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Chapter 9

Matrices

Class - 10 Concise Mathematics Selina

Exercise 9(A)

Question 1(a)

If [x+27y+3a−2]=[4b−343]\begin{bmatrix*}[r] x + 2 & 7 \\ y + 3 & a - 2 \end{bmatrix*} = \begin{bmatrix*}[r] 4 & b - 3 \\ 4 & 3 \end{bmatrix*}, the value of x, y, a and b are :

  1. x = 2, y = 1, a = 5 and b = 10

  2. x = -2, y = 1, a = 5 and b = 10

  3. x = 2, y = -1, a = 5 and b = 10

  4. x = 2, y = 1, a = -5 and b = 10

Answer

Given,

[x+27y+3a−2]=[4b−343]\begin{bmatrix*}[r] x + 2 & 7 \\ y + 3 & a - 2 \end{bmatrix*} = \begin{bmatrix*}[r] 4 & b - 3 \\ 4 & 3 \end{bmatrix*}.

∴ x + 2 = 4

⇒ x = 4 - 2 = 2.

∴ y + 3 = 4

⇒ y = 4 - 3 = 1.

∴ b - 3 = 7

⇒ b = 7 + 3 = 10.

∴ a - 2 = 3

⇒ a = 3 + 2 = 5.

Hence, Option 1 is the correct option.

Question 1(b)

If A = [5−53−3] and B=[−55−33]\begin{bmatrix*}[r] 5 & -5 \\ 3 & -3 \end{bmatrix*} \text{ and } B = \begin{bmatrix*}[r] -5 & 5 \\ -3 & 3 \end{bmatrix*}; the value of matrix (A - B) is :

  1. [0000]\begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*}

  2. [10−106−6]\begin{bmatrix*}[r] 10 & -10 \\ 6 & -6 \end{bmatrix*}

  3. [10−10−66]\begin{bmatrix*}[r] 10 & -10 \\ -6 & 6 \end{bmatrix*}

  4. [−1010−66]\begin{bmatrix*}[r] -10 & 10 \\ -6 & 6 \end{bmatrix*}

Answer

Given,

A = [5−53−3] and B=[−55−33]\begin{bmatrix*}[r] 5 & -5 \\ 3 & -3 \end{bmatrix*} \text{ and } B = \begin{bmatrix*}[r] -5 & 5 \\ -3 & 3 \end{bmatrix*}

A−B=[5−53−3]−[−55−33]=[5−(−5)−5−53−(−3)−3−3]=[10−106−6]A -B = \begin{bmatrix*}[r] 5 & -5 \\ 3 & -3 \end{bmatrix*} - \begin{bmatrix*}[r] -5 & 5 \\ -3 & 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 5 - (-5) & -5 - 5 \\ 3 - (-3) & -3 - 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 10 & -10 \\ 6 & -6 \end{bmatrix*}

Hence, Option 2 is the correct option.

Question 1(c)

If A = [5540],B=[3214] and C=[−2321]\begin{bmatrix*}[r] 5 & 5 \\ 4 & 0 \end{bmatrix*}, B = \begin{bmatrix*}[r] 3 & 2 \\ 1 & 4 \end{bmatrix*} \text{ and } C = \begin{bmatrix*}[r] -2 & 3 \\ 2 & 1 \end{bmatrix*} then matrix (A + B - C) is :

  1. [104−33]\begin{bmatrix*}[r] 10 & 4 \\ -3 & 3 \end{bmatrix*}

  2. [−1043−3]\begin{bmatrix*}[r] -10 & 4 \\ 3 & -3 \end{bmatrix*}

  3. [10433]\begin{bmatrix*}[r] 10 & 4 \\ 3 & 3 \end{bmatrix*}

  4. [10−433]\begin{bmatrix*}[r] 10 & -4 \\ 3 & 3 \end{bmatrix*}

Answer

Given,

A = [5540],B=[3214] and C=[−2321]\begin{bmatrix*}[r] 5 & 5 \\ 4 & 0 \end{bmatrix*}, B = \begin{bmatrix*}[r] 3 & 2 \\ 1 & 4 \end{bmatrix*} \text{ and } C = \begin{bmatrix*}[r] -2 & 3 \\ 2 & 1 \end{bmatrix*}

(A+B−C)=[5540]+[3214]−[−2321]=[5+3−(−2)5+2−34+1−20+4−1]=[8+27−35−24−1]=[10433].(A + B - C) = \begin{bmatrix*}[r] 5 & 5 \\ 4 & 0 \end{bmatrix*} + \begin{bmatrix*}[r] 3 & 2 \\ 1 & 4 \end{bmatrix*} - \begin{bmatrix*}[r] -2 & 3 \\ 2 & 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 5 + 3 - (-2) & 5 + 2 - 3 \\ 4 + 1 - 2 & 0 + 4 - 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 8 + 2 & 7 - 3 \\ 5 - 2 & 4 - 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 10 & 4 \\ 3 & 3 \end{bmatrix*}.

Hence, Option 3 is the correct option.

Question 1(d)

If A = [75−33] and B=[−2510]\begin{bmatrix*}[r] 7 & 5 \\ -3 & 3 \end{bmatrix*} \text{ and B} = \begin{bmatrix*}[r] -2 & 5 \\ 1 & 0 \end{bmatrix*}, then the matrix P (such that A + P = B) is :

  1. [409−3]\begin{bmatrix*}[r] 4 & 0 \\ 9 & -3 \end{bmatrix*}

  2. [904−2]\begin{bmatrix*}[r] 9 & 0 \\ 4 & -2 \end{bmatrix*}

  3. [−9043]\begin{bmatrix*}[r] -9 & 0 \\ 4 & 3 \end{bmatrix*}

  4. [−904−3]\begin{bmatrix*}[r] -9 & 0 \\ 4 & -3 \end{bmatrix*}

Answer

Given,

⇒ A + P = B

⇒ P = B - A

Substituting values we get :

⇒P=[−2510]−[75−33]⇒P=[−2−75−51−(−3)0−3]⇒P=[−904−3].\Rightarrow P = \begin{bmatrix*}[r] -2 & 5 \\ 1 & 0 \end{bmatrix*} - \begin{bmatrix*}[r] 7 & 5 \\ -3 & 3 \end{bmatrix*} \\[1em] \Rightarrow P = \begin{bmatrix*}[r] -2 - 7 & 5 - 5 \\ 1 - (-3) & 0 - 3 \end{bmatrix*} \\[1em] \Rightarrow P = \begin{bmatrix*}[r] -9 & 0 \\ 4 & -3 \end{bmatrix*}.

Hence, Option 4 is the correct option.

Question 1(e)

The additive inverse of matrix A + B, where

A = [427−2] and B=[−213−4]\begin{bmatrix*}[r] 4 & 2 \\ 7 & -2 \end{bmatrix*} \text{ and } B = \begin{bmatrix*}[r] -2 & 1 \\ 3 & -4 \end{bmatrix*} is :

  1. [−2−3−106]\begin{bmatrix*}[r] -2 & -3 \\ -10 & 6 \end{bmatrix*}

  2. [23−10−6]\begin{bmatrix*}[r] 2 & 3 \\ -10 & -6 \end{bmatrix*}

  3. [−2−3−10−6]\begin{bmatrix*}[r] -2 & -3 \\ -10 & -6 \end{bmatrix*}

  4. [−2310−6]\begin{bmatrix*}[r] -2 & 3 \\ 10 & -6 \end{bmatrix*}

Answer

Additive inverse of a matrix M is given by -M.

So, additive inverse of (A + B) = -(A + B).

Substituting values we get :

⇒−(A+B)=−([427−2]+[−213−4])=−([4+(−2)2+17+3−2+(−4)])=−([2310−6])=[−2−3−106]\Rightarrow -(A + B) = -\Big(\begin{bmatrix*}[r] 4 & 2 \\ 7 & -2 \end{bmatrix*} + \begin{bmatrix*}[r] -2 & 1 \\ 3 & -4 \end{bmatrix*}\Big) \\[1em] = -\Big(\begin{bmatrix*}[r] 4 + (-2) & 2 + 1 \\ 7 + 3 & -2 + (-4) \end{bmatrix*}\Big) \\[1em] = -\Big(\begin{bmatrix*}[r] 2 & 3 \\ 10 & -6 \end{bmatrix*}\Big) \\[1em] = \begin{bmatrix*}[r] -2 & -3 \\ -10 & 6 \end{bmatrix*}

Hence, Option 1 is the correct option.

Question 2

State, whether the following statements are true or false. If false, give a reason.

(i) If A and B are two matrices of orders 3 × 2 and 2 × 3 respectively; then their sum A + B is possible.

(ii) The matrices A2 × 3 and B2 × 3 are conformable for subtraction.

(iii) Transpose of a 2 × 1 matrix is a 2 × 1 matrix.

(iv) Transpose of a square matrix is a square matrix.

(v) A column matrix has many columns and only one row.

Answer

(i) False

Reason — For addition the order of both the matrices must be same.

Here, A and B have different orders.

Hence, the statement is false because the orders of both matrices are different.

(ii) True

Reason — For subtraction the order of both the matrices must be same.

Here, A and B have same orders.

Hence, the statement is true.

(iii) False

Reason — On transposing, the no. of rows and columns gets interchanged.

Hence, transpose of a 2 × 1 matrix will be of order 1 × 2.

Hence, the statement is false.

(iv) True

Reason — The transpose of a square matrix is also a square matrix.

Hence, the statement is true.

(v) False

Reason — A column matrix has only one column and can have many rows.

Hence, the statement is false.

Question 3(i)

Solve for a, b and c; if :

[−4a+532]=[b+423c−1]\begin{bmatrix} -4 & a + 5 \\ 3 & 2 \end{bmatrix} = \begin{bmatrix} b + 4 & 2 \\ 3 & c - 1 \end{bmatrix}

Answer

Given, [−4a+532]=[b+423c−1]\begin{bmatrix} -4 & a + 5 \\ 3 & 2 \end{bmatrix} = \begin{bmatrix} b + 4 & 2 \\ 3 & c - 1 \end{bmatrix}

By definition of equality of matrices we get,

-4 = b + 4 ⇒ b = -4 - 4 = -8,

a + 5 = 2 ⇒ a = 2 - 5 = -3,

2 = c - 1 ⇒ c = 2 + 1 = 3.

Hence, a = -3, b = -8 and c = 3.

Question 3(ii)

Solve for a, b and c; if :

[aa−bb+c0]=[3−120]\begin{bmatrix} a & a - b \\ b + c & 0 \end{bmatrix} = \begin{bmatrix} 3 & -1 \\ 2 & 0 \end{bmatrix}

Answer

Given, [aa−bb+c0]=[3−120]\begin{bmatrix} a & a - b \\ b + c & 0 \end{bmatrix} = \begin{bmatrix} 3 & -1 \\ 2 & 0 \end{bmatrix}

By definition of equality of matrices we get,

a = 3,

a - b = -1
⇒ 3 - b = -1 [∵ a = 3]
⇒ b = 3 + 1 = 4

b + c = 2
⇒ 4 + c = 2
⇒ c = -2.

Hence, a = 3, b = 4 and c = -2.

Question 4

Wherever possible, write each of the following as a single matrix.

(i) [1234]+[−1−21−7]\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} + \begin{bmatrix} -1 & -2 \\ 1 & -7 \end{bmatrix}

(ii) [234567]−[0236−10]\begin{bmatrix} 2 & 3 & 4 \\ 5 & 6 & 7 \end{bmatrix} - \begin{bmatrix} 0 & 2 & 3 \\ 6 & -1 & 0 \end{bmatrix}

(iii) [012467]+[3468]\begin{bmatrix} 0 & 1 & 2 \\ 4 & 6 & 7 \end{bmatrix} + \begin{bmatrix} 3 & 4 \\ 6 & 8 \end{bmatrix}

Answer

(i) Given,

⇒[1234]+[−1−21−7]⇒[1+(−1)2+(−2)3+14+(−7)]⇒[004−3].\phantom{\Rightarrow} \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} + \begin{bmatrix} -1 & -2 \\ 1 & -7 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 1 + (-1) & 2 + (-2) \\ 3 + 1 & 4 + (-7) \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 0 & 0 \\ 4 & -3 \end{bmatrix}.

Hence, resultant matrix = [004−3]\begin{bmatrix} 0 & 0 \\ 4 & -3 \end{bmatrix}.

(ii) Given,

⇒[234567]−[0236−10]⇒[2−03−24−35−66−(−1)7−0]⇒[211−177].\phantom{\Rightarrow} \begin{bmatrix} 2 & 3 & 4 \\ 5 & 6 & 7 \end{bmatrix} - \begin{bmatrix} 0 & 2 & 3 \\ 6 & -1 & 0 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2 - 0 & 3 - 2 & 4 - 3 \\ 5 - 6 & 6 - (-1) & 7 - 0 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2 & 1 & 1 \\ -1 & 7 & 7 \end{bmatrix}.

Hence, resultant matrix = [211−177]\begin{bmatrix} 2 & 1 & 1 \\ -1 & 7 & 7 \end{bmatrix}.

(iii) Given,

[012467]+[3468]\begin{bmatrix} 0 & 1 & 2 \\ 4 & 6 & 7 \end{bmatrix} + \begin{bmatrix} 3 & 4 \\ 6 & 8 \end{bmatrix}

The above calculation is not possible because for addition the order of both matrices must be equal.

Question 5(i)

Find, x and y from the following equations :

[52−1y−1]−[1x−12−3]=[47−32]\begin{bmatrix} 5 & 2 \\ -1 & y - 1 \end{bmatrix} - \begin{bmatrix} 1 & x - 1 \\ 2 & -3 \end{bmatrix} = \begin{bmatrix} 4 & 7 \\ -3 & 2 \end{bmatrix}

Answer

Given,

⇒[52−1y−1]−[1x−12−3]=[47−32]⇒[5−12−(x−1)−1−2y−1−(−3)]=[47−32]⇒[43−x−3y+2]=[47−32]\Rightarrow \begin{bmatrix} 5 & 2 \\ -1 & y - 1 \end{bmatrix} - \begin{bmatrix} 1 & x - 1 \\ 2 & -3 \end{bmatrix} = \begin{bmatrix} 4 & 7 \\ -3 & 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 5 - 1 & 2 - (x - 1) \\ -1 - 2 & y - 1 - (-3) \end{bmatrix} = \begin{bmatrix} 4 & 7 \\ -3 & 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 4 & 3 - x \\ -3 & y + 2 \end{bmatrix} = \begin{bmatrix} 4 & 7 \\ -3 & 2 \end{bmatrix} \\[1em]

By definition of equality of matrices we get,

⇒ 3 - x = 7
⇒ x = 3 - 7 = -4.

⇒ y + 2 = 2 ⇒ y = 0.

Hence, x = -4 and y = 0.

Question 5(ii)

Find, x and y from the following equations :

[−8x]+[y−2]=[−32]\begin{bmatrix} -8 & x \end{bmatrix} + \begin{bmatrix} y & -2 \end{bmatrix} = \begin{bmatrix} -3 & 2 \end{bmatrix}

Answer

Given,

⇒[−8x]+[y−2]=[−32]⇒[−8+yx+(−2)]=[−32]\Rightarrow \begin{bmatrix} -8 & x \end{bmatrix} + \begin{bmatrix} y & -2 \end{bmatrix} = \begin{bmatrix} -3 & 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -8 + y & x + (-2) \end{bmatrix} = \begin{bmatrix} -3 & 2 \end{bmatrix}

By definition of equality of matrices we get,

-8 + y = -3
⇒ y = -3 + 8 = 5.

x - 2 = 2
⇒ x = 2 + 2 = 4.

Hence, x = 4 and y = 5.

Question 6

Given : M = [5−3−24]\begin{bmatrix} 5 & -3 \\ -2 & 4 \end{bmatrix}, find its transpose matrix Mt. If possible, find :

(i) M + Mt

(ii) Mt - M

Answer

M = [5−3−24]\begin{bmatrix} 5 & -3 \\ -2 & 4 \end{bmatrix}

Mt = [5−2−34]\begin{bmatrix} 5 & -2 \\ -3 & 4 \end{bmatrix}

(i)

M+Mt=[5−3−24]+[5−2−34]=[5+5−3+(−2)−2+(−3)4+4]=[10−5−58].M + M^t = \begin{bmatrix} 5 & -3 \\ -2 & 4 \end{bmatrix} + \begin{bmatrix} 5 & -2 \\ -3 & 4 \end{bmatrix} \\[1em] = \begin{bmatrix} 5 + 5 & -3 + (-2) \\ -2 + (-3) & 4 + 4 \end{bmatrix} \\[1em] = \begin{bmatrix} 10 & -5 \\ -5 & 8 \end{bmatrix}.

Hence, M+Mt=[10−5−58].M + M^t = \begin{bmatrix} 10 & -5 \\ -5 & 8 \end{bmatrix}.

(ii)

Mt−M=[5−2−34]−[5−3−24]=[5−5−2−(−3)−3−(−2)4−4]=[01−10].M^t - M = \begin{bmatrix} 5 & -2 \\ -3 & 4 \end{bmatrix} - \begin{bmatrix} 5 & -3 \\ -2 & 4 \end{bmatrix} \\[1em] = \begin{bmatrix} 5 - 5 & -2 - (-3) \\ -3 - (-2) & 4 - 4 \end{bmatrix} \\[1em] = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}.

Hence, Mt - M = [01−10].\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}.

Question 7

Given A = [2−3],B=[02] and C=[−14];\begin{bmatrix} 2 & -3 \end{bmatrix}, B = \begin{bmatrix} 0 & 2 \end{bmatrix} \text{ and C} = \begin{bmatrix} -1 & 4 \end{bmatrix}; find the matrix X in each of the following :

(i) X + B = C - A

(ii) A - X = B + C

Answer

(i) Given,

⇒ X + B = C - A

⇒ X = C - A - B

Substituting values of A, B and C in above equation we get,

⇒X=[−14]−[2−3]−[02]=[−1−2−04−(−3)−2]=[−35].\Rightarrow X = \begin{bmatrix} -1 & 4 \end{bmatrix} - \begin{bmatrix} 2 & -3 \end{bmatrix} - \begin{bmatrix} 0 & 2 \end{bmatrix} \\[1em] = \begin{bmatrix} -1 - 2 - 0 & 4 - (-3) - 2 \end{bmatrix} \\[1em] = \begin{bmatrix} -3 & 5 \end{bmatrix}.

Hence, X = [−35].\begin{bmatrix} -3 & 5 \end{bmatrix}.

