmatrix*}[r] 1 & 2 \\ 2 & 4 \end{bmatrix*} + \begin{bmatrix*}[r] 4 & 1 \\ 1 & 5 \end{bmatrix*} = \begin{bmatrix*}[r] 5 & 3 \\ 3 & 9 \end{bmatrix*} [ 1 2 2 4 ] + [ 4 1 1 5 ] = [ 5 3 3 9 ]
Substituting values we get :
A ( B + C ) − 14 I = [ 1 3 2 4 ] [ 5 3 3 9 ] − 14 [ 1 0 0 1 ] = [ 1 × 5 + 3 × 3 1 × 3 + 3 × 9 2 × 5 + 4 × 3 2 × 3 + 4 × 9 ] − [ 14 0 0 14 ] = [ 5 + 9 3 + 27 10 + 12 6 + 36 ] − [ 14 0 0 14 ] = [ 14 30 22 42 ] − [ 14 0 0 14 ] = [ 14 − 14 30 − 0 22 − 0 42 − 14 ] = [ 0 30 22 28 ] . A(B + C) - 14I = \begin{bmatrix*}[r] 1 & 3 \\ 2 & 4 \end{bmatrix*}\begin{bmatrix*}[r] 5 & 3 \\ 3 & 9 \end{bmatrix*} - 14\begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 \times 5 + 3 \times 3 & 1 \times 3 + 3 \times 9 \\ 2 \times 5 + 4 \times 3 & 2 \times 3 + 4 \times 9 \end{bmatrix*} - \begin{bmatrix*}[r] 14 & 0 \\ 0 & 14 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 5 + 9 & 3 + 27 \\ 10 + 12 & 6 + 36 \end{bmatrix*} - \begin{bmatrix*}[r] 14 & 0 \\ 0 & 14 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 14 & 30 \\ 22 & 42 \end{bmatrix*} - \begin{bmatrix*}[r] 14 & 0 \\ 0 & 14 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 14 - 14 & 30 - 0 \\ 22 - 0 & 42 - 14 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 0 & 30 \\ 22 & 28 \end{bmatrix*}. A ( B + C ) − 14 I = [ 1 2 3 4 ] [ 5 3 3 9 ] − 14 [ 1 0 0 1 ] = [ 1 × 5 + 3 × 3 2 × 5 + 4 × 3 1 × 3 + 3 × 9 2 × 3 + 4 × 9 ] − [ 14 0 0 14 ] = [ 5 + 9 10 + 12 3 + 27 6 + 36 ] − [ 14 0 0 14 ] = [ 14 22 30 42 ] − [ 14 0 0 14 ] = [ 14 − 14 22 − 0 30 − 0 42 − 14 ] = [ 0 22 30 28 ] .
Hence, A(B + C) - 14I = [ 0 30 22 28 ] . \begin{bmatrix*}[r] 0 & 30 \\ 22 & 28 \end{bmatrix*}. [ 0 22 30 28 ] .
