If A = [ 3 5 4 − 2 ] and B = [ 2 4 ] , \begin{bmatrix*}[r] 3 & 5 \\ 4 & -2 \end{bmatrix*} \text{ and B } = \begin{bmatrix*}[r] 2 \\ 4 \end{bmatrix*}, [ 3 4 5 − 2 ] and B = [ 2 4 ] ,
Answer
The product is possible because number of rows in A = number of columns in B =
A B = [ 3 5 4 − 2 ] [ 2 4 ] = [ 3.2 + 5.4 4.2 + ( − 2 ) .4 ] = [ 26 0 ] AB = \begin{bmatrix*}[r] 3 & 5 \\ 4 & -2 \end{bmatrix*} \begin{bmatrix*}[r] 2 \\ 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 3.2 + 5.4 \\ 4.2 + (-2).4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 26 \\ 0 \end{bmatrix*} A B = [ 3 4 5 − 2 ] [ 2 4 ] = [ 3.2 + 5.4 4.2 + ( − 2 ) .4 ] = [ 26 0 ]
Hence, the matrix AB = [ 26 0 ] . \begin{bmatrix*}[r] 26 \\ 0 \end{bmatrix*}. [ 26 0 ] .
If A = [ 2 5 1 3 ] , B = [ 1 − 1 − 3 2 ] , \begin{bmatrix*}[r] 2 & 5 \\ 1 & 3 \end{bmatrix*}, \text{ B } = \begin{bmatrix*}[r] 1 & -1 \\ -3 & 2 \end{bmatrix*}, [ 2 1 5 3 ] , B = [ 1 − 3 − 1 2 ] ,
Answer
AB = [ 2 5 1 3 ] [ 1 − 1 − 3 2 ] = [ 2.1 + 5. ( − 3 ) 2. ( − 1 ) + 5.2 1.1 + 3. ( − 3 ) 1. ( − 1 ) + 3.2 ] = [ − 13 8 − 8 5 ] BA = [ 1 − 1 − 3 2 ] [ 2 5 1 3 ] = [ 1 × 2 + ( − 1 ) × 1 1 × 5 + ( − 1 ) × 3 − 3 × 2 + 2 × 1 − 3 × 5 + 2 × 3 ] = [ 2 − 1 5 − 3 − 6 + 2 − 15 + 6 ] = [ 1 2 − 4 − 9 ] \text{ AB } = \begin{bmatrix*}[r] 2 & 5 \\ 1 & 3 \end{bmatrix*} \begin{bmatrix*}[r] 1 & -1 \\ -3 & 2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2.1 + 5.(-3) & 2.(-1) + 5.2 \\ 1.1 + 3.(-3) & 1.(-1) + 3.2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -13 & 8 \\ -8 & 5 \end{bmatrix*} \\[1em] \text{ BA } = \begin{bmatrix*}[r] 1 & -1 \\ -3 & 2 \end{bmatrix*} \begin{bmatrix*}[r] 2 & 5 \\ 1 & 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 \times 2 + (-1) \times 1 & 1 \times 5 + (-1) \times 3 \\ -3 \times 2 + 2 \times 1 & -3 \times 5 + 2 \times 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 - 1 & 5 - 3 \\ -6 + 2 & -15 + 6 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 & 2 \\ -4 & -9 \end{bmatrix*} AB = [ 2 1 5 3 ] [ 1 − 3 − 1 2 ] = [ 2.1 + 5. ( − 3 ) 1.1 + 3. ( − 3 ) 2. ( − 1 ) + 5.2 1. ( − 1 ) + 3.2 ] = [ − 13 − 8 8 5 ] BA = [ 1 − 3 − 1 2 ] [ 2 1 5 3 ] = [ 1 × 2 + ( − 1 ) × 1 − 3 × 2 + 2 × 1 1 × 5 + ( − 1 ) × 3 − 3 × 5 + 2 × 3 ] = [ 2 − 1 − 6 + 2 5 − 3 − 15 + 6 ] = [ 1 − 4 2 − 9 ]
The matrix AB = [ − 13 8 − 8 5 ] and BA = [ 1 2 − 4 − 9 ] . \begin{bmatrix*}[r] -13 & 8 \\ -8 & 5 \end{bmatrix*} \text{ and BA } = \begin{bmatrix*}[r] 1 & 2 \\ -4 & -9 \end{bmatrix*}. [ − 13 − 8 8 5 ] and BA = [ 1 − 4 2 − 9 ] .