(ii) Given,

⇒ A - X = B + C

⇒ X = A - (B + C)

Substituting values of A, B and C in above equation we get,

⇒X=[2−3]−([02]+[−14])=[2−3]−([0+(−1)2+4])=[2−3]−[−16]=[2−(−1)−3−6]=[3−9].\Rightarrow X = \begin{bmatrix} 2 & -3 \end{bmatrix} - \Big(\begin{bmatrix} 0 & 2 \end{bmatrix} + \begin{bmatrix} -1 & 4 \end{bmatrix}\Big) \\[1em] = \begin{bmatrix} 2 & -3 \end{bmatrix} - \Big(\begin{bmatrix} 0 + (-1) & 2 + 4 \end{bmatrix}\Big) \\[1em] = \begin{bmatrix} 2 & -3 \end{bmatrix} - \begin{bmatrix} -1 & 6 \end{bmatrix} \\[1em] = \begin{bmatrix} 2 - (-1) & -3 - 6 \end{bmatrix} \\[1em] = \begin{bmatrix} 3 & -9 \end{bmatrix}.

Hence, X = [3−9]\begin{bmatrix} 3 & -9 \end{bmatrix}.

Question 8

Given A = [−102−4] and B=[3−3−20]\begin{bmatrix} -1 & 0 \\ 2 & -4 \end{bmatrix} \text{ and } B = \begin{bmatrix} 3 & -3 \\ -2 & 0 \end{bmatrix}; find the matrix X in each of the following :

(i) A + X = B

(ii) A - X = B

(iii) X - B = A

Answer

(i) Given,

⇒ A + X = B

⇒ X = B - A

⇒X=[3−3−20]−[−102−4]=[3−(−1)−3−0−2−20−(−4)]=[4−3−44].\Rightarrow X = \begin{bmatrix} 3 & -3 \\ -2 & 0 \end{bmatrix} - \begin{bmatrix} -1 & 0 \\ 2 & -4 \end{bmatrix} \\[1em] = \begin{bmatrix} 3 - (-1) & -3 - 0 \\ -2 - 2 & 0 - (-4) \end{bmatrix} \\[1em] = \begin{bmatrix} 4 & -3 \\ -4 & 4 \end{bmatrix}.

Hence, X = [4−3−44].\begin{bmatrix} 4 & -3 \\ -4 & 4 \end{bmatrix}.

(ii) Given,

⇒ A - X = B

⇒ X = A - B

⇒X=[−102−4]−[3−3−20]=[−1−30−(−3)2−(−2)−4−0]=[−434−4].\Rightarrow X = \begin{bmatrix} -1 & 0 \\ 2 & -4 \end{bmatrix} - \begin{bmatrix} 3 & -3 \\ -2 & 0 \end{bmatrix} \\[1em] = \begin{bmatrix} -1 - 3 & 0 - (-3) \\ 2 - (-2) & -4 - 0 \end{bmatrix} \\[1em] = \begin{bmatrix} -4 & 3 \\ 4 & -4 \end{bmatrix}.

Hence, X = [−434−4].\begin{bmatrix} -4 & 3 \\ 4 & -4 \end{bmatrix}.

(iii) Given,

⇒ X - B = A

⇒ X = A + B

⇒X=[−102−4]+[3−3−20]=[−1+30+(−3)2+(−2)−4+0]=[2−30−4].\Rightarrow X = \begin{bmatrix} -1 & 0 \\ 2 & -4 \end{bmatrix} + \begin{bmatrix} 3 & -3 \\ -2 & 0 \end{bmatrix} \\[1em] = \begin{bmatrix} -1 + 3 & 0 + (-3) \\ 2 + (-2) & -4 + 0 \end{bmatrix} \\[1em] = \begin{bmatrix} 2 & -3 \\ 0 & -4 \end{bmatrix}.

Hence, X = [2−30−4].\begin{bmatrix} 2 & -3 \\ 0 & -4 \end{bmatrix}.

Exercise 9(B)

Question 1(a)

If 4[5x]−5[y−2]=[1022]4\begin{bmatrix*}[r] 5 & x \end{bmatrix*} - 5\begin{bmatrix*}[r] y & -2 \end{bmatrix*} = \begin{bmatrix*}[r] 10 & 22 \end{bmatrix*}, the values of x and y are :

  1. x = 2 and y = 3

  2. x = 3 and y = 2

  3. x = -3 and y = 2

  4. x = 3 and y = -2

Answer

Given,

⇒4[5x]−5[y−2]=[1022]⇒[204x]−[5y−10]=[1022]⇒[20−5y4x−(−10)]=[1022]⇒[20−5y4x+10]=[1022]\Rightarrow 4\begin{bmatrix*}[r] 5 & x \end{bmatrix*} - 5\begin{bmatrix*}[r] y & -2 \end{bmatrix*} = \begin{bmatrix*}[r] 10 & 22 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 20 & 4x \end{bmatrix*} - \begin{bmatrix*}[r] 5y & -10 \end{bmatrix*} = \begin{bmatrix*}[r] 10 & 22 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 20 - 5y & 4x - (-10) \end{bmatrix*} = \begin{bmatrix*}[r] 10 & 22 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 20 - 5y & 4x + 10 \end{bmatrix*} = \begin{bmatrix*}[r] 10 & 22 \end{bmatrix*}

∴ 20 - 5y = 10 and 4x + 10 = 22

⇒ 5y = 20 - 10 and 4x = 22 - 10

⇒ 5y = 10 and 4x = 12

⇒ y = 2 and x = 3.

Hence, Option 2 is the correct option.

Question 1(b)

If A = [−3−70−8] and A−B=[64−30]\begin{bmatrix*}[r] -3 & -7 \\ 0 & -8 \end{bmatrix*}\text{ and } A - B = \begin{bmatrix*}[r] 6 & 4 \\ -3 & 0 \end{bmatrix*}, then matrix B is :

  1. [911−318]\begin{bmatrix*}[r] 9 & 11 \\ -3 & 18 \end{bmatrix*}

  2. [−9−1138]\begin{bmatrix*}[r] -9 & -11 \\ 3 & 8 \end{bmatrix*}

  3. [9−11−38]\begin{bmatrix*}[r] 9 & -11 \\ -3 & 8 \end{bmatrix*}

  4. [−9−11−3−8]\begin{bmatrix*}[r] -9 & -11 \\ -3 & -8 \end{bmatrix*}

Answer

Given,

A - B = [64−30]\begin{bmatrix*}[r] 6 & 4 \\ -3 & 0 \end{bmatrix*}

Substituting value of A in above equation we get :

⇒[−3−70−8]−B=[64−30]⇒B=[−3−70−8]−[64−30]⇒B=[−3−6−7−40−(−3)−8−0]⇒B=[−9−113−8]\Rightarrow \begin{bmatrix*}[r] -3 & -7 \\ 0 & -8 \end{bmatrix*} - B = \begin{bmatrix*}[r] 6 & 4 \\ -3 & 0 \end{bmatrix*} \\[1em] \Rightarrow B = \begin{bmatrix*}[r] -3 & -7 \\ 0 & -8 \end{bmatrix*} - \begin{bmatrix*}[r] 6 & 4 \\ -3 & 0 \end{bmatrix*} \\[1em] \Rightarrow B = \begin{bmatrix*}[r] -3 - 6 & -7 - 4 \\ 0 - (-3) & -8 - 0 \end{bmatrix*} \\[1em] \Rightarrow B = \begin{bmatrix*}[r] -9 & -11 \\ 3 & -8 \end{bmatrix*}

Hence, Option 4 is the correct option.

Question 1(c)

If I is a unit matrix of order 2 and M + 4I = [8−342]\begin{bmatrix*}[r] 8 & -3 \\ 4 & 2 \end{bmatrix*}, then matrix M is :

  1. [434−2]\begin{bmatrix*}[r] 4 & 3 \\ 4 & -2 \end{bmatrix*}

  2. [4342]\begin{bmatrix*}[r] 4 & 3 \\ 4 & 2 \end{bmatrix*}

  3. [4−3−42]\begin{bmatrix*}[r] 4 & -3 \\ -4 & 2 \end{bmatrix*}

  4. [4−34−2]\begin{bmatrix*}[r] 4 & -3 \\ 4 & -2 \end{bmatrix*}

Answer

As, I is a unit matrix of order 2.

∴ I = [1001]\begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*}

Given,

⇒M+4I=[8−342]⇒M+4[1001]=[8−342]⇒M+[4004]=[8−342]⇒M=[8−342]−[4004]⇒M=[8−4−3−04−02−4]⇒M=[4−34−2].\Rightarrow M + 4I = \begin{bmatrix*}[r] 8 & -3 \\ 4 & 2 \end{bmatrix*} \\[1em] \Rightarrow M + 4\begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} = \begin{bmatrix*}[r] 8 & -3 \\ 4 & 2 \end{bmatrix*} \\[1em] \Rightarrow M + \begin{bmatrix*}[r] 4 & 0 \\ 0 & 4 \end{bmatrix*} = \begin{bmatrix*}[r] 8 & -3 \\ 4 & 2 \end{bmatrix*} \\[1em] \Rightarrow M = \begin{bmatrix*}[r] 8 & -3 \\ 4 & 2 \end{bmatrix*} - \begin{bmatrix*}[r] 4 & 0 \\ 0 & 4 \end{bmatrix*} \\[1em] \Rightarrow M = \begin{bmatrix*}[r] 8 - 4 & -3 - 0 \\ 4 - 0 & 2 - 4 \end{bmatrix*} \\[1em] \Rightarrow M = \begin{bmatrix*}[r] 4 & -3 \\ 4 & -2 \end{bmatrix*}.

Hence, Option 4 is the correct option.

Question 1(d)

If 2[3x01]+3[13y2]=[z−7158]2\begin{bmatrix*}[r] 3 & x \\ 0 & 1 \end{bmatrix*} + 3\begin{bmatrix*}[r] 1 & 3 \\ y & 2 \end{bmatrix*} = \begin{bmatrix*}[r] z & -7 \\ 15 & 8 \end{bmatrix*}, the values of x, y and z are :

  1. x = 8, y = -5 and z = 9

  2. x = -8, y = 5 and z = 9

  3. x = -8, y = -5 and z = -9

  4. x = -8, y = 5 and z = -9

Answer

Given,

⇒2[3x01]+3[13y2]=[z−7158]⇒[62x02]+[393y6]=[z−7158]⇒[6+32x+90+3y2+6]=[z−7158]⇒[92x+93y8]=[z−7158]\Rightarrow 2\begin{bmatrix*}[r] 3 & x \\ 0 & 1 \end{bmatrix*} + 3\begin{bmatrix*}[r] 1 & 3 \\ y & 2 \end{bmatrix*} = \begin{bmatrix*}[r] z & -7 \\ 15 & 8 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 6 & 2x \\ 0 & 2 \end{bmatrix*} + \begin{bmatrix*}[r] 3 & 9 \\ 3y & 6 \end{bmatrix*} = \begin{bmatrix*}[r] z & -7 \\ 15 & 8 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 6 + 3 & 2x + 9 \\ 0 + 3y & 2 + 6 \end{bmatrix*} = \begin{bmatrix*}[r] z & -7 \\ 15 & 8 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 9 & 2x + 9 \\ 3y & 8 \end{bmatrix*} = \begin{bmatrix*}[r] z & -7 \\ 15 & 8 \end{bmatrix*} \\[1em]

From above equation we get :

⇒ z = 9, 3y = 15 and 2x + 9 = -7

⇒ z = 9, y = 153\dfrac{15}{3} and 2x = -7 - 9

⇒ z = 9, y = 5 and 2x = -16

⇒ z = 9, y = 5 and x = −162-\dfrac{16}{2}

⇒ z = 9, y = 5 and x = -8.

Hence, Option 2 is the correct option.

Question 1(e)

Given A = [473−2] and B=[12−14]\begin{bmatrix*}[r] 4 & 7 \\ 3 & -2 \end{bmatrix*} \text{ and } B = \begin{bmatrix*}[r] 1 & 2 \\ -1 & 4 \end{bmatrix*}, then A - 2B is :

  1. [−235−10]\begin{bmatrix*}[r] -2 & 3 \\ 5 & -10 \end{bmatrix*}

  2. [−2−3−510]\begin{bmatrix*}[r] -2 & -3 \\ -5 & 10 \end{bmatrix*}

  3. [235−10]\begin{bmatrix*}[r] 2 & 3 \\ 5 & -10 \end{bmatrix*}

  4. [23510]\begin{bmatrix*}[r] 2 & 3 \\ 5 & 10 \end{bmatrix*}

Answer

Substituting values of A and B in A - 2B, we get :

⇒A−2B=[473−2]−2[12−14]=[473−2]−[24−28]=[4−27−43−(−2)−2−8]=[233+2−10]=[235−10].\Rightarrow A - 2B = \begin{bmatrix*}[r] 4 & 7 \\ 3 & -2 \end{bmatrix*} - 2\begin{bmatrix*}[r] 1 & 2 \\ -1 & 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 4 & 7 \\ 3 & -2 \end{bmatrix*} - \begin{bmatrix*}[r] 2 & 4 \\ -2 & 8 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 4 - 2 & 7 - 4 \\ 3 - (-2) & -2 - 8 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 & 3 \\ 3 + 2 & -10 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 & 3 \\ 5 & -10 \end{bmatrix*}.

Hence, Option 3 is the correct option.

Question 2

Find x and y if :

(i) 3[4x]+2[y−3]=[100]3\begin{bmatrix*}[r] 4 & x \end{bmatrix*} + 2\begin{bmatrix*}[r] y & -3 \end{bmatrix*} = \begin{bmatrix*}[r] 10 & 0 \end{bmatrix*}

(ii) x[−12]−4[−2y]=[7−8]x\begin{bmatrix*}[r] -1 \\ 2 \end{bmatrix*} - 4\begin{bmatrix*}[r] -2 \\ y \end{bmatrix*} = \begin{bmatrix*}[r] 7 \\ -8 \end{bmatrix*}

Answer

(i) Given,

⇒3[4x]+2[y−3]=[100]⇒[123x]+[2y−6]=[100]⇒[12+2y3x+(−6)]=[100]⇒[12+2y3x−6]=[100]\Rightarrow 3\begin{bmatrix*}[r] 4 & x \end{bmatrix*} + 2\begin{bmatrix*}[r] y & -3 \end{bmatrix*} = \begin{bmatrix*}[r] 10 & 0 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 12 & 3x \end{bmatrix*} + \begin{bmatrix*}[r] 2y & -6 \end{bmatrix*} = \begin{bmatrix*}[r] 10 & 0 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 12 + 2y & 3x + (-6) \end{bmatrix*} = \begin{bmatrix*}[r] 10 & 0 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 12 + 2y & 3x - 6 \end{bmatrix*} = \begin{bmatrix*}[r] 10 & 0 \end{bmatrix*}

By equality of matrices we get,

12 + 2y = 10
⇒ 2y = 10 - 12
⇒ 2y = -2
⇒ y = -1.

3x - 6 = 0
⇒ 3x = 6
⇒ x = 2

Hence, x = 2 and y = -1.

(ii) Given,

⇒x[−12]−4[−2y]=[7−8]⇒[−x2x]−[−84y]=[7−8]⇒[−x−(−8)2x−4y]=[7−8]⇒[−x+82x−4y]=[7−8]\Rightarrow x\begin{bmatrix*}[r] -1 \\ 2 \end{bmatrix*} - 4\begin{bmatrix*}[r] -2 \\ y \end{bmatrix*} = \begin{bmatrix*}[r] 7 \\ -8 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] -x \\ 2x \end{bmatrix*} - \begin{bmatrix*}[r] -8 \\ 4y \end{bmatrix*} = \begin{bmatrix*}[r] 7 \\ -8 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] -x -(-8) \\ 2x - 4y \end{bmatrix*} = \begin{bmatrix*}[r] 7 \\ -8 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] -x + 8 \\ 2x - 4y \end{bmatrix*} = \begin{bmatrix*}[r] 7 \\ -8 \end{bmatrix*}

By definition of equality of matrices we get,

-x + 8 = 7 ........(i)

2x - 4y = -8 ......(ii)

Solving eq. (i) we get,

⇒ x = 8 - 7 = 1.

Substituting x = 1 in eq. (ii) we get,

⇒ 2x - 4y = -8
⇒ 2(1) - 4y = -8
⇒ 2 - 4y = -8
⇒ 4y = 2 + 8
⇒ 4y = 10
⇒ y = 104=52=2.5\dfrac{10}{4} = \dfrac{5}{2} = 2.5.

Hence, x = 1 and y = 2.5

Question 3

Given A = [2130],B=[1152] and C=[−3−100]\begin{bmatrix*}[r] 2 & 1 \\ 3 & 0 \end{bmatrix*}, B = \begin{bmatrix*}[r] 1 & 1 \\ 5 & 2 \end{bmatrix*} \text{ and } C = \begin{bmatrix*}[r] -3 & -1 \\ 0 & 0 \end{bmatrix*}; find :

(i) 2A - 3B + C

(ii) A + 2C - B

Answer

(i) Given,

2A - 3B + C

Substituting values of A, B and C in above equation we get,

⇒2A−3B+C=2[2130]−3[1152]+[−3−100]=[4260]−[33156]+[−3−100]=[4−3+(−3)2−3+(−1)6−15+00−6+0]=[−2−2−9−6].\Rightarrow 2A - 3B + C = 2\begin{bmatrix*}[r] 2 & 1 \\ 3 & 0 \end{bmatrix*} - 3\begin{bmatrix*}[r] 1 & 1 \\ 5 & 2 \end{bmatrix*} + \begin{bmatrix*}[r] -3 & -1 \\ 0 & 0 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 4 & 2 \\ 6 & 0 \end{bmatrix*} - \begin{bmatrix*}[r] 3 & 3 \\ 15 & 6 \end{bmatrix*} + \begin{bmatrix*}[r] -3 & -1 \\ 0 & 0 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 4 - 3 + (-3) & 2 - 3 + (-1) \\ 6 - 15 + 0 & 0 - 6 + 0 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -2 & -2 \\ -9 & -6 \end{bmatrix*}.

Hence 2A - 3B + C = [−2−2−9−6].\begin{bmatrix*}[r] -2 & -2 \\ -9 & -6 \end{bmatrix*}.

(ii) Given,

A + 2C - B

Substituting values of A, B and C in above equation we get,

⇒[2130]+2[−3−100]−[1152]=[2130]+[−6−200]−[1152]=[2+(−6)−11+(−2)−13+0−50+0−2]=[−5−2−2−2].\Rightarrow \begin{bmatrix*}[r] 2 & 1 \\ 3 & 0 \end{bmatrix*} + 2\begin{bmatrix*}[r] -3 & -1 \\ 0 & 0 \end{bmatrix*} - \begin{bmatrix*}[r] 1 & 1 \\ 5 & 2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 & 1 \\ 3 & 0 \end{bmatrix*} + \begin{bmatrix*}[r] -6 & -2 \\ 0 & 0 \end{bmatrix*} - \begin{bmatrix*}[r] 1 & 1 \\ 5 & 2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 + (-6) - 1 & 1 + (-2) - 1 \\ 3 + 0 - 5 & 0 + 0 - 2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -5 & -2 \\ -2 & -2 \end{bmatrix*}.