If A = [ 3 2 0 5 ] and B = [ 1 0 1 2 ] , \begin{bmatrix*}[r] 3 & 2 \\ 0 & 5 \end{bmatrix*} \text{ and B } = \begin{bmatrix*}[r] 1 & 0 \\ 1 & 2 \end{bmatrix*}, [ 3 0 2 5 ] and B = [ 1 1 0 2 ] ,
(i) (A + B)(A - B)
(ii) A2 - B2
Answer
(i) We need to find the value of (A + B)(A - B)
( A + B ) ( A − B ) = ( [ 3 2 0 5 ] + [ 1 0 1 2 ] ) ( [ 3 2 0 5 ] − [ 1 0 1 2 ] ) = [ 3 + 1 2 + 0 0 + 1 5 + 2 ] [ 3 − 1 2 − 0 0 − 1 5 − 2 ] = [ 4 2 1 7 ] [ 2 2 − 1 3 ] = [ 4 × 2 + 2 × ( − 1 ) 4 × 2 + 2 × 3 1 × 2 + 7 × ( − 1 ) 1 × 2 + 7 × 3 ] = [ 8 − 2 8 + 6 2 − 7 2 + 21 ] = [ 6 14 − 5 23 ] . (A + B)(A - B) = \Big(\begin{bmatrix*}[r] 3 & 2 \\ 0 & 5 \end{bmatrix*} + \begin{bmatrix*}[r] 1 & 0 \\ 1 & 2 \end{bmatrix*}\Big)\Big(\begin{bmatrix*}[r] 3 & 2 \\ 0 & 5 \end{bmatrix*} - \begin{bmatrix*}[r] 1 & 0 \\ 1 & 2 \end{bmatrix*}\Big) \\[0.5em] = \begin{bmatrix*}[r] 3 + 1 & 2 + 0 \\ 0 + 1 & 5 + 2 \end{bmatrix*} \begin{bmatrix*}[r] 3 - 1 & 2 - 0 \\ 0 - 1 & 5 - 2 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 4 & 2 \\ 1 & 7 \end{bmatrix*} \begin{bmatrix*}[r] 2 & 2 \\ -1 & 3 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 4 \times 2 + 2 \times (-1) & 4 \times 2 + 2 \times 3 \\ 1 \times 2 + 7 \times (-1) & 1 \times 2 + 7 \times 3 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 8 - 2 & 8 + 6 \\ 2 - 7 & 2 + 21 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 6 & 14 \\ -5 & 23 \end{bmatrix*}. ( A + B ) ( A − B ) = ( [ 3 0 2 5 ] + [ 1 1 0 2 ] ) ( [ 3 0 2 5 ] − [ 1 1 0 2 ] ) = [ 3 + 1 0 + 1 2 + 0 5 + 2 ] [ 3 − 1 0 − 1 2 − 0 5 − 2 ] = [ 4 1 2 7 ] [ 2 − 1 2 3 ] = [ 4 × 2 + 2 × ( − 1 ) 1 × 2 + 7 × ( − 1 ) 4 × 2 + 2 × 3 1 × 2 + 7 × 3 ] = [ 8 − 2 2 − 7 8 + 6 2 + 21 ] = [ 6 − 5 14 23 ] .
Hence, the value of (A + B)(A - B) = [ 6 14 − 5 23 ] \begin{bmatrix*}[r] 6 & 14 \\ -5 & 23 \end{bmatrix*} [ 6 − 5 14 23 ]
(ii) We need to find the value of A2 - B2
A 2 − B 2 = [ 3 2 0 5 ] [ 3 2 0 5 ] − [ 1 0 1 2 ] [ 1 0 1 2 ] = [ 3 × 3 + 2 × 0 3 × 2 + 2 × 5 0 × 3 + 5 × 0 0 × 2 + 5 × 5 ] − [ 1 × 1 + 0 × 1 1 × 0 + 0 × 2 1 × 1 + 2 × 1 1 × 0 + 2 × 2 ] = [ 9 16 0 25 ] − [ 1 0 3 4 ] = [ 9 − 1 16 − 0 0 − 3 25 − 4 ] = [ 8 16 − 3 21 ] . A^2 - B^2 = \begin{bmatrix*}[r] 3 & 2 \\ 0 & 5 \end{bmatrix*} \begin{bmatrix*}[r] 3 & 2 \\ 0 & 5 \end{bmatrix*} - \begin{bmatrix*}[r] 1 & 0 \\ 1 & 2 \end{bmatrix*} \begin{bmatrix*}[r] 1 & 0 \\ 1 & 2 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 3 \times 3 + 2 \times 0 & 3 \times 2 + 2 \times 5 \\ 0 \times 3 + 5 \times 0 & 0 \times 2 + 5 \times 5 \end{bmatrix*} - \begin{bmatrix*}[r] 1 \times 1 + 0 \times 1 & 1 \times 0 + 0 \times 2 \\ 1 \times 1 + 2 \times 1 & 1 \times 0 + 2 \times 2 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 9 & 16 \\ 0 & 25 \end{bmatrix*} - \begin{bmatrix*}[r] 1 & 0 \\ 3 & 4 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 9 - 1 & 16 - 0 \\ 0 - 3 & 25 - 4 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 8 & 16 \\ -3 & 21 \end{bmatrix*} . A 2 − B 2 = [ 3 0 2 5 ] [ 3 0 2 5 ] − [ 1 1 0 2 ] [ 1 1 0 2 ] = [ 3 × 3 + 2 × 0 0 × 3 + 5 × 0 3 × 2 + 2 × 5 0 × 2 + 5 × 5 ] − [ 1 × 1 + 0 × 1 1 × 1 + 2 × 1 1 × 0 + 0 × 2 1 × 0 + 2 × 2 ] = [ 9 0 16 25 ] − [ 1 3 0 4 ] = [ 9 − 1 0 − 3 16 − 0 25 − 4 ] = [ 8 − 3 16 21 ] .