If A = [ 3 7 2 4 ] , B = [ 0 2 5 3 ] and C = [ 1 − 5 − 4 6 ] , \begin{bmatrix*}[r] 3 & 7 \\ 2 & 4 \end{bmatrix*}, \text{ B } = \begin{bmatrix*}[r] 0 & 2 \\ 5 & 3 \end{bmatrix*} \text{ and C } = \begin{bmatrix*}[r] 1 & -5 \\ -4 & 6 \end{bmatrix*}, [ 3 2 7 4 ] , B = [ 0 5 2 3 ] and C = [ 1 − 4 − 5 6 ] ,
Answer
A B − 5 C = [ 3 7 2 4 ] [ 0 2 5 3 ] − 5 [ 1 − 5 − 4 6 ] = [ 3 × 0 + 7 × 5 3 × 2 + 7 × 3 2 × 0 + 4 × 5 2 × 2 + 4 × 3 ] − [ 5 − 25 − 20 30 ] = [ 0 + 35 6 + 21 0 + 20 4 + 12 ] − [ 5 − 25 − 20 30 ] = [ 35 − 5 27 − ( − 25 ) 20 − ( − 20 ) 16 − 30 ] = [ 30 52 40 − 14 ] AB - 5C = \begin{bmatrix*}[r] 3 & 7 \\ 2 & 4 \end{bmatrix*}\begin{bmatrix*}[r] 0 & 2 \\ 5 & 3 \end{bmatrix*} - 5\begin{bmatrix*}[r] 1 & -5 \\ -4 & 6 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 3 \times 0 + 7 \times 5 & 3 \times 2 + 7 \times 3 \\ 2 \times 0 + 4 \times 5 & 2 \times 2 + 4 \times 3 \end{bmatrix*} - \begin{bmatrix*}[r] 5 & -25 \\ -20 & 30 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 0 + 35 & 6 + 21 \\ 0 + 20 & 4 + 12 \end{bmatrix*} - \begin{bmatrix*}[r] 5 & -25 \\ -20 & 30 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 35 - 5 & 27 - (-25) \\ 20 - (-20) & 16 - 30 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 30 & 52 \\ 40 & -14 \end{bmatrix*} A B − 5 C = [ 3 2 7 4 ] [ 0 5 2 3 ] − 5 [ 1 − 4 − 5 6 ] = [ 3 × 0 + 7 × 5 2 × 0 + 4 × 5 3 × 2 + 7 × 3 2 × 2 + 4 × 3 ] − [ 5 − 20 − 25 30 ] = [ 0 + 35 0 + 20 6 + 21 4 + 12 ] − [ 5 − 20 − 25 30 ] = [ 35 − 5 20 − ( − 20 ) 27 − ( − 25 ) 16 − 30 ] = [ 30 40 52 − 14 ]
Hence, the matrix AB - 5C = [ 30 52 40 − 14 ] . \begin{bmatrix*}[r] 30 & 52 \\ 40 & -14 \end{bmatrix*}. [ 30 40 52 − 14 ] .