Hence, A + 2C - B = [−5−2−2−2].\begin{bmatrix*}[r] -5 & -2 \\ -2 & -2 \end{bmatrix*}.

Question 4

If [4−240]+3A=[−2−21−3]\begin{bmatrix*}[r] 4 & -2 \\ 4 & 0 \end{bmatrix*} + 3A = \begin{bmatrix*}[r] -2 & -2 \\ 1 & -3 \end{bmatrix*}; find A.

Answer

Given,

⇒[4−240]+3A=[−2−21−3]⇒3A=[−2−21−3]−[4−240]⇒3A=[−2−4−2−(−2)1−4−3−0]⇒3A=[−60−3−3]⇒A=13[−60−3−3]⇒A=[−20−1−1].\Rightarrow \begin{bmatrix*}[r] 4 & -2 \\ 4 & 0 \end{bmatrix*} + 3A = \begin{bmatrix*}[r] -2 & -2 \\ 1 & -3 \end{bmatrix*} \\[1em] \Rightarrow 3A = \begin{bmatrix*}[r] -2 & -2 \\ 1 & -3 \end{bmatrix*} - \begin{bmatrix*}[r] 4 & -2 \\ 4 & 0 \end{bmatrix*} \\[1em] \Rightarrow 3A = \begin{bmatrix*}[r] -2 - 4 & -2 - (-2) \\ 1 - 4 & -3 - 0 \end{bmatrix*} \\[1em] \Rightarrow 3A = \begin{bmatrix*}[r] -6 & 0 \\ -3 & -3 \end{bmatrix*} \\[1em] \Rightarrow A = \dfrac{1}{3}\begin{bmatrix*}[r] -6 & 0 \\ -3 & -3 \end{bmatrix*} \\[1em] \Rightarrow A = \begin{bmatrix*}[r] -2 & 0 \\ -1 & -1 \end{bmatrix*}.

Hence, A = [−20−1−1]\begin{bmatrix*}[r] -2 & 0 \\ -1 & -1 \end{bmatrix*}.

Question 5

Given A = [1423] and B=[−4−1−3−2]\begin{bmatrix*}[r] 1 & 4 \\ 2 & 3 \end{bmatrix*} \text{ and } B = \begin{bmatrix*}[r] -4 & -1 \\ -3 & -2 \end{bmatrix*}

(i) find the matrix 2A + B

(ii) find a matrix C such that :

C + B = [0000]\begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*}

Answer

(i)

2A+B=2[1423]+[−4−1−3−2]=[2846]+[−4−1−3−2]=[2+(−4)8+(−1)4+(−3)6+(−2)]=[−2714].2A + B = 2\begin{bmatrix*}[r] 1 & 4 \\ 2 & 3 \end{bmatrix*} + \begin{bmatrix*}[r] -4 & -1 \\ -3 & -2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 & 8 \\ 4 & 6 \end{bmatrix*} + \begin{bmatrix*}[r] -4 & -1 \\ -3 & -2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 + (-4) & 8 + (-1) \\ 4 + (-3) & 6 + (-2) \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -2 & 7 \\ 1 & 4 \end{bmatrix*}.

Hence, 2A + B = [−2714]\begin{bmatrix*}[r] -2 & 7 \\ 1 & 4 \end{bmatrix*}.

(ii) Given,

⇒C+B=[0000]⇒C=[0000]−B⇒C=[0000]−[−4−1−3−2]⇒C=[0−(−4)0−(−1)0−(−3)0−(−2)]⇒C=[4132].\Rightarrow C + B = \begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*} \\[1em] \Rightarrow C = \begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*} - B \\[1em] \Rightarrow C = \begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*} - \begin{bmatrix*}[r] -4 & -1 \\ -3 & -2 \end{bmatrix*} \\[1em] \Rightarrow C = \begin{bmatrix*}[r] 0 - (-4) & 0 - (-1) \\ 0 - (-3) & 0 - (-2) \end{bmatrix*} \\[1em] \Rightarrow C = \begin{bmatrix*}[r] 4 & 1 \\ 3 & 2 \end{bmatrix*}.

Hence, C = [4132]\begin{bmatrix*}[r] 4 & 1 \\ 3 & 2 \end{bmatrix*}.

Question 6

If 2[3x01]+3[13y2]=[z−7158]2\begin{bmatrix*}[r] 3 & x \\ 0 & 1 \end{bmatrix*} + 3\begin{bmatrix*}[r] 1 & 3 \\ y & 2 \end{bmatrix*} = \begin{bmatrix*}[r] z & -7 \\ 15 & 8 \end{bmatrix*}; find the values of x, y and z.

Answer

Given,

⇒2[3x01]+3[13y2]=[z−7158]⇒[62x02]+[393y6]=[z−7158]⇒[6+32x+90+3y2+6]=[z−7158]⇒[92x+93y8]=[z−7158]\Rightarrow 2\begin{bmatrix*}[r] 3 & x \\ 0 & 1 \end{bmatrix*} + 3\begin{bmatrix*}[r] 1 & 3 \\ y & 2 \end{bmatrix*} = \begin{bmatrix*}[r] z & -7 \\ 15 & 8 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 6 & 2x \\ 0 & 2 \end{bmatrix*} + \begin{bmatrix*}[r] 3 & 9 \\ 3y & 6 \end{bmatrix*} = \begin{bmatrix*}[r] z & -7 \\ 15 & 8 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 6 + 3 & 2x + 9 \\ 0 + 3y & 2 + 6 \end{bmatrix*} = \begin{bmatrix*}[r] z & -7 \\ 15 & 8 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 9 & 2x + 9 \\ 3y & 8 \end{bmatrix*} = \begin{bmatrix*}[r] z & -7 \\ 15 & 8 \end{bmatrix*}

By definition of equality of matrices we get,

z = 9,

2x + 9 = -7
⇒ 2x = -7 - 9
⇒ 2x = -16
⇒ x = -8,

3y = 15
⇒ y = 5.

Hence, x = -8, y = 5 and z = 9.

Question 7

Given A = [−360−9]\begin{bmatrix*}[r] -3 & 6 \\ 0 & -9 \end{bmatrix*} and At is its transpose matrix. Find :

(i) 2A + 3At

(ii) 2At - 3A

(iii) 12A−13At\dfrac{1}{2}A - \dfrac{1}{3}A^t

(iv) At−13AA^t - \dfrac{1}{3}A

Answer

A=[−360−9] and At=[−306−9]A = \begin{bmatrix*}[r] -3 & 6 \\ 0 & -9 \end{bmatrix*} \text{ and } A^t = \begin{bmatrix*}[r] -3 & 0 \\ 6 & -9 \end{bmatrix*}.

(i) Substituting value of A and At in 2A + 3At we get,

⇒2A+3At=2[−360−9]+3[−306−9]=[−6120−18]+[−9018−27]=[−6+(−9)12+00+18−18+(−27)]=[−151218−45].\Rightarrow 2A + 3A^t = 2\begin{bmatrix*}[r] -3 & 6 \\ 0 & -9 \end{bmatrix*} + 3\begin{bmatrix*}[r] -3 & 0 \\ 6 & -9 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -6 & 12 \\ 0 & -18 \end{bmatrix*} + \begin{bmatrix*}[r] -9 & 0 \\ 18 & -27 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -6 + (-9) & 12 + 0 \\ 0 + 18 & -18 + (-27) \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -15 & 12 \\ 18 & -45 \end{bmatrix*}.

Hence, 2A + 3At = [−151218−45].\begin{bmatrix*}[r] -15 & 12 \\ 18 & -45 \end{bmatrix*}.

(ii) Substituting value of A and At in 2At - 3A we get,

⇒2At−3A=2[−306−9]−3[−360−9]=[−6012−18]−[−9180−27]=[−6−(−9)0−1812−0−18−(−27)]=[3−18129].\Rightarrow 2A^t - 3A = 2\begin{bmatrix*}[r] -3 & 0 \\ 6 & -9 \end{bmatrix*} - 3\begin{bmatrix*}[r] -3 & 6 \\ 0 & -9 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -6 & 0 \\ 12 & -18 \end{bmatrix*} - \begin{bmatrix*}[r] -9 & 18 \\ 0 & -27 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -6 - (-9) & 0 - 18 \\ 12 - 0 & -18 - (-27) \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 3 & -18 \\ 12 & 9 \end{bmatrix*}.

Hence, 2At−3A=[3−18129].2A^t - 3A = \begin{bmatrix*}[r] 3 & -18 \\ 12 & 9 \end{bmatrix*}.

(iii) Substituting value of A and At in 12A−13At\dfrac{1}{2}A - \dfrac{1}{3}A^t we get,

⇒12A−13At=12[−360−9]−13[−306−9]=[−3230−92]−[−102−3]=[−32−(−1)3−00−2−92−(−3)]=[−32+13−2−92+3]=[−123−2−32]\Rightarrow \dfrac{1}{2}A - \dfrac{1}{3}A^t = \dfrac{1}{2}\begin{bmatrix*}[r] -3 & 6 \\ 0 & -9 \end{bmatrix*} - \dfrac{1}{3}\begin{bmatrix*}[r] -3 & 0 \\ 6 & -9 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -\dfrac{3}{2} & 3 \\ 0 & -\dfrac{9}{2} \end{bmatrix*} - \begin{bmatrix*}[r] -1 & 0 \\ 2 & -3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -\dfrac{3}{2} - (-1) & 3 - 0 \\ 0 - 2 & -\dfrac{9}{2} - (-3) \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -\dfrac{3}{2} + 1 & 3 \\ -2 & -\dfrac{9}{2} + 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -\dfrac{1}{2} & 3 \\ -2 & -\dfrac{3}{2} \end{bmatrix*}

Hence, 12A−13At=[−123−2−32].\dfrac{1}{2}A - \dfrac{1}{3}A^t = \begin{bmatrix*}[r] -\dfrac{1}{2} & 3 \\ -2 & -\dfrac{3}{2} \end{bmatrix*}.

(iv) Substituting value of A and At in At−13AA^t - \dfrac{1}{3}A we get,

⇒At−13A=[−306−9]−13[−360−9]=[−306−9]−[−120−3]=[−3−(−1)0−26−0−9−(−3)]=[−2−26−6].\Rightarrow A^t - \dfrac{1}{3}A = \begin{bmatrix*}[r] -3 & 0 \\ 6 & -9 \end{bmatrix*} - \dfrac{1}{3}\begin{bmatrix*}[r] -3 & 6 \\ 0 & -9 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -3 & 0 \\ 6 & -9 \end{bmatrix*} - \begin{bmatrix*}[r] -1 & 2 \\ 0 & -3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -3 - (-1) & 0 - 2 \\ 6 - 0 & -9 - (-3) \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -2 & -2 \\ 6 & -6 \end{bmatrix*}.

Hence, At−13A=[−2−26−6].A^t - \dfrac{1}{3}A = \begin{bmatrix*}[r] -2 & -2 \\ 6 & -6 \end{bmatrix*}.

Question 8

Given A = [11−20] and B=[2−111]\begin{bmatrix*}[r] 1 & 1 \\ -2 & 0 \end{bmatrix*} \text{ and } B = \begin{bmatrix*}[r] 2 & -1 \\ 1 & 1 \end{bmatrix*}.

Solve for matrix X :

(i) X + 2A = B

(ii) 3x + B + 2A = 0

(iii) 3A - 2X = X - 2B.

Answer

(i) Given,

⇒X+2A=B⇒X=B−2A⇒X=[2−111]−2[11−20]⇒X=[2−111]−[22−40]⇒X=[2−2−1−21−(−4)1−0]⇒X=[0−351]\Rightarrow X + 2A = B \\[1em] \Rightarrow X = B - 2A \\[1em] \Rightarrow X = \begin{bmatrix*}[r] 2 & -1 \\ 1 & 1 \end{bmatrix*} - 2\begin{bmatrix*}[r] 1 & 1 \\ -2 & 0 \end{bmatrix*} \\[1em] \Rightarrow X = \begin{bmatrix*}[r] 2 & -1 \\ 1 & 1 \end{bmatrix*} - \begin{bmatrix*}[r] 2 & 2 \\ -4 & 0 \end{bmatrix*} \\[1em] \Rightarrow X = \begin{bmatrix*}[r] 2 - 2 & -1 - 2 \\ 1 - (-4) & 1 - 0 \end{bmatrix*} \\[1em] \Rightarrow X = \begin{bmatrix*}[r] 0 & -3 \\ 5 & 1 \end{bmatrix*}

Hence, X = [0−351].\begin{bmatrix*}[r] 0 & -3 \\ 5 & 1 \end{bmatrix*}.

(ii) Given,

⇒3X+B+2A=0⇒3X=−(B+2A)⇒3X=−([2−111]+2[11−20])⇒3X=−([2−111]+[22−40])⇒3X=−([2+2−1+21+(−4)1+0])⇒3X=−([41−31])⇒X=−13([41−31])⇒X=[−43−131−13]\Rightarrow 3X + B + 2A = 0 \\[1em] \Rightarrow 3X = -(B + 2A) \\[1em] \Rightarrow 3X = -\Big(\begin{bmatrix*}[r] 2 & -1 \\ 1 & 1 \end{bmatrix*} + 2\begin{bmatrix*}[r] 1 & 1 \\ -2 & 0 \end{bmatrix*}\Big) \\[1em] \Rightarrow 3X = -\Big(\begin{bmatrix*}[r] 2 & -1 \\ 1 & 1 \end{bmatrix*} + \begin{bmatrix*}[r] 2 & 2 \\ -4 & 0 \end{bmatrix*}\Big) \\[1em] \Rightarrow 3X = -\Big(\begin{bmatrix*}[r] 2 + 2 & -1 + 2 \\ 1 + (-4) & 1 + 0 \end{bmatrix*}\Big) \\[1em] \Rightarrow 3X = -\Big(\begin{bmatrix*}[r] 4 & 1 \\ -3 & 1 \end{bmatrix*}\Big) \\[1em] \Rightarrow X = -\dfrac{1}{3}\Big(\begin{bmatrix*}[r] 4 & 1 \\ -3 & 1 \end{bmatrix*}\Big) \\[1em] \Rightarrow X = \begin{bmatrix*}[r] -\dfrac{4}{3} & -\dfrac{1}{3} \\ 1 & -\dfrac{1}{3} \end{bmatrix*}

Hence, X = [−43−131−13]\begin{bmatrix*}[r] -\dfrac{4}{3} & -\dfrac{1}{3} \\ 1 & -\dfrac{1}{3} \end{bmatrix*}.

(iii) Given,

⇒3A−2X=X−2B⇒X+2X=3A+2B⇒3X=3A+2B⇒3X=3[11−20]+2[2−111]⇒3X=[33−60]+[4−222]⇒3X=[3+43+(−2)−6+20+2]⇒3X=[71−42]⇒X=13[71−42]⇒X=[7313−4323].\Rightarrow 3A - 2X = X - 2B \\[1em] \Rightarrow X + 2X = 3A + 2B \\[1em] \Rightarrow 3X = 3A + 2B \\[1em] \Rightarrow 3X = 3\begin{bmatrix*}[r] 1 & 1 \\ -2 & 0 \end{bmatrix*} + 2\begin{bmatrix*}[r] 2 & -1 \\ 1 & 1 \end{bmatrix*} \\[1em] \Rightarrow 3X = \begin{bmatrix*}[r] 3 & 3 \\ -6 & 0 \end{bmatrix*} + \begin{bmatrix*}[r] 4 & -2 \\ 2 & 2 \end{bmatrix*} \\[1em] \Rightarrow 3X = \begin{bmatrix*}[r] 3 + 4 & 3 + (-2) \\ -6 + 2 & 0 + 2 \end{bmatrix*} \\[1em] \Rightarrow 3X = \begin{bmatrix*}[r] 7 & 1 \\ -4 & 2 \end{bmatrix*} \\[1em] \Rightarrow X = \dfrac{1}{3}\begin{bmatrix*}[r] 7 & 1 \\ -4 & 2 \end{bmatrix*} \\[1em] \Rightarrow X = \begin{bmatrix*}[r] \dfrac{7}{3} & \dfrac{1}{3} \\ -\dfrac{4}{3} & \dfrac{2}{3} \end{bmatrix*}.

Hence, X = [7313−4323].\begin{bmatrix*}[r] \dfrac{7}{3} & \dfrac{1}{3} \\ -\dfrac{4}{3} & \dfrac{2}{3} \end{bmatrix*}.

Question 9

If I is the unit matrix of order 2 × 2; find the matrix M such that :

5M + 3I = 4[2−50−3]4\begin{bmatrix*}[r] 2 & -5 \\ 0 & -3 \end{bmatrix*}

Answer

I = [1001]\begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*}

Given,

⇒5M+3I=4[2−50−3]⇒5M+3[1001]=[8−200−12]⇒5M+[3003]=[8−200−12]⇒5M=[8−200−12]−[3003]⇒5M=[8−3−20−00−0−12−3]⇒5M=[5−200−15]⇒M=15[5−200−15]⇒M=[1−40−3].\Rightarrow 5M + 3I = 4\begin{bmatrix*}[r] 2 & -5 \\ 0 & -3 \end{bmatrix*} \\[1em] \Rightarrow 5M + 3\begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} = \begin{bmatrix*}[r] 8 & -20 \\ 0 & -12 \end{bmatrix*} \\[1em] \Rightarrow 5M + \begin{bmatrix*}[r] 3 & 0 \\ 0 & 3 \end{bmatrix*} = \begin{bmatrix*}[r] 8 & -20 \\ 0 & -12 \end{bmatrix*} \\[1em] \Rightarrow 5M = \begin{bmatrix*}[r] 8 & -20 \\ 0 & -12 \end{bmatrix*} - \begin{bmatrix*}[r] 3 & 0 \\ 0 & 3 \end{bmatrix*} \\[1em] \\[1em] \Rightarrow 5M = \begin{bmatrix*}[r] 8 - 3 & -20 - 0 \\ 0 - 0 & -12 - 3 \end{bmatrix*} \\[1em] \Rightarrow 5M = \begin{bmatrix*}[r] 5 & -20 \\ 0 & -15 \end{bmatrix*} \\[1em] \Rightarrow M = \dfrac{1}{5}\begin{bmatrix*}[r] 5 & -20 \\ 0 & -15 \end{bmatrix*} \\[1em] \Rightarrow M = \begin{bmatrix*}[r] 1 & -4 \\ 0 & -3 \end{bmatrix*}.

Hence, M = [1−40−3].\begin{bmatrix*}[r] 1 & -4 \\ 0 & -3 \end{bmatrix*}.