Hence, the value of A 2 − B 2 = [ 8 16 − 3 21 ] and (A + B)(A - B) ≠ A 2 − B 2 . A^2 - B^2 = \begin{bmatrix*}[r] 8 & 16 \\ -3 & 21 \end{bmatrix*} \text{ and (A + B)(A - B) } \neq A^2 - B^2. A 2 − B 2 = [ 8 − 3 16 21 ] and (A + B)(A - B) = A 2 − B 2 .
If A = [ 3 − 5 − 4 2 ] , \begin{bmatrix*}[r] 3 & -5 \\ -4 & 2 \end{bmatrix*}, [ 3 − 4 − 5 2 ] , 2 - 5A - 14I, where I is unit matrix of order 2 × 2.
Answer
We need to find the value of A2 - 5A - 14I.
A 2 − 5 A − 14 I = [ 3 − 5 − 4 2 ] [ 3 − 5 − 4 2 ] − 5 [ 3 − 5 − 4 2 ] − 14 [ 1 0 0 1 ] = [ 3 × 3 + ( − 5 ) × ( − 4 ) 3 × ( − 5 ) + ( − 5 ) × 2 ( − 4 ) × 3 + 2 × ( − 4 ) ( − 4 ) × ( − 5 ) + 2 × 2 ] − [ 15 − 25 − 20 10 ] − [ 14 0 0 14 ] = [ 9 + 20 − 15 − 10 − 12 − 8 20 + 4 ] − [ 15 − 25 − 20 10 ] − [ 14 0 0 14 ] = [ 29 − 25 − 20 24 ] − [ 15 − 25 − 20 10 ] − [ 14 0 0 14 ] = [ 29 − 15 − 14 − 25 − ( − 25 ) − 0 − 20 − ( − 20 ) − 0 24 − 10 − 14 ] = [ 29 − 29 − 25 + 25 − 20 + 20 24 − 24 ] = [ 0 0 0 0 ] A^2 - 5A -14I = \begin{bmatrix*}[r] 3 & -5 \\ -4 & 2 \end{bmatrix*} \begin{bmatrix*}[r] 3 & -5 \\ -4 & 2 \end{bmatrix*} - 5\begin{bmatrix*}[r] 3 & -5 \\ -4 & 2 \end{bmatrix*} - 14\begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 3 \times 3 + (-5) \times (-4) & 3 \times (-5) + (-5) \times 2 \\ (-4) \times 3 + 2 \times (-4) & (-4) \times (-5) + 2 \times 2 \end{bmatrix*} - \begin{bmatrix*}[r] 15 & -25 \\ -20 & 10 \end{bmatrix*} - \begin{bmatrix*}[r] 14 & 0 \\ 0 & 14 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 9 + 20 & -15 - 10 \\ -12 - 8 & 20 + 4 \end{bmatrix*} - \begin{bmatrix*}[r] 15 & -25 \\ -20 & 10 \end{bmatrix*} - \begin{bmatrix*}[r] 14 & 0 \\ 0 & 14 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 29 & -25 \\ -20 & 24 \end{bmatrix*} - \begin{bmatrix*}[r] 15 & -25 \\ -20 & 10 \end{bmatrix*} - \begin{bmatrix*}[r] 14 & 0 \\ 0 & 14 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 29 - 15 - 14 & -25 - (-25) - 0 \\ -20 - (-20) - 0 & 24 - 10 - 14 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 29 - 29 & -25 + 25 \\ -20 + 20 & 24 - 24 