If A = [ 1 2 2 1 ] and B = [ 2 1 1 2 ] , \begin{bmatrix*}[r] 1 & 2 \\ 2 & 1 \end{bmatrix*} \text{ and B } = \begin{bmatrix*}[r] 2 & 1 \\ 1 & 2 \end{bmatrix*}, [ 1 2 2 1 ] and B = [ 2 1 1 2 ] ,
Answer
BA = [ 2 1 1 2 ] [ 1 2 2 1 ] = [ 2 × 1 + 1 × 2 2 × 2 + 1 × 1 1 × 1 + 2 × 2 1 × 2 + 2 × 1 ] = [ 2 + 2 4 + 1 1 + 4 2 + 2 ] = [ 4 5 5 4 ] ∴ A(BA) = A × BA = [ 1 2 2 1 ] [ 4 5 5 4 ] = [ 1 × 4 + 2 × 5 1 × 5 + 2 × 4 2 × 4 + 1 × 5 2 × 5 + 1 × 4 ] = [ 4 + 10 5 + 8 8 + 5 10 + 4 ] = [ 14 13 13 14 ] \text{BA } = \begin{bmatrix*}[r] 2 & 1 \\ 1 & 2 \end{bmatrix*} \begin{bmatrix*}[r] 1 & 2 \\ 2 & 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 \times 1 + 1 \times 2 & 2 \times 2 + 1 \times 1 \\ 1 \times 1 + 2 \times 2 & 1 \times 2 + 2 \times 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 + 2 & 4 + 1 \\ 1 + 4 & 2 + 2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 4 & 5 \\ 5 & 4 \end{bmatrix*} \\[1.5em] \therefore \text{A(BA) } = \text{A} \times \text{BA} \\[1em] = \begin{bmatrix*}[r] 1 & 2 \\ 2 & 1 \end{bmatrix*} \begin{bmatrix*}[r] 4 & 5 \\ 5 & 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 \times 4 + 2 \times 5 & 1 \times 5 + 2 \times 4 \\ 2 \times 4 + 1 \times 5 & 2 \times 5 + 1 \times 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 4 + 10 & 5 + 8 \\ 8 + 5 & 10 + 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 14 & 13 \\ 13 & 14 \end{bmatrix*} BA = [ 2 1 1 2 ] [ 1 2 2 1 ] = [ 2 × 1 + 1 × 2 1 × 1 + 2 × 2 2 × 2 + 1 × 1 1 × 2 + 2 × 1 ] = [ 2 + 2 1 + 4 4 + 1 2 + 2 ] = [ 4 5 5 4 ] ∴ A(BA) = A × BA = [ 1 2 2 1 ] [ 4 5 5 4 ] = [ 1 × 4 + 2 × 5 2 × 4 + 1 × 5 1 × 5 + 2 × 4 2 × 5 + 1 × 4 ] = [ 4 + 10 8 + 5 5 + 8 10 + 4 ] = [ 14 13 13 14 ]
Hence, the matrix A(BA) = [ 14 13 13 14 ] . \begin{bmatrix*}[r] 14 & 13 \\ 13 & 14 \end{bmatrix*}. [ 14 13 13 14 ] .
Given the matrices :
A = [ 2 1 4 2 ] , B = [ 3 4 − 1 − 2 ] and C = [ − 3 1 0 − 2 ] . \text { A } = \begin{bmatrix*}[r] 2 & 1 \\ 4 & 2 \end{bmatrix*}, \text{ B } = \begin{bmatrix*}[r] 3 & 4 \\ -1 & -2 \end{bmatrix*} \text{ and C } = \begin{bmatrix*}[r] -3 & 1 \\ 0 & -2 \end{bmatrix*} . A = [ 2 4 1 2 ] , B = [ 3 − 1 4 − 2 ] and C = [ − 3 0 1 − 2 ] .
Find the products of (i) ABC (ii) ACB and state whether they are equal.