Exercise 9(C)

Question 1(a)

If A is a matrix of order m × 3, B is a matrix of order 3 × 2 and R is a matrix of order 5 × n such that AB = R, the values of m and n are :

  1. m = -5 and n = -2

  2. m = 5 and n = 2

  3. m = 5 and n = -2

  4. m = 2 and n = 5

Answer

Given,

AB = R

Am × 3B3 × 2 = R5 × n

We know that,

For matrix multiplication :

No. of columns in A must be equal to the number of rows in B.

Resultant matrix order = No. of rows in A × No. of columns in B.

∴ m = 5 and n = 2.

Hence, Option 2 is the correct option.

Question 1(b)

If A = [4x01],B=[21201]\begin{bmatrix*}[r] 4 & x \\ 0 & 1 \end{bmatrix*}, B = \begin{bmatrix*}[r] 2 & 12 \\ 0 & 1 \end{bmatrix*} and A = B2, the value of x is :

  1. 38

  2. -6

  3. -36

  4. 36

Answer

Substituting value of A and B in A = B2, we get :

⇒[4x01]=[21201][21201]⇒[4x01]=[2×2+12×02×12+12×10×2+1×00×12+1×1]⇒[4x01]=[4+024+120+00+1]⇒[4x01]=[43601]\Rightarrow \begin{bmatrix*}[r] 4 & x \\ 0 & 1 \end{bmatrix*} = \begin{bmatrix*}[r] 2 & 12 \\ 0 & 1 \end{bmatrix*}\begin{bmatrix*}[r] 2 & 12 \\ 0 & 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 & x \\ 0 & 1 \end{bmatrix*} = \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*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 & x \\ 0 & 1 \end{bmatrix*} = \begin{bmatrix*}[r] 4 + 0 & 24 + 12 \\ 0 + 0 & 0 + 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 & x \\ 0 & 1 \end{bmatrix*} = \begin{bmatrix*}[r] 4 & 36 \\ 0 & 1 \end{bmatrix*}

∴ x = 36.

Hence, Option 4 is the correct option.

Question 1(c)

A, B and C are three matrices each of order 5; the order of matrix CA + B2 is :

  1. 5 × 4

  2. 5 × 5

  3. 4 × 5

  4. 5 × 3

Answer

Let resultant matrix of CA be R.

C5 × 5 × A5 × 5 = Rm × n

We know that,

For matrix multiplication :

No. of columns in C must be equal to the number of rows in A.

Resultant matrix order = No. of rows in C × No. of columns in A.

∴ m = 5 and n = 5.

∴ Order of matrix R will be 5 × 5.

Let resultant matrix of B2 be S.

B5 × 5 × B5 × 5 = Sg × h

Resultant matrix order = No. of rows in B × No. of columns in B.

∴ g = 5 and h = 5.

∴ Order of matrix S will be 5 × 5.

⇒ CA + B2

⇒ R + S

Addition of matrix possible between matrix of same order and same is the order of resultant matrix.

Hence, resultant matrix's order = 5 × 5

Hence, Option 2 is the correct option.

Question 1(d)

If A = [5−270] and B=[83]\begin{bmatrix*}[r] 5 & -2 \\ 7 & 0 \end{bmatrix*} \text{ and B} = \begin{bmatrix*}[r] 8 \\ 3 \end{bmatrix*}, then which of the following is not possible ?

  1. A2

  2. AB

  3. BA

  4. 15A

Answer

BA is not possible because no. of columns in B (1) is not equal to the no. of rows in A (2).

Hence, Option 3 is the correct option.

Question 1(e)

If A = [1011],B=[0110] and C=[1100]\begin{bmatrix*}[r] 1 & 0 \\ 1 & 1 \end{bmatrix*}, B = \begin{bmatrix*}[r] 0 & 1 \\ 1 & 0 \end{bmatrix*}\text{ and C} = \begin{bmatrix*}[r] 1 & 1 \\ 0 & 0 \end{bmatrix*}, the matrix A2 + 2B - 3C is :

  1. [−2−141]\begin{bmatrix*}[r] -2 & -1 \\ 4 & 1 \end{bmatrix*}

  2. [2−141]\begin{bmatrix*}[r] 2 & -1 \\ 4 & 1 \end{bmatrix*}

  3. [2141]\begin{bmatrix*}[r] 2 & 1 \\ 4 & 1 \end{bmatrix*}

  4. [21−4−1]\begin{bmatrix*}[r] 2 & 1 \\ -4 & -1 \end{bmatrix*}

Answer

Substituting values of A, B and C in A2 + 2B - 3C, we get :

⇒A2+2B−3C=[1011][1011]+2[0110]−3[1100]=[1×1+0×11×0+0×11×1+1×11×0+1×1]+[0220]−[3300]=[1+00+01+10+1]+[0220]−[3300]=[1021]+[0220]−[3300]=[1+0−30+2−32+2−01+0−0]=[−2−141]\Rightarrow A^2 + 2B - 3C = \begin{bmatrix*}[r] 1 & 0 \\ 1 & 1 \end{bmatrix*}\begin{bmatrix*}[r] 1 & 0 \\ 1 & 1 \end{bmatrix*} + 2\begin{bmatrix*}[r] 0 & 1 \\ 1 & 0 \end{bmatrix*} - 3\begin{bmatrix*}[r] 1 & 1 \\ 0 & 0 \end{bmatrix*} \\[1em] = \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*} + \begin{bmatrix*}[r] 0 & 2 \\ 2 & 0 \end{bmatrix*} - \begin{bmatrix*}[r] 3 & 3 \\ 0 & 0 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 + 0 & 0 + 0 \\ 1 + 1 & 0 + 1 \end{bmatrix*} + \begin{bmatrix*}[r] 0 & 2 \\ 2 & 0 \end{bmatrix*} - \begin{bmatrix*}[r] 3 & 3 \\ 0 & 0 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 & 0 \\ 2 & 1 \end{bmatrix*} + \begin{bmatrix*}[r] 0 & 2 \\ 2 & 0 \end{bmatrix*} - \begin{bmatrix*}[r] 3 & 3 \\ 0 & 0 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 + 0 - 3 & 0 + 2 - 3 \\ 2 + 2 - 0 & 1 + 0 - 0 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -2 & -1 \\ 4 & 1 \end{bmatrix*}

Hence, Option 1 is the correct option.

Question 2(i)

Evaluate : if possible :

[32][20]\begin{bmatrix*}[r] 3 & 2 \\ \end{bmatrix*}\begin{bmatrix*}[r] 2 \\ 0 \end{bmatrix*}

Answer

[32][20]\begin{bmatrix*}[r] 3 & 2 \\ \end{bmatrix*}\begin{bmatrix*}[r] 2 \\ 0 \end{bmatrix*}

For matrix multiplication, the no. of columns in first matrix should be equal to no. of rows in the second matrix.

Hence, matrix multiplication is possible.

⇒[32][20]=[3×2+2×0]=[6].\Rightarrow \begin{bmatrix*}[r] 3 & 2 \\ \end{bmatrix*}\begin{bmatrix*}[r] 2 \\ 0 \end{bmatrix*} = \begin{bmatrix*}[r] 3 \times 2 + 2 \times 0 \\ \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 6 \\ \end{bmatrix*}.

Hence, [32][20]=[6].\begin{bmatrix*}[r] 3 & 2 \\ \end{bmatrix*}\begin{bmatrix*}[r] 2 \\ 0 \end{bmatrix*} = \begin{bmatrix*}[r] 6 \end{bmatrix*}.

Question 2(ii)

Evaluate : if possible :

[1−2][−23−14]\begin{bmatrix*}[r] 1 & -2 \\ \end{bmatrix*}\begin{bmatrix*}[r] -2 & 3 \\ -1 & 4 \end{bmatrix*}

Answer

[1−2][−23−14]\begin{bmatrix*}[r] 1 & -2 \\ \end{bmatrix*}\begin{bmatrix*}[r] -2 & 3 \\ -1 & 4 \end{bmatrix*}

For matrix multiplication, the no. of columns in first matrix should be equal to no. of rows in the second matrix.

⇒[1−2][−23−14]=[1×(−2)+(−2)×(−1)1×3+(−2)×4]=[−2+23+(−8)]=[0−5].\Rightarrow \begin{bmatrix*}[r] 1 & -2 \\ \end{bmatrix*}\begin{bmatrix*}[r] -2 & 3 \\ -1 & 4 \end{bmatrix*} = \begin{bmatrix*}[r] 1 \times (-2) + (-2) \times (-1) & 1 \times 3 + (-2) \times 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -2 + 2 & 3 + (-8) \\ \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 0 & -5 \\ \end{bmatrix*}.

Hence, [1−2][−23−14]=[0−5].\begin{bmatrix*}[r] 1 & -2 \\ \end{bmatrix*}\begin{bmatrix*}[r] -2 & 3 \\ -1 & 4 \end{bmatrix*} = \begin{bmatrix*}[r] 0 & -5 \\ \end{bmatrix*}.

Question 2(iii)

Evaluate : if possible :

[643−1][−13]\begin{bmatrix*}[r] 6 & 4 \\ 3 & -1 \end{bmatrix*}\begin{bmatrix*}[r] -1 \\ 3 \end{bmatrix*}

Answer

[643−1][−13]\begin{bmatrix*}[r] 6 & 4 \\ 3 & -1 \end{bmatrix*}\begin{bmatrix*}[r] -1 \\ 3 \end{bmatrix*}

For matrix multiplication, the no. of columns in first matrix should be equal to no. of rows in the second matrix.

⇒[643−1][−13]=[6×(−1)+4×33×(−1)+(−1)×3]=[−6+12−3+(−3)]=[6−6].\Rightarrow \begin{bmatrix*}[r] 6 & 4 \\ 3 & -1 \end{bmatrix*}\begin{bmatrix*}[r] -1 \\ 3 \end{bmatrix*} = \begin{bmatrix*}[r] 6 \times (-1) + 4 \times 3 \\ 3 \times (-1) + (-1) \times 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -6 + 12 \\ -3 + (-3) \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 6 \\ -6 \end{bmatrix*}.

Hence, [643−1][−13]=[6−6].\begin{bmatrix*}[r] 6 & 4 \\ 3 & -1 \end{bmatrix*}\begin{bmatrix*}[r] -1 \\ 3 \end{bmatrix*} = \begin{bmatrix*}[r] 6 \\ -6 \end{bmatrix*}.

Question 2(iv)

Evaluate : if possible :

[643−1][−13]\begin{bmatrix*}[r] 6 & 4 \\ 3 & -1 \end{bmatrix*}\begin{bmatrix*}[r] -1 & 3 \\ \end{bmatrix*}

Answer

[643−1][−13]\begin{bmatrix*}[r] 6 & 4 \\ 3 & -1 \end{bmatrix*}\begin{bmatrix*}[r] -1 & 3 \\ \end{bmatrix*}

For matrix multiplication, the no. of columns in first matrix should be equal to no. of rows in the second matrix.

The above matrix multiplication is not possible as the no. of columns in first matrix is not equal to no. of rows in the second matrix.

Question 3

If A = [025−2],B=[1−132]\begin{bmatrix*}[r] 0 & 2 \\ 5 & -2 \end{bmatrix*}, B = \begin{bmatrix*}[r] 1 & -1 \\ 3 & 2 \end{bmatrix*} and I is a unit matrix of order 2 × 2, find :

(i) AB

(ii) BA

(iii) AI

Answer

(i) Substituting value in AB,

⇒AB=[025−2][1−132]⇒[0×1+2×30×(−1)+2×25×1+(−2)×35×(−1)+(−2)×2]⇒[0+60+45−6−5−4]⇒[64−1−9].\Rightarrow AB = \begin{bmatrix*}[r] 0 & 2 \\ 5 & -2 \end{bmatrix*}\begin{bmatrix*}[r] 1 & -1 \\ 3 & 2 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 0 \times 1 + 2 \times 3 & 0 \times (-1) + 2 \times 2 \\ 5 \times 1 + (-2) \times 3 & 5 \times (-1) + (-2) \times 2 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 0 + 6 & 0 + 4 \\ 5 - 6 & -5 - 4 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 6 & 4 \\ -1 & -9 \end{bmatrix*}.

Hence, AB = [64−1−9].\begin{bmatrix*}[r] 6 & 4 \\ -1 & -9 \end{bmatrix*}.

(ii) Substituting value in BA,

⇒BA=[1−132][025−2]=[1×0+(−1)×51×2+(−1)×(−2)3×0+2×53×2+2×(−2)]=[0−52+20+106−4]=[−54102].\Rightarrow BA = \begin{bmatrix*}[r] 1 & -1 \\ 3 & 2 \end{bmatrix*}\begin{bmatrix*}[r] 0 & 2 \\ 5 & -2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 \times 0 + (-1) \times 5 & 1 \times 2 + (-1) \times (-2) \\ 3 \times 0 + 2 \times 5 & 3 \times 2 + 2 \times (-2) \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 0 - 5 & 2 + 2 \\ 0 + 10 & 6 - 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -5 & 4 \\ 10 & 2 \end{bmatrix*}.

Hence, BA = [−54102].\begin{bmatrix*}[r] -5 & 4 \\ 10 & 2 \end{bmatrix*}.

(iii) Substituting value in AI,

⇒AI=[025−2][1001]=[0×1+2×00×0+2×15×1+(−2)×05×0+(−2)×1]=[025−2]=A\Rightarrow AI = \begin{bmatrix*}[r] 0 & 2 \\ 5 & -2 \end{bmatrix*}\begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 0 \times 1 + 2 \times 0 & 0 \times 0 + 2 \times 1 \\ 5 \times 1 + (-2) \times 0 & 5 \times 0 + (-2) \times 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 0 & 2 \\ 5 & -2 \end{bmatrix*} = A

Hence, AI = matrix A.

Question 4

If A = [3x01] and B =[9160−y]\begin{bmatrix*}[r] 3 & x \\ 0 & 1 \end{bmatrix*} \text{ and B }= \begin{bmatrix*}[r] 9 & 16 \\ 0 & -y \end{bmatrix*}, find x and y when A2 = B.

Answer

Given,

⇒A2=B⇒[3x01][3x01]=[9160−y]⇒[3×3+x×03×x+x×10×3+1×00×x+1×1]=[9160−y]⇒[9+03x+x0+00+1]=[9160−y]⇒[94x01]=[9160−y]\Rightarrow A^2 = B \\[1em] \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*}

By definition of equality of matrices we get,

4x = 16
⇒ x = 4.

1 = -y
⇒ y = -1.

Hence, x = 4 and y = -1.

Question 5

Find x and y, if :

[x0−31][110y]=[22−3−2]\begin{bmatrix*}[r] x & 0 \\ -3 & 1 \end{bmatrix*}\begin{bmatrix*}[r] 1 & 1 \\ 0 & y \end{bmatrix*} = \begin{bmatrix*}[r] 2 & 2 \\ -3 & -2 \end{bmatrix*}

Answer

Given,

⇒[x0−31][110y]=[22−3−2]⇒[x×1+0×0x×1+0×y−3×1+1×0−3×1+1×y]=[22−3−2]⇒[x+0x+0−3+0−3+y]=[22−3−2]⇒[xx−3−3+y]=[22−3−2]\Rightarrow \begin{bmatrix*}[r] x & 0 \\ -3 & 1 \end{bmatrix*}\begin{bmatrix*}[r] 1 & 1 \\ 0 & y \end{bmatrix*} = \begin{bmatrix*}[r] 2 & 2 \\ -3 & -2 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] x \times 1 + 0 \times 0 & x \times 1 + 0 \times y \\ -3 \times 1 + 1 \times 0 & -3 \times 1 + 1 \times y \end{bmatrix*} = \begin{bmatrix*}[r] 2 & 2 \\ -3 & -2 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] x + 0 & x + 0 \\ -3 + 0 & -3 + y \end{bmatrix*} = \begin{bmatrix*}[r] 2 & 2 \\ -3 & -2 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] x & x \\ -3 & -3 + y \end{bmatrix*} = \begin{bmatrix*}[r] 2 & 2 \\ -3 & -2 \end{bmatrix*}

By definition of equality of matrices we get,

x = 2

-3 + y = -2
⇒ y = -2 + 3
⇒ y = 1.

Hence, x = 2 and y = 1.

Question 6

If A = [1324],B=[1243] and C=[4312]\begin{bmatrix*}[r] 1 & 3 \\ 2 & 4 \end{bmatrix*}, B = \begin{bmatrix*}[r] 1 & 2 \\ 4 & 3 \end{bmatrix*} \text{ and } C = \begin{bmatrix*}[r] 4 & 3 \\ 1 & 2 \end{bmatrix*}, find :

(i) (AB)C

(ii) A(BC)

Is A(BC) = (AB)C ?

Answer

(i) Substituting value of A, B and C in (AB)C we get,

⇒([1324][1243])[4312]=([1×1+3×41×2+3×32×1+4×42×2+4×3])[4312]=([1+122+92+164+12])[4312]=[13111816][4312]=[13×4+11×113×3+11×218×4+16×118×3+16×2]=[52+1139+2272+1654+32]=[63618886].\Rightarrow \Big(\begin{bmatrix*}[r] 1 & 3 \\ 2 & 4 \end{bmatrix*}\begin{bmatrix*}[r] 1 & 2 \\ 4 & 3 \end{bmatrix*}\Big)\begin{bmatrix*}[r] 4 & 3 \\ 1 & 2 \end{bmatrix*} \\[1em] = \Big(\begin{bmatrix*}[r] 1 \times 1 + 3 \times 4 & 1 \times 2 + 3 \times 3 \\ 2 \times 1 + 4 \times 4 & 2 \times 2 + 4 \times 3 \end{bmatrix*}\Big)\begin{bmatrix*}[r] 4 & 3 \\ 1 & 2 \end{bmatrix*} \\[1em] = \Big(\begin{bmatrix*}[r] 1 + 12 & 2 + 9 \\ 2 + 16 & 4 + 12 \end{bmatrix*}\Big)\begin{bmatrix*}[r] 4 & 3 \\ 1 & 2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 13 & 11 \\ 18 & 16 \end{bmatrix*}\begin{bmatrix*}[r] 4 & 3 \\ 1 & 2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 13 \times 4 + 11 \times 1 & 13 \times 3 + 11 \times 2 \\ 18 \times 4 + 16 \times 1 & 18 \times 3 + 16 \times 2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 52 + 11 & 39 + 22 \\ 72 + 16 & 54 + 32 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 63 & 61 \\ 88 & 86 \end{bmatrix*}.

Hence, (AB)C = [63618886].\begin{bmatrix*}[r] 63 & 61 \\ 88 & 86 \end{bmatrix*}.