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*} \\[0.5em] A 2 − 5 A − 14 I = [ 3 − 4 − 5 2 ] [ 3 − 4 − 5 2 ] − 5 [ 3 − 4 − 5 2 ] − 14 [ 1 0 0 1 ] = [ 3 × 3 + ( − 5 ) × ( − 4 ) ( − 4 ) × 3 + 2 × ( − 4 ) 3 × ( − 5 ) + ( − 5 ) × 2 ( − 4 ) × ( − 5 ) + 2 × 2 ] − [ 15 − 20 − 25 10 ] − [ 14 0 0 14 ] = [ 9 + 20 − 12 − 8 − 15 − 10 20 + 4 ] − [ 15 − 20 − 25 10 ] − [ 14 0 0 14 ] = [ 29 − 20 − 25 24 ] − [ 15 − 20 − 25 10 ] − [ 14 0 0 14 ] = [ 29 − 15 − 14 − 20 − ( − 20 ) − 0 − 25 − ( − 25 ) − 0 24 − 10 − 14 ] = [ 29 − 29 − 20 + 20 − 25 + 25 24 − 24 ] = [ 0 0 0 0 ]
Hence, the value of A 2 − 5 A − 14 I = [ 0 0 0 0 ] . A^2 - 5A - 14I = \begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*}. A 2 − 5 A − 14 I = [ 0 0 0 0 ] .
If A = [ 3 3 p q ] and A 2 = O , A = \begin{bmatrix*}[r] 3 & 3 \\ p & q \end{bmatrix*} \text{ and } A^2 = O, A = [ 3 p 3 q ] and A 2 = O ,
Answer
Given, A2 = O.
⇒ [ 3 3 p q ] [ 3 3 p q ] = [ 0 0 0 0 ] ⇒ [ 3 × 3 + 3 × p 3 × 3 + 3 × q p × 3 + q × p p × 3 + q × q ] = [ 0 0 0 0 ] ⇒ [ 9 + 3 p 9 + 3 q 3 p + q p 3 p + q 2 ] = [ 0 0 0 0 ] \Rightarrow \begin{bmatrix*}[r] 3 & 3 \\ p & q \end{bmatrix*} \begin{bmatrix*}[r] 3 & 3 \\ p & q \end{bmatrix*} = \begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*} \\[0.5em] \Rightarrow \begin{bmatrix*}[r] 3 \times 3 + 3 \times p & 3 \times 3 + 3 \times q \\ p \times 3 + q \times p & p \times 3 + q \times q \end{bmatrix*} = \begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*} \\[0.5em] \Rightarrow \begin{bmatrix*}[r] 9 + 3p & 9 + 3q \\ 3p + qp & 3p + q^2 \end{bmatrix*} = \begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*} \\[0.5em] ⇒ [ 3 p 3 q ] [ 3 p 3 q ] = [ 0 0 0 0 ] ⇒ [ 3 × 3 + 3 × p p × 3 + q × p 3 × 3 + 3 × q p × 3 + q × q ] = [ 0 0 0 0 ] ⇒ [ 9 + 3 p 3 p + qp 9 + 3 q 3 p + q 2 ] = [ 0 0 0 0 ]
By definition of equality of matrices we get,
⇒ 9 + 3p = 0 or 3p = -9 or p = -3
⇒ 9 + 3q = 0 or 3q = -9 or q = -3
⇒ 3p + qp = 0 (Eq 1)
⇒ 3p + q2 = 0 (Eq 2)
Checking whether p = -3 and q = -3 satisfy Eq 1,
⇒ 3p + qp = 0
L.H.S. = 3(-3) + (-3)(-3) = -9 + 9 = 0 = R.H.S.