Answer
(i)
ABC = [ 2 1 4 2 ] [ 3 4 − 1 − 2 ] [ − 3 1 0 − 2 ] = [ 2 × 3 + 1 × ( − 1 ) 2 × 4 + 1 × ( − 2 ) 4 × 3 + 2 × ( − 1 ) 4 × 4 + 2 × ( − 2 ) ] [ − 3 1 0 − 2 ] = [ 6 − 1 8 − 2 12 − 2 16 − 4 ] [ − 3 1 0 − 2 ] = [ 5 6 10 12 ] [ − 3 1 0 − 2 ] = [ 5 × ( − 3 ) + 6 × 0 5 × 1 + 6 × ( − 2 ) 10 × ( − 3 ) + 12 × 0 10 × 1 + 12 × ( − 2 ) ] = [ − 15 + 0 5 − 12 − 30 + 0 10 + ( − 24 ) ] = [ − 15 − 7 − 30 − 14 ] \text{ABC } = \begin{bmatrix} 2 & 1 \\ 4 & 2 \end{bmatrix}\begin{bmatrix*}[r] 3 & 4 \\ -1 & -2 \end{bmatrix*}\begin{bmatrix*}[r] -3 & 1 \\ 0 & -2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 \times 3 + 1 \times (-1) & 2 \times 4 + 1 \times (-2) \\ 4 \times 3 + 2 \times (-1) & 4 \times 4 + 2 \times (-2) \end{bmatrix*} \begin{bmatrix*}[r] -3 & 1 \\ 0 & -2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 6 - 1 & 8 - 2 \\ 12 - 2 & 16 - 4 \end{bmatrix*} \begin{bmatrix*}[r] -3 & 1 \\ 0 & -2 \end{bmatrix*} \\[1em] = \begin{bmatrix} 5 & 6 \\ 10 & 12 \end{bmatrix} \begin{bmatrix*}[r] -3 & 1 \\ 0 & -2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 5 \times (-3) + 6 \times 0 & 5 \times 1 + 6 \times (-2) \\ 10 \times (-3) + 12 \times 0 & 10 \times 1 + 12 \times (-2) \end{bmatrix*} \\[1em] = \begin{bmatrix} -15 + 0 & 5 - 12 \\ -30 + 0 & 10 + (-24) \end{bmatrix} \\[1em] = \begin{bmatrix} -15 & -7 \\ -30 & -14 \end{bmatrix} \\[1em] ABC = [ 2 4 1 2 ] [ 3 − 1 4 − 2 ] [ − 3 0 1 − 2 ] = [ 2 × 3 + 1 × ( − 1 ) 4 × 3 + 2 × ( − 1 ) 2 × 4 + 1 × ( − 2 ) 4 × 4 + 2 × ( − 2 ) ] [ − 3 0 1 − 2 ] = [ 6 − 1 12 − 2 8 − 2 16 − 4 ] [ − 3 0 1 − 2 ] = [ 5 10 6 12 ] [ − 3 0 1 − 2 ] = [ 5 × ( − 3 ) + 6 × 0 10 × ( − 3 ) + 12 × 0 5 × 1 + 6 × ( − 2 ) 10 × 1 + 12 × ( − 2 ) ] = [ − 15 + 0 − 30 + 0 5 − 12 10 + ( − 24 ) ] = [ − 15 − 30 − 7 − 14 ]
Hence, the matrix ABC = [ − 15 − 7 − 30 − 14 ] . \begin{bmatrix} -15 & -7 \\ -30 & -14 \end{bmatrix} . [ − 15 − 30 − 7 − 14 ] .
(ii)
ACB = [ 2 1 4 2 ] [ − 3 1 0 − 2 ] [ 3 4 − 1 − 2 ] = [ 2 × ( − 3 ) + 1 × 0 2 × 1 + 1 × ( − 2 ) 4 × ( − 3 ) + 2 × 0 4 × 1 + 2 × ( − 2 ) ] [ 3 4 − 1 − 2 ] = [ − 6 + 0 2 − 2 − 12 + 0 4 − 4 ] [ 3 4 − 1 − 2 ] = [ − 6 0 − 12 0 ] [ 3 4 − 1 − 2 ] = [ − 6 × 3 + 0 × ( − 1 ) − 6 × 4 + 0 × ( − 2 ) − 12 × 3 + 0 × ( − 1 ) − 12 × 4 + 0 × ( − 2 ) ] = [ − 18 + 0 − 24 + 0 − 36 + 0 − 48 + 0 ] = [ − 18 − 24 − 36 − 48 ] \text{ACB } = \begin{bmatrix*}[r] 2 & 1 \\ 4 & 2 \end{bmatrix*} \begin{bmatrix*}[r] -3 & 1 \\ 0 & -2 \end{bmatrix*} \begin{bmatrix*}[r] 3 & 4 \\ -1 & -2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 \times (-3) + 1 \times 0 & 2 \times 1 + 1 \times (-2) \\ 4 \times (-3) + 2 \times 0 & 4 \times 1 + 2 \times (-2) \end{bmatrix*} \begin{bmatrix*}[r] 3 & 4 \\ -1 & -2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -6 + 0 & 2 - 2 \\ -12 + 0 & 4 - 4 \end{bmatrix*} \begin{bmatrix*}[r] 3 & 4 \\ -1 & -2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -6 & 0 \\ -12 & 0 \end{bmatrix*} \begin{bmatrix*}[r] 3 & 4 \\ -1 & -2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -6 \times 3 + 0 \times (-1) & -6 \times 4 + 0 \times (-2) \\ -12 \times 3 + 0 \times (-1) & -12 \times 4 + 0 \times (-2) \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -18 + 0 & -24 + 0 \\ -36 + 0 & -48 + 0 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -18 & -24 \\ -36 & -48 \end{bmatrix*} \\[1em] ACB = [ 2 4 1 2 ] [ − 3 0 1 − 2 ] [ 3 − 1 4 − 2 ] = [ 2 × ( − 3 ) + 1 × 0 4 × ( − 3 ) + 2 × 0 2 × 1 + 1 × ( − 2 ) 4 × 1 + 2 × ( − 2 ) ] [ 3 − 1 4 − 2 ] = [ − 6 + 0 − 12 + 0 2 − 2 4 − 4 ] [ 3 − 1 4 − 2 ] = [ − 6 − 12 0 0 ] [ 3 − 1 4 − 2 ] = [ − 6 × 3 + 0 × ( − 1 ) − 12 × 3 + 0 × ( − 1 ) − 6 × 4 + 0 × ( − 2 ) − 12 × 4 + 0 × ( − 2 ) ] = [ − 18 + 0 − 36 + 0 − 24 + 0 − 48 + 0 ] = [ − 18 − 36 − 24 − 48 ]
Hence, the matrix ACB = [ − 18 − 24 − 36 − 48 ] , \begin{bmatrix} -18 & -24 \\ -36 & -48 \end{bmatrix} , [ − 18 − 36 − 24 − 48 ] ,
Evaluate : [ 4 sin 30° 2 cos 60° sin 90° 2 cos 0° ] [ 4 5 5 4 ] . \begin{bmatrix} \text{4 sin 30° } & \text {2 cos 60°} \\ \text{ sin 90° } & \text{ 2 cos 0°} \end{bmatrix} \begin{bmatrix} 4 & 5 \\ 5 & 4 \end{bmatrix}. [ 4 sin 30° sin 90° 2 cos 60° 2 cos 0° ] [ 4 5 5 4 ] .
Answer
[ 4 sin 30° 2 cos 60° sin 90° 2 cos 0° ] [ 4 5 5 4 ] Since, sin 30° = cos 60° = 1 2 , sin 90° = cos 0° = 1. ⇒ [ 4 × 1 2 2 × 1 2 1 2 × 1 ] [ 4 5 5 4 ] = [ 2 1 1 2 ] [ 4 5 5 4 ] = [ 2 × 4 + 1 × 5 2 × 5 + 1 × 4 1 × 4 + 2 × 5 1 × 5 + 2 × 4 ] = [ 8 + 5 10 + 4 4 + 10 5 + 8 ] = [ 13 14 14 13 ] . \begin{bmatrix} \text{4 sin 30° } & \text {2 cos 60°} \\ \text{ sin 90° } & \text{ 2 cos 0°} \end{bmatrix} \begin{bmatrix} 4 & 5 \\ 5 & 4 \end{bmatrix} \\[1em] \text{ Since, sin 30° = cos 60° } = \dfrac{1}{2}, \text { sin 90° = cos 0° = 1.} \\[1em] \Rightarrow \begin{bmatrix} 4 \times \dfrac{1}{2} & 2 \times \dfrac{1}{2} \\ 1 & 2 \times 1 \end{bmatrix} \begin{bmatrix} 4 & 5 \\ 5 & 4 \end{bmatrix} \\[1em] = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} 4 & 5 \\ 5 & 4 \end{bmatrix} \\[1em] = \begin{bmatrix} 2 \times 4 + 1 \times 5 & 2 \times 5 + 1 \times 4 \\ 1 \times 4 + 2 \times 5 & 1 \times 5 + 2 \times 4 \end{bmatrix} \\[1em] = \begin{bmatrix} 8 + 5 & 10 + 4 \\ 4 + 10 & 5 + 8 \end{bmatrix} \\[1em] = \begin{bmatrix} 13 & 14 \\ 14 & 13 \end{bmatrix} . [ 4 sin 30° sin 90° 2 cos 60° 2 cos 0° ] [ 4 5 5 4 ] Since, sin 30° = cos 60° = 2 1 , sin 90° = cos 0° = 1. ⇒ [ 4 × 2 1 1 2 × 2 1 2 × 1 ] [ 4 5 5 4 ] = [ 2 1 1 2 ] [ 4 5 5 4 ] = [ 2 × 4 + 1 × 5 1 × 4 + 2 × 5 2 × 5 + 1 × 4 1 × 5 + 2 × 4 ] = [ 8 + 5 4 + 10 10 + 4 5 + 8 ] = [ 13 14 14 13 ] .