(ii) Substituting value of A, B and C in A(BC) we get,

⇒[1324]([1243][4312])=[1324]([1×4+2×11×3+2×24×4+3×14×3+3×2])=[1324][4+23+416+312+6]=[1324][671918]=[1×6+3×191×7+3×182×6+4×192×7+4×18]=[6+577+5412+7614+72]=[63618886]\Rightarrow \begin{bmatrix*}[r] 1 & 3 \\ 2 & 4 \end{bmatrix*}\Big(\begin{bmatrix*}[r] 1 & 2 \\ 4 & 3 \end{bmatrix*}\begin{bmatrix*}[r] 4 & 3 \\ 1 & 2 \end{bmatrix*}\Big) \\[1em] = \begin{bmatrix*}[r] 1 & 3 \\ 2 & 4 \end{bmatrix*}\Big(\begin{bmatrix*}[r] 1 \times 4 + 2\times 1 & 1 \times 3 + 2 \times 2 \\ 4 \times 4 + 3 \times 1 & 4 \times 3 + 3 \times 2 \end{bmatrix*}\Big) \\[1em] = \begin{bmatrix*}[r] 1 & 3 \\ 2 & 4 \end{bmatrix*}\begin{bmatrix*}[r] 4 + 2 & 3 + 4 \\ 16 + 3 & 12 + 6 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 & 3 \\ 2 & 4 \end{bmatrix*}\begin{bmatrix*}[r] 6 & 7 \\ 19 & 18 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 \times 6 + 3 \times 19 & 1 \times 7 + 3 \times 18 \\ 2 \times 6 + 4 \times 19 & 2 \times 7 + 4 \times 18 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 6 + 57 & 7 + 54 \\ 12 + 76 & 14 + 72 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 63 & 61 \\ 88 & 86 \end{bmatrix*}

Hence, A(BC) = [63618886]\begin{bmatrix*}[r] 63 & 61 \\ 88 & 86 \end{bmatrix*}.

Hence, A(BC) = (AB)C.

Question 7

Let A = [210−2],B=[41−3−2] and C =[−32−14]\begin{bmatrix*}[r] 2 & 1 \\ 0 & -2 \end{bmatrix*}, B = \begin{bmatrix*}[r] 4 & 1 \\ -3 & -2 \end{bmatrix*} \text{ and C }= \begin{bmatrix*}[r] -3 & 2 \\ -1 & 4 \end{bmatrix*}. Find A2 + AC - 5B.

Answer

Substituting values of A, B and C in A2 + AC - 5B we get,

⇒[210−2][210−2]+[210−2][−32−14]−5[41−3−2]=[2×2+1×02×1+1×(−2)0×2+(−2)×00×1+(−2)×(−2)]+[2×(−3)+1×(−1)2×2+1×40×(−3)+(−2)×(−1)0×2+(−2)×4]−[205−15−10]=[4+02−20+00+4]+[−6−14+40+20−8]−[205−15−10]=[4004]+[−782−8]−[205−15−10]=[4+(−7)−200+8−50+2−(−15)4+(−8)−(−10)]=[−233176]\Rightarrow \begin{bmatrix*}[r] 2 & 1 \\ 0 & -2 \end{bmatrix*}\begin{bmatrix*}[r] 2 & 1 \\ 0 & -2 \end{bmatrix*} + \begin{bmatrix*}[r] 2 & 1 \\ 0 & -2 \end{bmatrix*}\begin{bmatrix*}[r] -3 & 2 \\ -1 & 4 \end{bmatrix*} - 5\begin{bmatrix*}[r] 4 & 1 \\ -3 & -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] 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] 20 & 5 \\ -15 & -10 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 4 + 0 & 2 - 2 \\ 0 + 0 & 0 + 4 \end{bmatrix*} + \begin{bmatrix*}[r] -6 - 1 & 4 + 4 \\ 0 + 2 & 0 - 8 \end{bmatrix*} - \begin{bmatrix*}[r] 20 & 5 \\ -15 & -10 \end{bmatrix*} \\[1em] = \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*}

Hence, A2 + AC - 5B = [−233176]\begin{bmatrix*}[r] -23 & 3 \\ 17 & 6 \end{bmatrix*}.

Question 8

If M = [1221]\begin{bmatrix*}[r] 1 & 2 \\ 2 & 1 \end{bmatrix*} and I is a unit matrix of the same order as that of M; show that :

M2 = 2M + 3I

Answer

I = [1001]\begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*}

M = [1221]\begin{bmatrix*}[r] 1 & 2 \\ 2 & 1 \end{bmatrix*}

L.H.S. =M2=[1221][1221]=[1×1+2×21×2+2×12×1+1×22×2+1×1]=[1+42+22+24+1]=[5445]R.H.S.=2M+3I=2[1221]+3[1001]=[2442]+[3003]=[2+34+04+02+3]=[5445].\text{L.H.S. }= M^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] 5 & 4 \\ 4 & 5 \end{bmatrix*} \\[1em] \text{R.H.S.} = 2M + 3I \\[1em] = 2\begin{bmatrix*}[r] 1 & 2 \\ 2 & 1 \end{bmatrix*} + 3\begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 & 4 \\ 4 & 2 \end{bmatrix*} + \begin{bmatrix*}[r] 3 & 0 \\ 0 & 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 + 3 & 4 + 0 \\ 4 + 0 & 2 + 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 5 & 4 \\ 4 & 5 \end{bmatrix*}.

Since, L.H.S. = R.H.S.

Hence, proved that M2 = 2M + 3I.

Question 9

If A = [a002],B=[0−b10],M=[1−111]\begin{bmatrix*}[r] a & 0 \\ 0 & 2 \end{bmatrix*}, B = \begin{bmatrix*}[r] 0 & -b \\ 1 & 0 \end{bmatrix*}, M = \begin{bmatrix*}[r] 1 & -1 \\ 1 & 1 \end{bmatrix*} and BA = M2, find the values of a and b.

Answer

Given,

⇒BA=M2⇒[0−b10][a002]=[1−111][1−111]⇒[0×a+(−b)×00×0+(−b)×21×a+0×01×0+0×2]=[1×1+(−1)×11×(−1)+(−1)×11×1+1×11×(−1)+1×1]⇒[00−2ba+00]=[1−1−1−11+1−1+1]⇒[0−2ba0]=[0−220]\Rightarrow BA = M^2 \\[1em] \Rightarrow \begin{bmatrix*}[r] 0 & -b \\ 1 & 0 \end{bmatrix*}\begin{bmatrix*}[r] a & 0 \\ 0 & 2 \end{bmatrix*} = \begin{bmatrix*}[r] 1 & -1 \\ 1 & 1 \end{bmatrix*}\begin{bmatrix*}[r] 1 & -1 \\ 1 & 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 0 \times a + (-b) \times 0 & 0 \times 0 + (-b) \times 2 \\ 1 \times a + 0 \times 0 & 1 \times 0 + 0 \times 2 \end{bmatrix*} = \begin{bmatrix*}[r] 1 \times 1 + (-1) \times 1 & 1 \times (-1) + (-1) \times 1 \\ 1 \times 1 + 1 \times 1 & 1 \times (-1) + 1 \times 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 0 & 0 - 2b \\ a + 0 & 0 \end{bmatrix*} = \begin{bmatrix*}[r] 1 - 1 & -1 - 1 \\ 1 + 1 & -1 + 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 0 & -2b \\ a & 0 \end{bmatrix*} = \begin{bmatrix*}[r] 0 & -2 \\ 2 & 0 \end{bmatrix*}

By definition of equality of matrices we get,

-2b = -2
⇒ b = 1.

a = 2.

Hence, a = 2 and b = 1.

Question 10

Find the matrix A, if B = [2101] and B2=B+12A\begin{bmatrix*}[r] 2 & 1 \\ 0 & 1 \end{bmatrix*} \text{ and } B^2 = B + \dfrac{1}{2}A.

Answer

Given,

⇒B2=B+12A\Rightarrow B^2 = B + \dfrac{1}{2}A

Substituting value of B in above equation we get,

⇒[2101][2101]=[2101]+12A⇒[2×2+1×02×1+1×1 0×2+1×00×1+1×1]=[2101]+12A⇒[4+02+100+1]=[2101]+12A⇒[4301]=[2101]+12A⇒12A=[4301]−[2101]⇒12A=[4−23−10−01−1]⇒12A=[2200]⇒A=2[2200]⇒A=[4400].\Rightarrow \begin{bmatrix*}[r] 2 & 1 \\ 0 & 1 \end{bmatrix*}\begin{bmatrix*}[r] 2 & 1 \\ 0 & 1 \end{bmatrix*} = \begin{bmatrix*}[r] 2 & 1 \\ 0 & 1 \end{bmatrix*} + \dfrac{1}{2}A \\[1em] \Rightarrow \begin{bmatrix*}[r] 2 \times 2 + 1 \times 0 & 2 \times 1 + 1 \times 1\ 0 \times 2 + 1 \times 0 & 0 \times 1 + 1 \times 1 \end{bmatrix*} = \begin{bmatrix*}[r] 2 & 1 \\ 0 & 1 \end{bmatrix*} + \dfrac{1}{2}A \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 + 0 & 2 + 1 \\ 0 & 0 + 1 \end{bmatrix*} = \begin{bmatrix*}[r] 2 & 1 \\ 0 & 1 \end{bmatrix*} + \dfrac{1}{2}A \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 & 3 \\ 0 & 1 \end{bmatrix*} = \begin{bmatrix*}[r] 2 & 1 \\ 0 & 1 \end{bmatrix*} + \dfrac{1}{2}A \\[1em] \Rightarrow \dfrac{1}{2}A = \begin{bmatrix*}[r] 4 & 3 \\ 0 & 1 \end{bmatrix*} - \begin{bmatrix*}[r] 2 & 1 \\ 0 & 1 \end{bmatrix*} \\[1em] \Rightarrow \dfrac{1}{2}A = \begin{bmatrix*}[r] 4 - 2 & 3 - 1 \\ 0 - 0 & 1 - 1 \end{bmatrix*} \\[1em] \Rightarrow \dfrac{1}{2}A = \begin{bmatrix*}[r] 2 & 2 \\ 0 & 0 \end{bmatrix*} \\[1em] \Rightarrow A = 2\begin{bmatrix*}[r] 2 & 2 \\ 0 & 0 \end{bmatrix*} \\[1em] \Rightarrow A = \begin{bmatrix*}[r] 4 & 4 \\ 0 & 0 \end{bmatrix*}.

Hence, A = [4400].\begin{bmatrix*}[r] 4 & 4 \\ 0 & 0 \end{bmatrix*}.

Question 11

If A = [−11ab]\begin{bmatrix*}[r] -1 & 1 \\ a & b \end{bmatrix*} and A2 = I, find a and b.

Answer

Given, A2 = I

∴[−11ab][−11ab]=[1001]⇒[−1×−1+1×a−1×1+1×ba×(−1)+b×aa×1+b×b]=[1001]⇒[1+a−1+b−a+aba+b2]=[1001]\therefore \begin{bmatrix*}[r] -1 & 1 \\ a & b \end{bmatrix*}\begin{bmatrix*}[r] -1 & 1 \\ a & b \end{bmatrix*} = \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] -1 \times -1 + 1 \times a & -1 \times 1 + 1 \times b \\ a \times (-1) + b \times a & a \times 1 + b \times b \end{bmatrix*} = \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 1 + a & -1 + b \\ -a + ab & a + b^2 \end{bmatrix*} = \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*}

By definition of equality of matrices we get,

1 + a = 1
⇒ a = 0.

-1 + b = 0
⇒ b = 1.

Hence, a = 0 and b = 1

Question 12(i)

Solve for x and y :

[2552][xy]=[−714]\begin{bmatrix*}[r] 2 & 5 \\ 5 & 2 \end{bmatrix*}\begin{bmatrix*}[r] x \\ y \end{bmatrix*} = \begin{bmatrix*}[r] -7 \\ 14 \end{bmatrix*}

Answer

Given,

⇒[2552][xy]=[−714]⇒[2×x+5×y5×x+2×y]=[−714]⇒[2x+5y5x+2y]=[−714]\Rightarrow \begin{bmatrix*}[r] 2 & 5 \\ 5 & 2 \end{bmatrix*}\begin{bmatrix*}[r] x \\ y \end{bmatrix*} = \begin{bmatrix*}[r] -7 \\ 14 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2 \times x + 5 \times y \\ 5 \times x + 2 \times y \end{bmatrix*} = \begin{bmatrix*}[r] -7 \\ 14 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2x + 5y \\ 5x + 2y \end{bmatrix*} = \begin{bmatrix*}[r] -7 \\ 14 \end{bmatrix*}

By definition of equality of matrices we get,

2x + 5y = -7

⇒ 2x = -(7 + 5y)

⇒ x = −(7+5y)2\dfrac{-(7 + 5y)}{2} ......(i)

5x + 2y = 14

Substituting value of x from (i) in above equation we get,

⇒5×−(7+5y)2+2y=14⇒−35−25y2+2y=14⇒−35−25y+4y2=14⇒−35−21y=28⇒−21y=28+35⇒−21y=63⇒y=−3.\Rightarrow 5 \times \dfrac{-(7 + 5y)}{2} + 2y = 14 \\[1em] \Rightarrow \dfrac{-35 - 25y}{2} + 2y = 14 \\[1em] \Rightarrow \dfrac{-35 - 25y + 4y}{2} = 14 \\[1em] \Rightarrow -35 - 21y = 28 \\[1em] \Rightarrow -21y = 28 + 35 \\[1em] \Rightarrow -21y = 63 \\[1em] \Rightarrow y = -3.

Substituting y = -3 in (i) we get,

x=−(7+5(−3))2=−(7−15)2=82=4x = \dfrac{-(7 + 5(-3))}{2} = \dfrac{-(7 - 15)}{2} = \dfrac{8}{2} = 4.

Hence, x = 4 and y = -3.

Question 12(ii)

Solve for x and y :

[x+yx−4][−1−222]=[−7−11]\begin{bmatrix*}[r] x + y & x - 4 \end{bmatrix*}\begin{bmatrix*}[r] -1 & -2 \\ 2 & 2 \end{bmatrix*} = \begin{bmatrix*}[r] -7 & -11 \end{bmatrix*}

Answer

Given,

⇒[x+yx−4][−1−222]=[−7−11]⇒[(x+y)×−1+(x−4)×2(x+y)×−2+(x−4)×2]=[−7−11]⇒[−x−y+2x−8−2x−2y+2x−8]=[−7−11]⇒[x−y−8−2y−8]=[−7−11]\Rightarrow \begin{bmatrix*}[r] x + y & x - 4 \end{bmatrix*}\begin{bmatrix*}[r] -1 & -2 \\ 2 & 2 \end{bmatrix*} = \begin{bmatrix*}[r] -7 & -11 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] (x + y) \times -1 + (x - 4) \times 2 & (x + y) \times -2 + (x - 4) \times 2 \end{bmatrix*} = \begin{bmatrix*}[r] -7 & -11 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] -x - y + 2x - 8 & -2x - 2y + 2x - 8 \end{bmatrix*} = \begin{bmatrix*}[r] -7 & -11 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] x - y - 8 & - 2y - 8 \end{bmatrix*} = \begin{bmatrix*}[r] -7 & -11 \end{bmatrix*}

By definition of equality of matrices we get,

-2y - 8 = -11

⇒ -2y = -3

⇒ y = 32\dfrac{3}{2}.

x - y - 8 = -7

⇒ x - y = 1

⇒ x = 1 + y

⇒ x = 1+32=521 + \dfrac{3}{2} = \dfrac{5}{2}.

Hence, x=52,y=32.x = \dfrac{5}{2}, y = \dfrac{3}{2}.

Question 12(iii)

Solve for x and y :

[−2031][−12x]+3[−21]=2[y3]\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*}.

Answer

Given,

⇒[−2031][−12x]+3[−21]=2[y3]⇒[−2×−1+0×2x3×(−1)+1×2x]+[−63]=[2y6]⇒[2−3+2x]+[−63]=[2y6][2+(−6)−3+2x+3]=[2y6][−42x]=[2y6]\Rightarrow \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] \begin{bmatrix*}[r] 2 + (-6) \\ -3 + 2x + 3 \end{bmatrix*} = \begin{bmatrix*}[r] 2y \\ 6 \end{bmatrix*} \\[1em] \begin{bmatrix*}[r] -4 \\ 2x \end{bmatrix*} = \begin{bmatrix*}[r] 2y \\ 6 \end{bmatrix*}

By definition of equality of matrices we get,

2y = -4
⇒ y = -2.

2x = 6
⇒ x = 3.

Hence, x = 3 and y = -2.

Question 13

In each case given below, find :

(a) the order of matrix M.

(b) the matrix M.

(i) M×[1102]=[12]M \times \begin{bmatrix*}[r] 1 & 1 \\ 0 & 2 \end{bmatrix*} =\begin{bmatrix*}[r] 1 & 2 \end{bmatrix*}

(ii) [1421]×M=[135]\begin{bmatrix*}[r] 1 & 4 \\ 2 & 1 \end{bmatrix*} \times M = \begin{bmatrix*}[r] 13 \\ 5 \end{bmatrix*}

Answer

(i) Let order of matrix M be a × b.

Ma×b×[1102]2×2=[12]1×2M_{a \times b} \times \begin{bmatrix*}[r] 1 & 1 \\ 0 & 2 \end{bmatrix*}_{2 \times 2} = \begin{bmatrix*}[r] 1 & 2 \end{bmatrix*}_{1 \times 2}

Since, the product of matrices is possible, only when the number of columns in the first matrix is equal to the number of rows in the second.

∴ b = 2

Also, the no. of rows of product (resulting) matrix is equal to no. of rows of first matrix.

∴ a = 1

Order of matrix M = a × b = 1 × 2.

Let M = [xy]\begin{bmatrix*}[r] x & y \end{bmatrix*}.

⇒[xy]×[1102]=[12]⇒[x×1+y×0x×1+y×2]=[12]⇒[xx+2y]=[12]\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*}

By definition of equality of matrices we get,

x = 1

x + 2y = 2
⇒ 1 + 2y = 2
⇒ 2y = 1
⇒ y = 12\dfrac{1}{2}.

∴ M = [xy]=[112]\begin{bmatrix*}[r] x & y \end{bmatrix*} = \begin{bmatrix*}[r] 1 & \dfrac{1}{2} \end{bmatrix*}.

Hence, M = [112].\begin{bmatrix*}[r] 1 & \dfrac{1}{2} \end{bmatrix*}.

(ii) Let order of matrix M be a × b.

i.e. [1421]2×2×Ma×b=[135]2×1\begin{bmatrix*}[r] 1 & 4 \\ 2 & 1 \end{bmatrix*}_{2 \times 2} \times M_{a \times b} = \begin{bmatrix*}[r] 13 \\ 5 \end{bmatrix*}_{2 \times 1}

Since product of matrix is possible, only when the number of columns in the first matrix is equal to no. of rows in second.

∴ a = 2.