Checking whether p = -3 and q = -3 satisfy Eq 2,
⇒ 3p + q2 = 0
L.H.S. = 3(-3) + (-3)2 = -9 + 9 = 0 = R.H.S.
Since, p = -3 and q = -3 satisfies Eq 1 and Eq 2,
∴ p = -3 and q = -3.
Hence, the values are p = -3 and q = -3.
If [ − 1 0 0 1 ] [ a b c d ] = [ 1 0 0 − 1 ] , \begin{bmatrix*}[r] -1 & 0 \\ 0 & 1 \end{bmatrix*} \begin{bmatrix*}[r] a & b \\ c & d \end{bmatrix*} = \begin{bmatrix*}[r] 1 & 0 \\ 0 & -1 \end{bmatrix*}, [ − 1 0 0 1 ] [ a c b d ] = [ 1 0 0 − 1 ] ,
Answer
Given,
[ − 1 0 0 1 ] [ a b c d ] = [ 1 0 0 − 1 ] ⇒ [ − 1 × a + 0 × c − 1 × b + 0 × d 0 × a + 1 × c 0 × b + 1 × d ] = [ 1 0 0 − 1 ] ⇒ [ − a − b c d ] = [ 1 0 0 − 1 ] \begin{bmatrix*}[r] -1 & 0 \\ 0 & 1 \end{bmatrix*} \begin{bmatrix*}[r] a & b \\ c & d \end{bmatrix*} = \begin{bmatrix*}[r] 1 & 0 \\ 0 & -1 \end{bmatrix*} \\[0.5em] \Rightarrow \begin{bmatrix*}[r] -1 \times a + 0 \times c & -1 \times b + 0 \times d \\ 0 \times a + 1 \times c & 0 \times b + 1 \times d \end{bmatrix*} = \begin{bmatrix*}[r] 1 & 0 \\ 0 & -1 \end{bmatrix*} \\[0.5em] \Rightarrow \begin{bmatrix*}[r] -a & -b \\ c & d \end{bmatrix*} = \begin{bmatrix*}[r] 1 & 0 \\ 0 & -1 \end{bmatrix*} \\[0.5em] [ − 1 0 0 1 ] [ a c b d ] = [ 1 0 0 − 1 ] ⇒ [ − 1 × a + 0 × c 0 × a + 1 × c − 1 × b + 0 × d 0 × b + 1 × d ] = [ 1 0 0 − 1 ] ⇒ [ − a c − b d ] = [ 1 0 0 − 1 ]
By definition of equality of matrices we get,
⇒ -a = 1 or a = -1
⇒ -b = 0 or b = 0
⇒ c = 0
⇒ d = -1.
Hence, the value of a = -1, b = 0, c = 0 and d = -1.
Find a and b if [ a − b b − 4 b + 4 a − 2 ] [ 2 0 0 2 ] = [ − 2 − 2 14 0 ] . \begin{bmatrix*}[r] a - b & b - 4 \\ b + 4 & a - 2 \end{bmatrix*} \begin{bmatrix*}[r] 2 & 0 \\ 0 & 2 \end{bmatrix*} = \begin{bmatrix*}[r] -2 & -2 \\ 14 & 0 \end{bmatrix*}. [ a − b b + 4 b − 4 a − 2 ] [ 2 0 0 2 ] = [ − 2 14 − 2 0 ] .