Also the no. of columns of product (resulting matrix) is equal to no. of columns of second matrix.

∴ b = 1.

Hence, order of matrix = 2 × 1.

Let M = [xy]\begin{bmatrix*}[r] x \\ y \end{bmatrix*}

⇒[1421]×[xy]=[135]⇒[1×x+4×y2×x+1×y]=[135]⇒[x+4y2x+y]=[135]\Rightarrow \begin{bmatrix*}[r] 1 & 4 \\ 2 & 1 \end{bmatrix*} \times \begin{bmatrix*}[r] x \\ y \end{bmatrix*} = \begin{bmatrix*}[r] 13 \\ 5 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 1 \times x + 4 \times y \\ 2 \times x + 1 \times y \end{bmatrix*} = \begin{bmatrix*}[r] 13 \\ 5 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] x + 4y \\ 2x + y \end{bmatrix*} = \begin{bmatrix*}[r] 13 \\ 5 \end{bmatrix*}

By definition of equality of matrices we get,

x + 4y = 13
⇒ x = 13 - 4y ......(i)

2x + y = 5
⇒ 2(13 - 4y) + y = 5
⇒ 26 - 8y + y = 5
⇒ -7y = -21
⇒ y = 3.

⇒ x = 13 - 4y = 13 - 4(3) = 13 - 12 = 1.

∴M=[xy]=[13].\therefore M = \begin{bmatrix*}[r] x \\ y \end{bmatrix*} = \begin{bmatrix*}[r] 1 \\ 3 \end{bmatrix*}.

Hence, M=[13].M = \begin{bmatrix*}[r] 1 \\ 3 \end{bmatrix*}.

Question 14

If A = [2x01]and B=[43601];\begin{bmatrix*}[r] 2 & x \\ 0 & 1 \end{bmatrix*} \text{and } B = \begin{bmatrix*}[r] 4 & 36 \\ 0 & 1 \end{bmatrix*}; find the value of x, given that : A2 = B.

Answer

Given,

⇒ A2 = B

⇒[2x01][2x01]=[43601]⇒[2×2+x×02×x+x×10×2+1×00×x+1×1]=[43601]⇒[4+02x+x00+1]=[43601]⇒[43x01]=[43601]\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 + 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*}

By definition of equality of matrices we get,

3x = 36
⇒ x = 12.

Hence, x = 12.

Question 15

If A and B are any two 2 × 2 matrices such that AB = BA = B and B is not a zero matrix, what can you say about the matrix A?

Answer

Since, AB = BA = B and B is not a zero matrix.

It means that A is a unit or identity matrix.

Question 16

Given A = [3004],B=[ab0c]\begin{bmatrix*}[r] 3 & 0 \\ 0 & 4 \end{bmatrix*}, B = \begin{bmatrix*}[r] a & b \\ 0 & c \end{bmatrix*} and that AB = A + B; find the values of a, b and c.

Answer

Given,

⇒AB=A+B⇒[3004][ab0c]=[3004]+[ab0c]⇒[3×a+0×03×b+0×c0×a+4×00×b+4×c]=[3+ab04+c]⇒[3a3b04c]=[3+ab04+c]\Rightarrow AB = A + B \\[1em] \Rightarrow \begin{bmatrix*}[r] 3 & 0 \\ 0 & 4 \end{bmatrix*}\begin{bmatrix*}[r] a & b \\ 0 & c \end{bmatrix*} = \begin{bmatrix*}[r] 3 & 0 \\ 0 & 4 \end{bmatrix*} + \begin{bmatrix*}[r] a & b \\ 0 & c \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 3 \times a + 0 \times 0 & 3 \times b + 0 \times c \\ 0 \times a + 4 \times 0 & 0 \times b + 4 \times c \end{bmatrix*} = \begin{bmatrix*}[r] 3 + a & b \\ 0 & 4 + c \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 3a & 3b \\ 0 & 4c \end{bmatrix*} = \begin{bmatrix*}[r] 3 + a & b \\ 0 & 4 + c \end{bmatrix*}

By definition of equality of matrices we get,

3a = 3 + a
⇒ 3a - a = 3
⇒ 2a = 3
⇒ a = 32\dfrac{3}{2}.

3b = b
⇒ b = 0.

4c = 4 + c
⇒ 4c - c = 4
⇒ 3c = 4
⇒ c = 43\dfrac{4}{3}

Hence, a = 32,b=0 and c=43\dfrac{3}{2}, b = 0 \text{ and c} = \dfrac{4}{3}.

Question 17

If A = [2113] and B=[3−11]\begin{bmatrix*}[r] 2 & 1 \\ 1 & 3 \end{bmatrix*} \text{ and } B = \begin{bmatrix*}[r] 3 \\ -11 \end{bmatrix*} find the matrix X such that AX = B.

Answer

Given,

⇒AX=B⇒[2113]X=[3−11]\Rightarrow AX = B \\[1em] \Rightarrow \begin{bmatrix*}[r] 2 & 1 \\ 1 & 3 \end{bmatrix*}X = \begin{bmatrix*}[r] 3 \\ -11 \end{bmatrix*}

X will be a matrix of order 2 × 1. So, let X = [ab]\begin{bmatrix*}[r] a \\ b \end{bmatrix*}.

⇒[2113][ab]=[3−11]⇒[2×a+1×b1×a+3×b]=[3−11]⇒[2a+ba+3b]=[3−11]\Rightarrow \begin{bmatrix*}[r] 2 & 1 \\ 1 & 3 \end{bmatrix*}\begin{bmatrix*}[r] a \\ b \end{bmatrix*} = \begin{bmatrix*}[r] 3 \\ -11 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2 \times a + 1 \times b \\ 1 \times a + 3 \times b \end{bmatrix*} = \begin{bmatrix*}[r] 3 \\ -11 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2a + b \\ a + 3b \end{bmatrix*} = \begin{bmatrix*}[r] 3 \\ -11 \end{bmatrix*}

By definition of equality of matrices we get,

2a + b = 3
⇒ b = 3 - 2a .......(i)

a + 3b = -11

Substituting value of b from (i) in above equation we get,

⇒ a + 3(3 - 2a) = -11
⇒ a + 9 - 6a = -11
⇒ -5a = -11 - 9
⇒ -5a = -20
⇒ a = 4.

b = 3 - 2a = 3 - 2(4) = 3 - 8 = -5.

Hence, X = [ab]=[4−5].\begin{bmatrix*}[r] a \\ b \end{bmatrix*} = \begin{bmatrix*}[r] 4 \\ -5 \end{bmatrix*}.

Question 18

If M = [41−12]\begin{bmatrix*}[r] 4 & 1 \\ -1 & 2 \end{bmatrix*}, show that : 6M - M2 = 9I; where I is a 2 × 2 unit matrix.

Answer

M2=[41−12][41−12]=[4×4+1×(−1)4×1+1×2−1×4+2×(−1)−1×1+2×2]=[16−14+2−4+(−2)−1+4]=[156−63].M^2 = \begin{bmatrix*}[r] 4 & 1 \\ -1 & 2 \end{bmatrix*}\begin{bmatrix*}[r] 4 & 1 \\ -1 & 2 \end{bmatrix*} \\[1em] = \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] 16 - 1 & 4 + 2 \\ -4 + (-2) & -1 + 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 15 & 6 \\ -6 & 3 \end{bmatrix*}.

Substituting value of M2 in L.H.S. of 6M - M2 = 9I,

=6[41−12]−[156−63]=[246−612]−[156−63]=[24−156−6−6−(−6)12−3]=[9009]=9[1001]=9I\phantom{=} 6\begin{bmatrix*}[r] 4 & 1 \\ -1 & 2 \end{bmatrix*} - \begin{bmatrix*}[r] 15 & 6 \\ -6 & 3 \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*} \\[1em] = 9\begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em] = 9\text{I}

Since, L.H.S. = R.H.S.

Hence, proved that 6M - M2 = 9I.

Question 19

If P = [2639] and Q=[3xy2]\begin{bmatrix*}[r] 2 & 6 \\ 3 & 9 \end{bmatrix*} \text{ and Q} = \begin{bmatrix*}[r] 3 & x \\ y & 2 \end{bmatrix*}, find x and y such that PQ = null matrix.

Answer

Given,

PQ = null matrix.

∴[2639][3xy2]=[0000]⇒[2×3+6×y2×x+6×23×3+9×y3×x+9×2]=[0000]⇒[6+6y2x+129+9y3x+18]=[0000].\therefore \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*}.

By definition of equality of matrices we get,

6 + 6y = 0
⇒ 6y = -6
⇒ y = -1.

2x + 12 = 0
⇒ 2x = -12
⇒ x = -6.

Hence, x = -6 and y = -1.

Question 20

Evaluate :

[2cos60°−2sin30°−tan45°cos0°][cot45°cosec30°sec60°sin90°]\begin{bmatrix*}[r] 2cos60° & -2sin30° \\ -tan 45° & cos0° \end{bmatrix*}\begin{bmatrix*}[r] cot 45° & cosec 30° \\ sec60° & sin 90° \end{bmatrix*}

Answer

Given,

⇒[2cos60°−2sin30°−tan45°cos0°][cot45°cosec30°sec60°sin90°]⇒[2×12−2×12−11][1221]⇒[1−1−11][1221]⇒[1×1+(−1)×21×2+(−1)×1−1×1+1×2−1×2+1×1]⇒[1−22−1−1+2−2+1]⇒[−111−1].\Rightarrow \begin{bmatrix*}[r] 2cos60° & -2sin30° \\ -tan 45° & cos0° \end{bmatrix*}\begin{bmatrix*}[r] cot 45° & cosec 30° \\ sec60° & sin 90° \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2 \times \dfrac{1}{2} & -2 \times \dfrac{1}{2} \\ -1 & 1 \end{bmatrix*}\begin{bmatrix*}[r] 1 & 2 \\ 2 & 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 1 & -1 \\ -1 & 1 \end{bmatrix*}\begin{bmatrix*}[r] 1 & 2 \\ 2 & 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 1 \times 1 + (-1) \times 2 & 1 \times 2 + (-1) \times 1 \\ -1 \times 1 + 1 \times 2 & -1 \times 2 + 1 \times 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 1 - 2 & 2 - 1 \\ -1 + 2 & -2 + 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] -1 & 1 \\ 1 & -1 \end{bmatrix*}.

Hence, [2cos60°−2sin30°−tan45°cos0°][cot45°cosec30°sec60°sin90°]=[−111−1].\begin{bmatrix*}[r] 2cos60° & -2sin30° \\ -tan 45° & cos0° \end{bmatrix*}\begin{bmatrix*}[r] cot 45° & cosec 30° \\ sec60° & sin 90° \end{bmatrix*} = \begin{bmatrix*}[r] -1 & 1 \\ 1 & -1 \end{bmatrix*}.

Question 21

State, with reason, whether the following are true or false. A, B and C are matrices of order 2 × 2.

(i) A + B = B + A

(ii) A - B = B - A

(iii) (B.C).A = B.(C.A)

(iv) (A + B).C = A.C + B.C

(v) A.(B - C) = A.B - A.C

(vi) (A - B).C = A.C - B.C

(vii) A2 - B2 = (A + B)(A - B)

(viii) (A - B)2 = A2 - 2A.B + B2.

Answer

(i) A + B = B + A

The above statement is true because addition of matrices is commutative.

(ii) A - B = B - A

The above statement is false because subtraction of matrices is not commutative.

(iii) (B.C).A = B.(C.A)

The above statement is true because multiplication of matrices is associative.

(iv) (A + B).C = A.C + B.C

The above statement is true because multiplication of matrices is distributive over addition.

(v) A.(B - C) = A.B - A.C

The above statement is true because multiplication of matrices is distributive over subtraction.

(vi) (A - B).C = A.C - B.C

The above statement is true because multiplication of matrices is distributive over subtraction.

(vii) A2 - B2 = (A + B)(A - B)

The above statement is false because laws of algebra for factorization and expansion are not applicable to matrices.

(viii) (A - B)2 = A2 - 2A.B + B2

The above statement is false because laws of algebra for factorization and expansion are not applicable to matrices.

Test Yourself

Question 1(a)

If a matrix A = [012−1]\begin{bmatrix*}[r] 0 & 1 \\ 2 & -1 \end{bmatrix*} and matrix B = [31]\begin{bmatrix*}[r] 3 \\ 1 \end{bmatrix*}, then which of the following is possible :

  1. A + B

  2. A - B

  3. AB

  4. BA

Answer

Since, addition and subtraction of matrices requires each matrix to be of same order.

∴ A + B and A - B are not possible.

AB is possible because no. of columns in A (2) is equal to no. of rows in B (2).

BA is not possible because no. of columns in B (1) is not equal to no. of rows in A (2).

Hence, Option 3 is the correct option.

Question 1(b)

If M × [32−10]=[3−1]\begin{bmatrix*}[r] 3 & 2 \\ -1 & 0 \end{bmatrix*} = \begin{bmatrix*}[r] 3 & -1 \end{bmatrix*}, the order of matrix M is :

  1. 2 × 2

  2. 2 × 1

  3. 1 × 2

  4. 1 × 3

Answer

We know that,

For matrix multiplication :

No. of columns in 1st matrix must be equal to the number of rows in 2nd matrix.

Resultant matrix order = No. of rows in 1st matrix × No. of columns in 2nd matrix.

Let order of matrix M be a × b.

⇒ Ma × b × [32−10]<em>2×2=[3−1]</em>1×2\begin{bmatrix*}[r] 3 & 2 \\ -1 & 0 \end{bmatrix*}<em>{2 \times 2} = \begin{bmatrix*}[r] 3 & -1 \end{bmatrix*}</em>{1 \times 2}

b = 2 and a = 1.

Order of matrix M = 1 × 2.

Hence, Option 3 is the correct option.

Question 1(c)

If [2x−yx+y]=[99]\begin{bmatrix*}[r] 2x - y \\ x + y \end{bmatrix*} = \begin{bmatrix*}[r] 9 \\ 9 \end{bmatrix*}, the value of x and y are :

  1. x = 3 and y = 3

  2. x = 3 and y = 9

  3. x = 3 and y = 6

  4. x = 6 and y = 3

Answer

Given,

[2x−yx+y]=[99]\begin{bmatrix*}[r] 2x - y \\ x + y \end{bmatrix*} = \begin{bmatrix*}[r] 9 \\ 9 \end{bmatrix*}

⇒ 2x - y = 9 .......(1)

⇒ x + y = 9 ........(2)

Adding equation (1) and (2), we get :

⇒ 2x - y + x + y = 9 + 9

⇒ 3x = 18

⇒ x = 183\dfrac{18}{3} = 6.

Substituting value of x in equation (2), we get :

⇒ 6 + y = 9

⇒ y = 9 - 6 = 3.

Hence, Option 4 is the correct option.

Question 1(d)

If matrix A = [x−yx+yy−xy+x]\begin{bmatrix*}[r] x - y & x + y \\ y - x & y + x \end{bmatrix*} and matrix B = [x+yy−xx−yy+x]\begin{bmatrix*}[r] x + y & y - x \\ x - y & y + x \end{bmatrix*}, then A + B is :

  1. [2y2x02(x+y)]\begin{bmatrix*}[r] 2y & 2x \\ 0 & 2(x + y) \end{bmatrix*}

  2. [2x2(x+y)00]\begin{bmatrix*}[r] 2x & 2(x + y) \\ 0 & 0 \end{bmatrix*}

  3. [2x2y02(x+y)]\begin{bmatrix*}[r] 2x & 2y \\ 0 & 2(x + y) \end{bmatrix*}

  4. [2x−2y2y00]\begin{bmatrix*}[r] 2x - 2y & 2y \\ 0 & 0 \end{bmatrix*}

Answer

Substituting values of A and B in A + B, we get :

⇒A+B=[x−yx+yy−xy+x]+[x+yy−xx−yy+x]=[x−y+x+yx+y+y−xy−x+x−yy+x+y+x]=[2x2y02(x+y)].\Rightarrow A + B = \begin{bmatrix*}[r] x - y & x + y \\ y - x & y + x \end{bmatrix*} + \begin{bmatrix*}[r] x + y & y - x \\ x - y & y + x \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] x - y + x + y & x + y + y - x \\ y - x + x - y & y + x + y + x \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2x & 2y \\ 0 & 2(x + y) \end{bmatrix*}.

Hence, Option 3 is the correct option.

Question 1(e)

Event A : Order of matrix A is 3 × 5.

Event B : Order of matrix B is 5 × 3.

Event C : Order of matrix C is 3 × 3.

Product of which two matrices gives a square matrix.

  1. AB and AC

  2. AB and BC

  3. BA and BC

  4. AB and BA

Answer

We know that,

Resultant matrix order = No. of rows in 1st matrix × No. of columns in 2nd matrix.

Resultant matrix (P) on multiplication of AB has order = 3 × 3.

Resultant matrix (Q) on multiplication of BA has order = 5 × 5.

Both P and Q are square matrix of order 3 and 5 respectively.

Hence, Option 4 is the correct option.

Question 1(f)

Two matrices A and B each of order 2 x 2.

Assertion (A) : A X B = 0 ⇒ A = 0 or B = 0.

Reason (R) : Let A=[2255]A = \begin{bmatrix*}[r] 2 & 2 \\ 5 & 5 \end{bmatrix*} ≠ 0 and B=[−434−3]B = \begin{bmatrix*}[r] -4 & 3 \\ 4 & -3 \end{bmatrix*} ≠ 0 but A x B = [2255][−434−3]\begin{bmatrix*}[r] 2 & 2 \\ 5 & 5 \end{bmatrix*}\begin{bmatrix*}[r] -4 & 3 \\ 4 & -3 \end{bmatrix*} = 0.

  1. A is true, R is false.

  2. A is false, R is true.

  3. Both A and R are true and R is correct reason for A.

  4. Both A and R are true and R is incorrect reason for A.

Answer

It is not necessarily true in case of matrices that if A X B = 0

Then, either A = 0 or B = 0.

So, assertion (A) is false.

According to reason, A=[2255]A = \begin{bmatrix*}[r] 2 & 2 \\ 5 & 5 \end{bmatrix*} and B=[−434−3]B = \begin{bmatrix*}[r] -4 & 3 \\ 4 & -3 \end{bmatrix*}

⇒AB=[2255].[−434−3]=[2×(−4)+2×42×3+2×(−3) 5×(−4)+5×45×3+5×(−3)]=[−8+86−6 −20+2015−15]=[00 00]\Rightarrow AB = \begin{bmatrix*}[r] 2 & 2 \\ 5 & 5 \end{bmatrix*}.\begin{bmatrix*}[r] -4 & 3 \\ 4 & -3 \end{bmatrix*}\\[1em] = \begin{bmatrix*}[r] 2 \times (-4) + 2 \times 4 & 2 \times 3 + 2 \times (-3)\ 5 \times (-4) + 5 \times 4 & 5 \times 3 + 5 \times (-3) \end{bmatrix*}\\[1em] = \begin{bmatrix*}[r] -8 + 8 & 6 - 6\ -20 + 20 & 15 - 15 \end{bmatrix*}\\[1em] = \begin{bmatrix*}[r] 0 & 0\ 0 & 0 \end{bmatrix*}\\[1em]

So, reason (R) is true.

Hence, option 2 is the correct option.

Question 1(g)

Matrix A = [xy]\begin{bmatrix*}[r] x & y \end{bmatrix*} and Matrix B = [ab]\begin{bmatrix*}[r] a \\ b \end{bmatrix*}.

Assertion (A) : Product BA is possible and order of resulting matrix is 2 x 2.

Reason (R) : The product BA of two matrices A and B is possible only if number of rows in matrix B Is same as number of columns in matrix A.

  1. A is true, R is false.

  2. A is false, R is true.

  3. Both A and R are true and R is correct reason for A.

  4. Both A and R are true and R is incorrect reason for A.

Answer

∴ Order of matrix A = 1 x 2

∴ Order of matrix B = 2 x 1

Since product of matrix is possible, only when the number of columns in the first matrix is equal to number of rows in second.

Since,

Number of columns in B = 1

Number of rows in A = 1

∴ Product BA is possible.

We know that,

The no. of rows in the resulting matrix is equal to the no. of rows in first matrix and no. of columns equals to the no. of columns in second matrix.

∴ Order of matrix BA = 2 x 2

So, assertion is true. But reason says the product BA is possible only if number of rows in matrix B is same as number of columns in matrix A, which is false.

Hence, option 1 is the correct option.

Question 1(h)

A, B and C are three matrices each of order 2 x 2.

Statement 1 : If A x B = A x C ⇒ B = C

Statement 2 : Cancellation law is applicable in matrix multiplication.

  1. Both the statements are true.

  2. Both the statements are false.

  3. Statement 1 is true, and statement 2 is false.

  4. Statement 1 is false, and statement 2 is true.

Answer

We know that,

Matrix multiplication does not satisfy the cancellation law.

∴ Statement 2 is false.

Thus, if A x B = A x C, we cannot conclude that B = C.

∴ Statement 1 is false.

Hence, option 2 is the correct option.

Question 1(i)

Matrix A = [2−2 −20]\begin{bmatrix*}[r] 2 & -2\ -2 & 0 \end{bmatrix*} and matrix B = [55 55]\begin{bmatrix*}[r] 5 & 5\ 5 & 5 \end{bmatrix*}

Statement 1 : AB = 0

Statement 2 : AB = 0, even if A ≠ 0 and B ≠ 0.

  1. Both the statements are true.

  2. Both the statements are false.

  3. Statement 1 is true, and statement 2 is false.

  4. Statement 1 is false, and statement 2 is true.

Answer

Given,

Matrix A = [2−2 −20]\begin{bmatrix*}[r] 2 & -2\ -2 & 0 \end{bmatrix*} and matrix B = [55 55]\begin{bmatrix*}[r] 5 & 5\ 5 & 5 \end{bmatrix*}

⇒AB=[2−2 −20].[55 55]=[2×5+(−2)×52×5+(−2)×5 (−2)×5+0×5(−2)×5+0×5]=[10−1010−10 −10+0−10+0]=[00 −10−10]\Rightarrow AB = \begin{bmatrix*}[r] 2 & -2\ -2 & 0 \end{bmatrix*}. \begin{bmatrix*}[r] 5 & 5\ 5 & 5 \end{bmatrix*}\\[1em] = \begin{bmatrix*}[r] 2 \times 5 + (-2) \times 5 & 2 \times 5 + (-2) \times 5\ (-2) \times 5 + 0 \times 5 & (-2) \times 5 + 0 \times 5 \end{bmatrix*}\\[1em] = \begin{bmatrix*}[r] 10 - 10 & 10 - 10\ -10 + 0 & -10 + 0 \end{bmatrix*}\\[1em] = \begin{bmatrix*}[r] 0 & 0\ -10 & -10 \end{bmatrix*}

So, AB ≠ 0

∴ Statement 1 is false.

Lets take  A =[1−1 2−2] and B =[3333]\text{ A } = \begin{bmatrix*}[r] 1 & -1\ 2 & -2 \end{bmatrix*} \text{ and B } = \begin{bmatrix*}[r] 3 & 3 \\ 3 & 3 \end{bmatrix*}

AB=[1−1 2−2].[3333]=[1×3+(−1)×31×3+(−1)×3 2×3+(−2)×32×3+(−2)×3]=[3−33−3 6−66−6]=[00 00]AB = \begin{bmatrix*}[r] 1 & -1\ 2 & -2 \end{bmatrix*} . \begin{bmatrix*}[r] 3 & 3 \\ 3 & 3 \end{bmatrix*}\\[1em] = \begin{bmatrix*}[r] 1 \times 3 + (-1) \times 3 & 1 \times 3 + (-1) \times 3\ 2 \times 3 + (-2) \times 3 & 2 \times 3 + (-2) \times 3\end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 3 - 3 & 3 - 3\ 6 - 6 & 6 - 6 \end{bmatrix*}\\[1em] = \begin{bmatrix*}[r] 0 & 0\ 0 & 0 \end{bmatrix*}\\[1em]

Hence, it is proved that AB can be equal to 0, even if A ≠ 0 and B ≠ 0.

∴ Statement 2 is true.

Hence, option 4 is the correct option.

Question 2

Find x and y, if :

[3−2−14][2x1]+2[−45]=4[2y]\begin{bmatrix*}[r] 3 & -2 \\ -1 & 4 \end{bmatrix*}\begin{bmatrix*}[r] 2x \\ 1 \end{bmatrix*} + 2\begin{bmatrix*}[r] -4 \\ 5 \end{bmatrix*} = 4\begin{bmatrix*}[r] 2 \\ y \end{bmatrix*}

Answer

Given,

⇒[3−2−14][2x1]+2[−45]=4[2y]⇒[3×2x+(−2)×1−1×2x+4×1]+[−810]=[84y]⇒[6x−2−2x+4]+[−810]=[84y]⇒[6x−2+(−8)−2x+4+10]=[84y]⇒[6x−10−2x+14]=[84y]\Rightarrow \begin{bmatrix*}[r] 3 & -2 \\ -1 & 4 \end{bmatrix*}\begin{bmatrix*}[r] 2x \\ 1 \end{bmatrix*} + 2\begin{bmatrix*}[r] -4 \\ 5 \end{bmatrix*} = 4\begin{bmatrix*}[r] 2 \\ y \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 3 \times 2x + (-2) \times 1 \\ -1 \times 2x + 4 \times 1 \end{bmatrix*} + \begin{bmatrix*}[r] -8 \\ 10 \end{bmatrix*} = \begin{bmatrix*}[r] 8 \\ 4y \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 6x - 2 \\ -2x + 4 \end{bmatrix*} + \begin{bmatrix*}[r] -8 \\ 10 \end{bmatrix*} = \begin{bmatrix*}[r] 8 \\ 4y \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 6x - 2 + (-8) \\ -2x + 4 + 10 \end{bmatrix*} = \begin{bmatrix*}[r] 8 \\ 4y \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 6x - 10 \\ -2x + 14 \end{bmatrix*} = \begin{bmatrix*}[r] 8 \\ 4y \end{bmatrix*}

By definition of equality of matrices we get,

6x - 10 = 8
⇒ 6x = 18
⇒ x = 3

-2x + 14 = 4y
⇒ -2(3) + 14 = 4y
⇒ -6 + 14 = 4y
⇒ 8 = 4y
⇒ y = 2.

Hence, x = 3 and y = 2.

Question 3

Find x and y, if :

[3x8][1437]−3[2−7]=5[32y]\begin{bmatrix*}[r] 3x & 8 \end{bmatrix*}\begin{bmatrix*}[r] 1 & 4 \\ 3 & 7 \end{bmatrix*} - 3\begin{bmatrix*}[r] 2 & -7 \end{bmatrix*} = 5\begin{bmatrix*}[r] 3 & 2y \end{bmatrix*}

Answer

Given,

⇒[3x×1+8×33x×4+8×7]−[6−21]=[1510y]⇒[3x+2412x+56]−[6−21]=[1510y]⇒[3x+24−612x+56−(−21)]=[1510y]⇒[3x+1812x+77]=[1510y]\Rightarrow \begin{bmatrix*}[r] 3x \times 1 + 8 \times 3 & 3x \times 4 + 8 \times 7 \end{bmatrix*} - \begin{bmatrix*}[r] 6 & -21 \end{bmatrix*} = \begin{bmatrix*}[r] 15 & 10y \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 3x + 24 & 12x + 56 \end{bmatrix*} - \begin{bmatrix*}[r] 6 & -21 \end{bmatrix*} = \begin{bmatrix*}[r] 15 & 10y \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 3x + 24 - 6 & 12x + 56 - (-21) \end{bmatrix*} = \begin{bmatrix*}[r] 15 & 10y \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 3x + 18 & 12x + 77 \end{bmatrix*} = \begin{bmatrix*}[r] 15 & 10y \end{bmatrix*}

By definition of equality of matrices we get,

3x + 18 = 15
⇒ 3x = -3
⇒ x = -1.

12x + 77 = 10y
⇒ 12(-1) + 77 = 10y
⇒ 65 = 10y
⇒ y = 6.5

Hence, x = -1 and y = 6.5

Question 4

If [xy][xy]=[25] and [−xy][2xy]=[−2]\begin{bmatrix*}[r] x & y \end{bmatrix*}\begin{bmatrix*}[r] x \\ y \end{bmatrix*} = \begin{bmatrix*}[r] 25 \end{bmatrix*} \text{ and } \begin{bmatrix*}[r] -x & y \end{bmatrix*}\begin{bmatrix*}[r] 2x \\ y \end{bmatrix*} = \begin{bmatrix*}[r] -2 \end{bmatrix*};

find x and y, if :

(i) x, y ∈ W (whole numbers)

(ii) x, y ∈ Z (integers)

Answer

Given,

⇒[xy][xy]=[25]⇒[x×x+y×y]=[25]⇒[x2+y2]=[25]\Rightarrow \begin{bmatrix*}[r] x & y \end{bmatrix*}\begin{bmatrix*}[r] x \\ y \end{bmatrix*} = \begin{bmatrix*}[r] 25 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] x \times x + y \times y \end{bmatrix*} = \begin{bmatrix*}[r] 25 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] x^2 + y^2 \end{bmatrix*} = \begin{bmatrix*}[r] 25 \end{bmatrix*}

By definition of equality of matrices we get,

x2 + y2 = 25
⇒ x2 = 25 - y2 ......(i)

Given,

⇒[−xy][2xy]=[−2]⇒[−x×2x+y×y]=[−2]⇒[−2x2+y2]=[−2]\Rightarrow \begin{bmatrix*}[r] -x & y \end{bmatrix*}\begin{bmatrix*}[r] 2x \\ y \end{bmatrix*} = \begin{bmatrix*}[r] -2 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] -x \times 2x + y \times y \end{bmatrix*} = \begin{bmatrix*}[r] -2 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] -2x^2 + y^2 \end{bmatrix*} = \begin{bmatrix*}[r] -2 \end{bmatrix*}

By definition of equality of matrices we get,

-2x2 + y2 = -2 ......(ii)

Substituting value of x2 from (i) in (ii) we get,

⇒ -2(25 - y2) + y2 = -2
⇒ -50 + 2y2 + y2 = -2
⇒ 3y2 = -2 + 50
⇒ 3y2 = 48
⇒ y2 = 16
⇒ y = ± 4.

⇒ x2 = 25 - y2
⇒ x2 = 25 - 16
⇒ x2 = 9
⇒ x = ± 3.

(i) Since, x, y ∈ W

∴ x = 3, y = 4.

Hence, x = 3 and y = 4.

(ii) Since, x, y ∈ Z

∴ x = ±3, y = ±4.

Hence, x = ±3 and y = ±4.

Question 5

Evaluate :

[cos45°sin30°2cos0°sin0°][sin45°cos90°sin90°cot45°]\begin{bmatrix*}[r] cos 45° & sin 30° \\ \sqrt{2}cos 0° & sin 0° \end{bmatrix*}\begin{bmatrix*}[r] sin 45° & cos 90° \\ sin 90° & cot 45° \end{bmatrix*}

Answer

Given,

⇒[cos45°sin30°2cos0°sin0°][sin45°cos90°sin90°cot45°]⇒[12122(1)0][12011]⇒[121220][12011]⇒[12×12+12×112×0+12×12×12+0×12×0+0×1]⇒[12+120+121+00+0]⇒[11210]⇒[10.510]\Rightarrow \begin{bmatrix*}[r] cos 45° & sin 30° \\ \sqrt{2}cos 0° & sin 0° \end{bmatrix*}\begin{bmatrix*}[r] sin 45° & cos 90° \\ sin 90° & cot 45° \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] \dfrac{1}{\sqrt{2}} & \dfrac{1}{2} \\ \sqrt{2}(1) & 0 \end{bmatrix*}\begin{bmatrix*}[r] \dfrac{1}{\sqrt{2}} & 0 \\ 1 & 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] \dfrac{1}{\sqrt{2}} & \dfrac{1}{2} \\ \sqrt{2} & 0 \end{bmatrix*}\begin{bmatrix*}[r] \dfrac{1}{\sqrt{2}} & 0 \\ 1 & 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] \dfrac{1}{\sqrt{2}} \times \dfrac{1}{\sqrt{2}} + \dfrac{1}{2} \times 1 & \dfrac{1}{\sqrt{2}} \times 0 + \dfrac{1}{2} \times 1 \\ \sqrt{2} \times \dfrac{1}{\sqrt{2}} + 0 \times 1 & \sqrt{2} \times 0 + 0 \times 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] \dfrac{1}{2} + \dfrac{1}{2} & 0 + \dfrac{1}{2} \\ 1 + 0 & 0 + 0 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 1 & \dfrac{1}{2} \\ 1 & 0 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 1 & 0.5 \\ 1 & 0 \end{bmatrix*}

Hence, [cos45°sin30°2cos0°sin0°][sin45°cos90°sin90°cot45°]=[10.510].\begin{bmatrix*}[r] cos 45° & sin 30° \\ \sqrt{2}cos 0° & sin 0° \end{bmatrix*}\begin{bmatrix*}[r] sin 45° & cos 90° \\ sin 90° & cot 45° \end{bmatrix*} = \begin{bmatrix*}[r] 1 & 0.5 \\ 1 & 0 \end{bmatrix*}.

Question 6

If A = [0−14−3],B=[−56]\begin{bmatrix*}[r] 0 & -1 \\ 4 & -3 \end{bmatrix*}, B = \begin{bmatrix*}[r] -5 \\ 6 \end{bmatrix*} and 3A × M = 2B; find matrix M.

Answer

Let order of matrix M be a × b.

Given,

⇒3A×M=2B.......(i)⇒3[0−14−3]<em>2×2×M</em>a×b=2[−56]2×1\Rightarrow 3A \times M = 2B .......(i) \\[1em] \Rightarrow 3\begin{bmatrix*}[r] 0 & -1 \\ 4 & -3 \end{bmatrix*}<em>{2 \times 2} \times M</em>{a \times b} = 2\begin{bmatrix*}[r] -5 \\ 6 \end{bmatrix*}_{2 \times 1}

Since product of matrix is possible, only when the number of columns in the first matrix is equal to no. of rows in second.

∴ a = 2.

Also the no. of columns of product (resulting matrix) is equal to no. of columns of second matrix.

∴ b = 1.

Hence, order of matrix M = 2 × 1.

Let M = [ab]\begin{bmatrix*}[r] a \\ b \end{bmatrix*}

Substituting value of A, M and B in (i) we get,

⇒3[0−14−3]×[ab]=2[−56]⇒[0−312−9]×[ab]=[−1012]⇒[0×a+(−3)×b12×a+(−9)×b]=[−1012]⇒[−3b12a−9b]=[−1012]\Rightarrow 3\begin{bmatrix*}[r] 0 & -1 \\ 4 & -3 \end{bmatrix*} \times \begin{bmatrix*}[r] a \\ b \end{bmatrix*} = 2\begin{bmatrix*}[r] -5 \\ 6 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 0 & -3 \\ 12 & -9 \end{bmatrix*} \times \begin{bmatrix*}[r] a \\ b \end{bmatrix*} = \begin{bmatrix*}[r] -10 \\ 12 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 0 \times a + (-3) \times b \\ 12 \times a + (-9) \times b \end{bmatrix*} = \begin{bmatrix*}[r] -10 \\ 12 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] -3b \\ 12a - 9b \end{bmatrix*} = \begin{bmatrix*}[r] -10 \\ 12 \end{bmatrix*}

By definition of equality of matrices we get,

-3b = -10
⇒ b = 103\dfrac{10}{3} ....(i)

12a - 9b = 12

Substituting value of b from (i) in above equation we get,

⇒12a−9×103=12⇒12a−30=12⇒12a=42⇒a=72.\Rightarrow 12a - 9 \times \dfrac{10}{3} = 12 \\[1em] \Rightarrow 12a - 30 = 12 \\[1em] \Rightarrow 12a = 42 \\[1em] \Rightarrow a = \dfrac{7}{2}.

∴[ab]=[72103].\therefore \begin{bmatrix*}[r] a \\ b \end{bmatrix*} = \begin{bmatrix*}[r] \dfrac{7}{2} \\ \dfrac{10}{3} \end{bmatrix*}.

Hence, M = [72103].\begin{bmatrix*}[r] \dfrac{7}{2} \\ \dfrac{10}{3} \end{bmatrix*}.

Question 7

Find x and y if : [x3xy4y][21]=[512]\begin{bmatrix*}[r] x & 3x \\ y & 4y \end{bmatrix*}\begin{bmatrix*}[r] 2 \\ 1 \end{bmatrix*} = \begin{bmatrix*}[r] 5 \\ 12 \end{bmatrix*}.

Answer

⇒[x3xy4y][21]=[512]⇒[x×2+3x×1y×2+4y×1]=[512]⇒[5x6y]=[512]\Rightarrow \begin{bmatrix*}[r] x & 3x \\ y & 4y \end{bmatrix*}\begin{bmatrix*}[r] 2 \\ 1 \end{bmatrix*} = \begin{bmatrix*}[r] 5 \\ 12 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] x \times 2 + 3x \times 1 \\ y \times 2 + 4y \times 1 \end{bmatrix*}= \begin{bmatrix*}[r] 5 \\ 12 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 5x \\ 6y \end{bmatrix*}= \begin{bmatrix*}[r] 5 \\ 12 \end{bmatrix*} \\[1em]

By definition of equality of matrices we get,

5x = 5
⇒ x = 1.