Answer
Given,
[ a − b b − 4 b + 4 a − 2 ] [ 2 0 0 2 ] = [ − 2 − 2 14 0 ] ⇒ [ ( a − b ) × 2 + ( b − 4 ) × 0 ( a − b ) × 0 + ( b − 4 ) × 2 ( b + 4 ) × 2 + ( a − 2 ) × 0 ( b + 4 ) × 0 + ( a − 2 ) × 2 ] = [ − 2 − 2 14 0 ] ⇒ [ 2 a − 2 b 2 b − 8 2 b + 8 2 a − 4 ] = [ − 2 − 2 14 0 ] \begin{bmatrix*}[r] a - b & b - 4 \\ b + 4 & a - 2 \end{bmatrix*} \begin{bmatrix*}[r] 2 & 0 \\ 0 & 2 \end{bmatrix*} = \begin{bmatrix*}[r] -2 & -2 \\ 14 & 0 \end{bmatrix*} \\[0.5em] \Rightarrow \begin{bmatrix*}[r] (a - b) \times 2 + (b - 4) \times 0 & (a - b) \times 0 + (b - 4) \times 2 \\ (b + 4) \times 2 + (a - 2) \times 0 & (b + 4) \times 0 + (a - 2) \times 2 \end{bmatrix*} = \begin{bmatrix*}[r] -2 & -2 \\ 14 & 0 \end{bmatrix*} \\[0.5em] \Rightarrow \begin{bmatrix*}[r] 2a - 2b & 2b - 8 \\ 2b + 8 & 2a - 4 \end{bmatrix*} = \begin{bmatrix*}[r] -2 & -2 \\ 14 & 0 \end{bmatrix*} \\[0.5em] [ a − b b + 4 b − 4 a − 2 ] [ 2 0 0 2 ] = [ − 2 14 − 2 0 ] ⇒ [ ( a − b ) × 2 + ( b − 4 ) × 0 ( b + 4 ) × 2 + ( a − 2 ) × 0 ( a − b ) × 0 + ( b − 4 ) × 2 ( b + 4 ) × 0 + ( a − 2 ) × 2 ] = [ − 2 14 − 2 0 ] ⇒ [ 2 a − 2 b 2 b + 8 2 b − 8 2 a − 4 ] = [ − 2 14 − 2 0 ]
By definition of equality of matrices we get,
⇒ 2a - 4 = 0 or 2a = 4 or a = 2
⇒ 2b - 8 = -2 or 2b = -2 + 8 = 6 or b = 3
⇒ 2a - 2b = -2 (Eq 1)
Checking whether a = 2 and b = 3 satisfies Eq 1,
⇒ 2a - 2b = -2
L.H.S. = 2a - 2b = 2(2) - 2(3) = 4 - 6 = -2 = R.H.S.
∴ a = 2 and b = 3.
Hence, the values are a = 2 and b = 3.
If A = [ sec 60° cos 90° -3 tan 45° sin 90° ] and B = [ 0 cot 45° − 2 3 sin 90° ] , \begin{bmatrix*}[r] \text{sec 60°} & \text{cos 90°} \\ \text{-3 tan 45°} & \text{sin 90°} \end{bmatrix*} \text{ and B } = \begin{bmatrix*}[r] 0 & \text{cot 45°} \\ -2 & \text{3 sin 90°} \end{bmatrix*}, [ sec 60° -3 tan 45° cos 90° sin 90° ] and B = [ 0 − 2 cot 45° 3 sin 90° ] ,
(i) 2A - 3B
(ii) A2
(iii) BA
Answer
(i) Given,
A = [ sec 60° cos 90° -3 tan 45° sin 90° ] = [ 2 0 − 3 1 ] B = [ 0 cot 45° − 2 3 sin 90° ] = [ 0 1 − 2 3 ] 2A - 3B = 2 [ 2 0 − 3 1 ] − 3 [ 0 1 − 2 3 ] = [ 4 0 − 6 2 ] − [ 0 3 − 6 9 ] = [ 4 − 0 0 − 3 − 6 − ( − 6 ) 2 − 9 ] = [ 4 − 3 0 − 7 ] . \text{A} = \begin{bmatrix*}[r] \text{sec 60°} & \text{cos 90°} \\ \text{-3 tan 45°} & \text{sin 90°} \end{bmatrix*} = \begin{bmatrix*}[r] 2 & 0 \\ -3 & 1 \end{bmatrix*} \\[1em] \text{B} = \begin{bmatrix*}[r] 0 & \text{cot 45°} \\ -2 & \text{3 sin 90°} \end{bmatrix*} = \begin{bmatrix*}[r] 0 & 1 \\ -2 & 3 \end{bmatrix*} \\[1em] \text{2A - 3B} = 2\begin{bmatrix*}[r] 2 & 0 \\ -3 & 1 \end{bmatrix*} - 3 \begin{bmatrix*}[r] 0 & 1 \\ -2 & 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 4 & 0 \\ -6 & 2 \end{bmatrix*} - \begin{bmatrix*}[r] 0 & 3 \\ -6 & 9 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 4 - 0 & 0 - 3 \\ -6 - (-6) & 2 - 9 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 4 & -3 \\ 0 & -7 \end{bmatrix*}. A = [ sec 60° -3 tan 45° cos 90° sin 90° ] = [ 2 − 3 0 1 ] B = [ 0 − 2 cot 45° 3 sin 90° ] = [ 0 − 2 1 3 ] 2A - 3B = 2 [ 2 − 3 0 1 ] − 3 [ 0 − 2 1 3 ] = [ 4 − 6 0 2 ] − [ 0 − 6 3 9 ] = [ 4 − 0 − 6 − ( − 6 ) 0 − 3 2 − 9 ] = [ 4 0 − 3 − 7 ] .
Hence, the value of 2A - 3B = [ 4 − 3 0 − 7 ] . = \begin{bmatrix*}[r] 4 & -3 \\ 0 & -7 \end{bmatrix*}. = [ 4 0 − 3 − 7 ] .
(ii) Given,
A 2 = [ 2 0 − 3 1 ] [ 2 0 − 3 1 ] = [ 2 × 2 + 0 × ( − 3 ) 2 × 0 + 0 × 1 ( − 3 ) × 2 + 1 × ( − 3 ) ( − 3 ) × 0 + 1 × 1 ] = [ 4 + 0 0 + 0 − 6 − 3 0 + 1 ] = [ 4 0 − 9 1 ] . \text{A}^2 = \begin{bmatrix*}[r] 2 & 0 \\ -3 & 1 \end{bmatrix*} \begin{bmatrix*}[r] 2 & 0 \\ -3 & 1 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 2 \times 2 + 0 \times (-3) & 2 \times 0 + 0 \times 1 \\ (-3) \times 2 + 1 \times (-3) & (-3) \times 0 + 1 \times 1 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 4 + 0 & 0 + 0 \\ -6 - 3 & 0 + 1 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 4 & 0 \\ -9 & 1 \end{bmatrix*}. A 2 = [ 2 − 3 0 1 ] [ 2 − 3 0 1 ] = [ 2 × 2 + 0 × ( − 3 ) ( − 3 ) × 2 + 1 × ( − 3 ) 2 × 0 + 0 × 1 ( − 3 ) × 0 + 1 × 1 ] = [ 4 + 0 − 6 − 3 0 + 0 0 + 1 ] = [ 4 − 9 0 1 ] .
Hence, the value of A 2 = [ 4 0 − 9 1 ] . \text{A}^2 = \begin{bmatrix*}[r] 4 & 0 \\ -9 & 1 \end{bmatrix*}. A 2 = [ 4 − 9 0 1 ] .