6y = 12
⇒ y = 2.

Hence, x = 1 and y = 2.

Question 8

If matrix X = [−342−3][2−2] and 2X - 3Y=[10−8]\begin{bmatrix*}[r] -3 & 4 \\ 2 & -3 \end{bmatrix*}\begin{bmatrix*}[r] 2 \\ -2 \end{bmatrix*} \text{ and 2X - 3Y} = \begin{bmatrix*}[r] 10 \\ -8 \end{bmatrix*}, find the matrix 'X' and matrix 'Y'.

Answer

Given,

X=[−342−3][2−2]=[−3×2+4×(−2)2×2+(−3)×(−2)]=[−6+(−8)4+6]=[−1410].X = \begin{bmatrix*}[r] -3 & 4 \\ 2 & -3 \end{bmatrix*}\begin{bmatrix*}[r] 2 \\ -2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -3 \times 2 + 4 \times (-2) \\ 2 \times 2 + (-3) \times (-2) \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -6 + (-8) \\ 4 + 6 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -14 \\ 10 \end{bmatrix*}.

Given,

⇒2X−3Y=[10−8]⇒2[−1410]−3Y=[10−8]⇒[−2820]−3Y=[10−8]⇒3Y=[−2820]−[10−8]⇒3Y=[−28−1020−(−8)]⇒3Y=[−3828]⇒Y=13[−3828].\Rightarrow 2X - 3Y = \begin{bmatrix*}[r] 10 \\ -8 \end{bmatrix*} \\[1em] \Rightarrow 2\begin{bmatrix*}[r] -14 \\ 10 \end{bmatrix*} - 3Y = \begin{bmatrix*}[r] 10 \\ -8 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] -28 \\ 20 \end{bmatrix*} - 3Y = \begin{bmatrix*}[r] 10 \\ -8 \end{bmatrix*} \\[1em] \Rightarrow 3Y = \begin{bmatrix*}[r] -28 \\ 20 \end{bmatrix*} - \begin{bmatrix*}[r] 10 \\ -8 \end{bmatrix*} \\[1em] \Rightarrow 3Y = \begin{bmatrix*}[r] -28 - 10 \\ 20 - (-8) \end{bmatrix*} \\[1em] \Rightarrow 3Y = \begin{bmatrix*}[r] -38 \\ 28 \end{bmatrix*} \\[1em] \Rightarrow Y = \dfrac{1}{3}\begin{bmatrix*}[r] -38 \\ 28 \end{bmatrix*}.

Hence, X = [−1410] and Y=13[−3828]\begin{bmatrix*}[r] -14 \\ 10 \end{bmatrix*} \text{ and Y} = \dfrac{1}{3}\begin{bmatrix*}[r] -38 \\ 28 \end{bmatrix*}.

Question 9

If A = [2513],B=[4−2−13]\begin{bmatrix*}[r] 2 & 5 \\ 1 & 3 \end{bmatrix*}, B = \begin{bmatrix*}[r] 4 & -2 \\ -1 & 3 \end{bmatrix*} and I is the identity matric of same order and At is transpose of matrix A, find At.B + BI.

Answer

A=[2513],At=[2153],I=[1001]A = \begin{bmatrix*}[r] 2 & 5 \\ 1 & 3 \end{bmatrix*}, A^t = \begin{bmatrix*}[r] 2 & 1 \\ 5 & 3 \end{bmatrix*}, I = \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*}.

⇒At.B+BI=[2153][4−2−13]+[4−2−13][1001]=[2×4+1×(−1)2×(−2)+1×35×4+3×(−1)5×(−2)+3×3]+[4×1+(−2)×04×0+(−2)×1−1×1+3×0−1×0+3×1]=[8−1−4+320−3−10+9]+[4+00−2−1+00+3]=[7−117−1]+[4−2−13]=[7+4−1+(−2)17+(−1)−1+3]=[11−3162].\Rightarrow A^t.B + BI = \begin{bmatrix*}[r] 2 & 1 \\ 5 & 3 \end{bmatrix*}\begin{bmatrix*}[r] 4 & -2 \\ -1 & 3 \end{bmatrix*} + \begin{bmatrix*}[r] 4 & -2 \\ -1 & 3 \end{bmatrix*} \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 \times 4 + 1 \times (-1) & 2 \times (-2) + 1 \times 3 \\ 5 \times 4 + 3 \times (-1) & 5 \times (-2) + 3 \times 3 \end{bmatrix*} + \begin{bmatrix*}[r] 4 \times 1 + (-2) \times 0 & 4 \times 0 + (-2) \times 1 \\ -1 \times 1 + 3 \times 0 & -1 \times 0 + 3 \times 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 8 - 1 & -4 + 3 \\ 20 - 3 & -10 + 9 \end{bmatrix*} + \begin{bmatrix*}[r] 4 + 0 & 0 - 2 \\ -1 + 0 & 0 + 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 7 & -1 \\ 17 & -1 \end{bmatrix*} + \begin{bmatrix*}[r] 4 & -2 \\ -1 & 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 7 + 4 & -1 + (-2) \\ 17 + (-1) & -1 + 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 11 & -3 \\ 16 & 2 \end{bmatrix*}.

Hence, At.B+BI=[11−3162].A^t.B + BI = \begin{bmatrix*}[r] 11 & -3 \\ 16 & 2 \end{bmatrix*}.

Question 10

Let A = [1021],B=[23−10].\begin{bmatrix*}[r] 1 & 0 \\ 2 & 1 \end{bmatrix*}, B = \begin{bmatrix*}[r] 2 & 3 \\ -1 & 0 \end{bmatrix*}. Find A2 + AB + B2.

Answer

A2=[1021][1021]=[1×1+0×21×0+0×12×1+1×22×0+1×1]=[1+00+02+20+1]=[1041]B2=[23−10][23−10]=[2×2+3×(−1)2×3+3×0−1×2+0×(−1)−1×3+0×0]=[4−36+0−2+0−3+0]=[16−2−3].A^2 = \begin{bmatrix*}[r] 1 & 0 \\ 2 & 1 \end{bmatrix*}\begin{bmatrix*}[r] 1 & 0 \\ 2 & 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 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*}[r] 1 + 0 & 0 + 0 \\ 2 + 2 & 0 + 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 & 0 \\ 4 & 1 \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*}[r] 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*}.

Substituting value of A2 and B2 in A2 + AB + B2 we get,

A2+AB+B2=[1041]+[1021][23−10]+[16−2−3]=[1041]+[1×2+0×(−1)1×3+0×02×2+1×(−1)2×3+1×0]+[16−2−3]=[1041]+[2+03+04−16+0]+[16−2−3]=[1041]+[2336]+[16−2−3]=[1+2+10+3+64+3+(−2)1+6+(−3)]=[4954].A^2 + AB + B^2 = \begin{bmatrix*}[r] 1 & 0 \\ 4 & 1 \end{bmatrix*} + \begin{bmatrix*}[r] 1 & 0 \\ 2 & 1 \end{bmatrix*}\begin{bmatrix*}[r] 2 & 3 \\ -1 & 0 \end{bmatrix*} + \begin{bmatrix*}[r] 1 & 6 \\ -2 & -3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 & 0 \\ 4 & 1 \end{bmatrix*} + \begin{bmatrix*}[r] 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*} + \begin{bmatrix*}[r] 1 & 6 \\ -2 & -3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 & 0 \\ 4 & 1 \end{bmatrix*} + \begin{bmatrix*}[r] 2 + 0 & 3 + 0 \\ 4 - 1 & 6 + 0 \end{bmatrix*} + \begin{bmatrix*}[r] 1 & 6 \\ -2 & -3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 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*}[r] 4 & 9 \\ 5 & 4 \end{bmatrix*}.

Hence, A2 + AB + B2 = [4954].\begin{bmatrix*}[r] 4 & 9 \\ 5 & 4 \end{bmatrix*}.

Question 11

If A = [3a−48],B=[c4−30],C=[−143b]\begin{bmatrix*}[r] 3 & a \\ -4 & 8 \end{bmatrix*}, B = \begin{bmatrix*}[r] c & 4 \\ -3 & 0 \end{bmatrix*}, C = \begin{bmatrix*}[r] -1 & 4 \\ 3 & b \end{bmatrix*} and 3A - 2C = 6B, find the values of a, b and c.

Answer

Given,

⇒3A−2C=6B⇒3[3a−48]−2[−143b]=6[c4−30]⇒[93a−1224]−[−2862b]=[6c24−180]⇒[9−(−2)3a−8−12−624−2b]=[6c24−180]⇒[113a−8−1824−2b]=[6c24−180]\Rightarrow 3A - 2C = 6B \\[1em] \Rightarrow 3\begin{bmatrix*}[r] 3 & a \\ -4 & 8 \end{bmatrix*} - 2\begin{bmatrix*}[r] -1 & 4 \\ 3 & b \end{bmatrix*} = 6\begin{bmatrix*}[r] c & 4 \\ -3 & 0 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 9 & 3a \\ -12 & 24 \end{bmatrix*} - \begin{bmatrix*}[r] -2 & 8 \\ 6 & 2b \end{bmatrix*} = \begin{bmatrix*}[r] 6c & 24 \\ -18 & 0 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 9 - (-2) & 3a - 8 \\ -12 - 6 & 24 - 2b \end{bmatrix*} = \begin{bmatrix*}[r] 6c & 24 \\ -18 & 0 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 11 & 3a - 8 \\ -18 & 24 - 2b \end{bmatrix*} = \begin{bmatrix*}[r] 6c & 24 \\ -18 & 0 \end{bmatrix*}

By definition of equality of matrices we get,

6c = 11
⇒ c = 116=156.\dfrac{11}{6} = 1\dfrac{5}{6}.

3a - 8 = 24
⇒ 3a = 32
⇒ a = 323=1023\dfrac{32}{3} = 10\dfrac{2}{3}

24 - 2b = 0
⇒ 2b = 24
⇒ b = 12.

Hence, a = 1023,b=12,c=156.10\dfrac{2}{3}, b = 12, c = 1\dfrac{5}{6}.

Question 12

Given A = [p002],B=[0−q10],C=[2−222]\begin{bmatrix*}[r] p & 0 \\ 0 & 2 \end{bmatrix*}, B = \begin{bmatrix*}[r] 0 & -q \\ 1 & 0 \end{bmatrix*}, C = \begin{bmatrix*}[r] 2 & -2 \\ 2 & 2 \end{bmatrix*} and BA = C2. Find the values of p and q.

Answer

Given,

⇒BA=C2⇒[0−q10][p002]=[2−222][2−222]⇒[0×p+(−q)×00×0+(−q)×21×p+0×01×0+0×2]=[2×2+(−2)×22×(−2)+(−2)×22×2+2×22×(−2)+2×2]⇒[0−2qp0]=[0−880]\Rightarrow BA = C^2 \\[1em] \Rightarrow \begin{bmatrix*}[r] 0 & -q \\ 1 & 0 \end{bmatrix*}\begin{bmatrix*}[r] p & 0 \\ 0 & 2 \end{bmatrix*} = \begin{bmatrix*}[r] 2 & -2 \\ 2 & 2 \end{bmatrix*}\begin{bmatrix*}[r] 2 & -2 \\ 2 & 2 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 0 \times p + (-q) \times 0 & 0 \times 0 + (-q) \times 2 \\ 1 \times p + 0 \times 0 & 1 \times 0 + 0 \times 2 \end{bmatrix*} = \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*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 0 & -2q \\ p & 0 \end{bmatrix*} = \begin{bmatrix*}[r] 0 & -8 \\ 8 & 0 \end{bmatrix*}

By definition of equality of matrices we get,

-2q = -8
⇒ q = 4.

p = 8.

Hence, p = 8 and q = 4.

Question 13

Evaluate : [4sin30°2cos60°sin90°2cos0°][4554]\begin{bmatrix*}[r] 4 sin 30° & 2cos 60° \\ sin 90° & 2 cos 0° \end{bmatrix*}\begin{bmatrix*}[r] 4 & 5 \\ 5 & 4 \end{bmatrix*}.

Answer

⇒[4sin30°2cos60°sin90°2cos0°][4554]=[4×122×1212×1][4554]=[2112][4554]=[2×4+1×52×5+1×41×4+2×51×5+2×4]=[8+510+44+105+8]=[13141413].\Rightarrow \begin{bmatrix*}[r] 4 sin 30° & 2cos 60° \\ sin 90° & 2 cos 0° \end{bmatrix*}\begin{bmatrix*}[r] 4 & 5 \\ 5 & 4 \end{bmatrix*} = \begin{bmatrix*}[r] 4 \times \dfrac{1}{2} & 2 \times \dfrac{1}{2} \\ 1 & 2 \times 1 \end{bmatrix*}\begin{bmatrix*}[r] 4 & 5 \\ 5 & 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 & 1 \\ 1 & 2 \end{bmatrix*}\begin{bmatrix*}[r] 4 & 5 \\ 5 & 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 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*}[r] 8 + 5 & 10 + 4 \\ 4 + 10 & 5 + 8 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 13 & 14 \\ 14 & 13 \end{bmatrix*}.

Hence, [4sin30°2cos60°sin90°2cos0°][4554]=[13141413].\begin{bmatrix*}[r] 4 sin 30° & 2cos 60° \\ sin 90° & 2 cos 0° \end{bmatrix*}\begin{bmatrix*}[r] 4 & 5 \\ 5 & 4 \end{bmatrix*} = \begin{bmatrix*}[r] 13 & 14 \\ 14 & 13 \end{bmatrix*}.

Question 14

Given A = [20−17] and I=[1001]\begin{bmatrix*}[r] 2 & 0 \\ -1 & 7 \end{bmatrix*} \text{ and } I = \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} and A2 = 9A + mI. Find m.

Answer

Given,

⇒A2=9A+mI⇒[20−17][20−17]=9[20−17]+m[1001]⇒[2×2+0×(−1)2×0+0×7−1×2+7×(−1)−1×0+7×7]=[180−963]+[m00m]⇒[40−949]=[180−963]+[m00m]⇒[m00m]=[40−949]−[180−963]⇒[m00m]=[4−180−0−9−(−9)49−63]⇒[m00m]=[−1400−14]\Rightarrow A^2 = 9A + mI \\[1em] \Rightarrow \begin{bmatrix*}[r] 2 & 0 \\ -1 & 7 \end{bmatrix*}\begin{bmatrix*}[r] 2 & 0 \\ -1 & 7 \end{bmatrix*} = 9\begin{bmatrix*}[r] 2 & 0 \\ -1 & 7 \end{bmatrix*} + m\begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2 \times 2 + 0 \times (-1) & 2 \times 0 + 0 \times 7 \\ -1 \times 2 + 7 \times (-1) & -1 \times 0 + 7 \times 7 \end{bmatrix*} = \begin{bmatrix*}[r] 18 & 0 \\ -9 & 63 \end{bmatrix*} + \begin{bmatrix*}[r] m & 0 \\ 0 & m \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 & 0 \\ -9 & 49 \end{bmatrix*} = \begin{bmatrix*}[r] 18 & 0 \\ -9 & 63 \end{bmatrix*} + \begin{bmatrix*}[r] m & 0 \\ 0 & m \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] m & 0 \\ 0 & m \end{bmatrix*} = \begin{bmatrix*}[r] 4 & 0 \\ -9 & 49 \end{bmatrix*} - \begin{bmatrix*}[r] 18 & 0 \\ -9 & 63 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] m & 0 \\ 0 & m \end{bmatrix*} = \begin{bmatrix*}[r] 4 - 18 & 0 - 0 \\ -9 - (-9) & 49 - 63 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] m & 0 \\ 0 & m \end{bmatrix*} = \begin{bmatrix*}[r] -14 & 0 \\ 0 & -14 \end{bmatrix*}

By definition of equality of matrices we get,

⇒ m = -14.

Hence, m = -14.

Question 15

Given matrix A = [4sin30°cos0°cos0°4sin30°] and B=[45]\begin{bmatrix*}[r] 4 sin 30° & cos 0° \\ cos 0° & 4 sin 30° \end{bmatrix*}\text{ and } B = \begin{bmatrix*}[r] 4 \\ 5 \end{bmatrix*}. If AX = B,

(i) write the order of matrix X.

(ii) find the matrix 'X'.

Answer

(i) Let order of matrix X be a × b.

i.e. [4sin30°cos0°cos0°4sin30°]2×2×Xa×b=[45]2×1\begin{bmatrix*}[r] 4 sin 30° & cos 0° \\ cos 0° & 4 sin 30° \end{bmatrix*}_{2 \times 2} \times X_{a \times b} = \begin{bmatrix*}[r] 4 \\ 5 \end{bmatrix*}_{2 \times 1}

Since product of matrix is possible, only when the number of columns in the first matrix is equal to no. of rows in second.

∴ a = 2.

Also the no. of columns of product (resulting matrix) is equal to no. of columns of second matrix.

∴ b = 1.

Hence, order of matrix X = 2 × 1.

(ii) Let matrix X = [xy]\begin{bmatrix*}[r] x \\ y \end{bmatrix*}

Given,

⇒AX=B⇒[4sin30°cos0°cos0°4sin30°][xy]=[45]⇒[4×12114×12][xy]=[45]⇒[2112][xy]=[45]⇒[2x+yx+2y]=[45]\Rightarrow AX = B \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 sin 30° & cos 0° \\ cos 0° & 4 sin 30° \end{bmatrix*}\begin{bmatrix*}[r] x \\ y \end{bmatrix*} = \begin{bmatrix*}[r] 4 \\ 5 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 \times \dfrac{1}{2} & 1 \\ 1 & 4 \times \dfrac{1}{2} \end{bmatrix*}\begin{bmatrix*}[r] x \\ y \end{bmatrix*} = \begin{bmatrix*}[r] 4 \\ 5 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2 & 1 \\ 1 & 2 \end{bmatrix*}\begin{bmatrix*}[r] x \\ y \end{bmatrix*} = \begin{bmatrix*}[r] 4 \\ 5 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2x + y \\ x + 2y \end{bmatrix*} = \begin{bmatrix*}[r] 4 \\ 5 \end{bmatrix*}

By definition of equality of matrices we get,

2x + y = 4
⇒ y = 4 - 2x .......(i)

x + 2y = 5

Substituting value of y from (i) in above equation we get,

⇒ x + 2(4 - 2x) = 5
⇒ x + 8 - 4x = 5
⇒ -3x = 5 - 8
⇒ -3x = -3
⇒ x = 1.

⇒ y = 4 - 2x = 4 - 2(1) = 2.

∴X=[xy]=[12]\therefore X = \begin{bmatrix*}[r] x \\ y \end{bmatrix*} = \begin{bmatrix*}[r] 1 \\ 2 \end{bmatrix*}

Hence, X = [12].\begin{bmatrix*}[r] 1 \\ 2 \end{bmatrix*}.

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