(iii)
BA = [ 0 1 − 2 3 ] [ 2 0 − 3 1 ] = [ 0 × 2 + 1 × ( − 3 ) 0 × 0 + 1 × 1 − 2 × 2 + 3 × ( − 3 ) − 2 × 0 + 3 × 1 ] = [ − 3 1 − 13 3 ] . \text{BA } = \begin{bmatrix*}[r] 0 & 1 \\ -2 & 3 \end{bmatrix*} \begin{bmatrix*}[r] 2 & 0 \\ -3 & 1 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 0 \times 2 + 1 \times (-3) & 0 \times 0 + 1 \times 1 \\ -2 \times 2 + 3 \times (-3) & -2 \times 0 + 3 \times 1 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] -3 & 1 \\ -13 & 3 \end{bmatrix*}. BA = [ 0 − 2 1 3 ] [ 2 − 3 0 1 ] = [ 0 × 2 + 1 × ( − 3 ) − 2 × 2 + 3 × ( − 3 ) 0 × 0 + 1 × 1 − 2 × 0 + 3 × 1 ] = [ − 3 − 13 1 3 ] .
Hence, the value of matrix BA = [ − 3 1 − 13 3 ] . \begin{bmatrix*}[r] -3 & 1 \\ -13 & 3 \end{bmatrix*}. [ − 3 − 13 1 3 ] .
Given matrix, X = [ 1 1 8 3 ] and I = [ 1 0 0 1 ] , \begin{bmatrix*}[r] 1 & 1 \\ 8 & 3 \end{bmatrix*} \text{ and } I = \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*}, [ 1 8 1 3 ] and I = [ 1 0 0 1 ] , 2 = 4X + 5I.
Answer
Given,
X2 = 4X + 5I
Solving for L.H.S.,
X 2 = [ 1 1 8 3 ] [ 1 1 8 3 ] = [ 1 × 1 + 1 × 8 1 × 1 + 1 × 3 8 × 1 + 3 × 8 8 × 1 + 3 × 3 ] = [ 1 + 8 1 + 3 8 + 24 8 + 9 ] = [ 9 4 32 17 ] . X^2 = \begin{bmatrix*}[r] 1 & 1 \\ 8 & 3 \end{bmatrix*}\begin{bmatrix*}[r] 1 & 1 \\ 8 & 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 \times 1 + 1 \times 8 & 1 \times 1 + 1\times 3 \\ 8 \times 1 + 3 \times 8 & 8 \times 1 + 3 \times 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 + 8 & 1 + 3 \\ 8 + 24 & 8 + 9 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 9 & 4 \\ 32 & 17 \end{bmatrix*}. X 2 = [ 1 8 1 3 ] [ 1 8 1 3 ] = [ 1 × 1 + 1 × 8 8 × 1 + 3 × 8 1 × 1 + 1 × 3 8 × 1 + 3 × 3 ] = [ 1 + 8 8 + 24 1 + 3 8 + 9 ] = [ 9 32 4 17 ] .
Solving for R.H.S.,
4 X + 5 I = 4 [ 1 1 8 3 ] + 5 [ 1 0 0 1 ] = [ 4 4 32 12 ] + [ 5 0 0 5 ] = [ 4 + 5 4 + 0 32 + 0 12 + 5 ] = [ 9 4 32 17 ] . 4X + 5I = 4\begin{bmatrix*}[r] 1 & 1 \\ 8 & 3 \end{bmatrix*} + 5\begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 4 & 4 \\ 32 & 12 \end{bmatrix*} + \begin{bmatrix*}[r] 5 & 0 \\ 0 & 5 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 4 + 5 & 4 + 0 \\ 32 + 0 & 12 + 5 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 9 & 4 \\ 32 & 17 \end{bmatrix*}. 4 X + 5 I = 4 [ 1 8 1 3 ] + 5 [ 1 0 0 1 ] = [ 4 32 4 12 ] + [ 5 0 0 5 ] = [ 4 + 5 32 + 0 4 + 0 12 + 5 ] = [ 9 32 4 17 ] .
Since, L.H.S. = R.H.S.
Hence, proved that X2 = 4X + 5I.