Hence, the resultant matrix is [ 13 14 14 13 ] . \begin{bmatrix} 13 & 14 \\ 14 & 13 \end{bmatrix}. [ 13 14 14 13 ] .
If A = [ − 1 3 2 4 ] and B = [ 2 − 3 − 4 − 6 ] , \begin{bmatrix*}[r] -1 & 3 \\ 2 & 4 \end{bmatrix*} \text{ and B } = \begin{bmatrix*}[r] 2 & -3 \\ -4 & -6 \end{bmatrix*}, [ − 1 2 3 4 ] and B = [ 2 − 4 − 3 − 6 ] ,
Answer
AB = [ − 1 3 2 4 ] [ 2 − 3 − 4 − 6 ] = [ − 1 × 2 + 3 × − 4 − 1 × ( − 3 ) + 3 × ( − 6 ) 2 × 2 + 4 × ( − 4 ) 2 × ( − 3 ) + 4 × ( − 6 ) ] = [ − 2 − 12 3 − 18 4 − 16 − 6 − 24 ] = [ − 14 − 15 − 12 − 30 ] BA = [ 2 − 3 − 4 − 6 ] [ − 1 3 2 4 ] = [ 2 × ( − 1 ) + ( − 3 ) × 2 2 × 3 + ( − 3 ) × 4 ( − 4 ) × ( − 1 ) + ( − 6 ) × 2 ( − 4 ) × 3 + ( − 6 ) × 4 ] = [ − 2 − 6 6 − 12 4 − 12 − 12 − 24 ] = [ − 8 − 6 − 8 − 36 ] Given, AB + BA = [ − 14 − 15 − 12 − 30 ] + [ − 8 − 6 − 8 − 36 ] = [ − 14 + ( − 8 ) − 15 + ( − 6 ) − 12 + ( − 8 ) − 30 + ( − 36 ) ] = [ − 22 − 21 − 20 − 66 ] . \text{ AB } = \begin{bmatrix*}[r] -1 & 3 \\ 2 & 4 \end{bmatrix*} \begin{bmatrix*}[r] 2 & -3 \\ -4 & -6 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -1 \times 2 + 3 \times -4 & -1 \times (-3) + 3 \times (-6) \\ 2 \times 2 + 4 \times (-4) & 2 \times (-3) + 4 \times (-6) \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -2 - 12 & 3 - 18 \\ 4 - 16 & -6 - 24 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -14 & -15 \\ -12 & -30 \end{bmatrix*} \\[1em] \text{BA } = \begin{bmatrix*}[r] 2 & -3 \\ -4 & -6 \end{bmatrix*} \begin{bmatrix*}[r] -1 & 3 \\ 2 & 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 \times (-1) + (-3) \times 2 & 2 \times 3 + (-3) \times 4 \\ (-4) \times (-1) + (-6) \times 2 & (-4) \times 3 + (-6) \times 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -2 - 6 & 6 - 12 \\ 4 - 12 & -12 - 24 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -8 & -6 \\ -8 & -36 \end{bmatrix*} \\[1em] \text{Given, AB + BA } = \begin{bmatrix*}[r] -14 & -15 \\ -12 & -30 \end{bmatrix*} + \begin{bmatrix*}[r] -8 & -6 \\ -8 & -36 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -14 + (-8) & -15 + (-6) \\ -12 + (-8) & -30 + (-36) \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -22 & -21 \\ -20 & -66 \end{bmatrix*}. AB = [ − 1 2 3 4 ] [ 2 − 4 − 3 − 6 ] = [ − 1 × 2 + 3 × − 4 2 × 2 + 4 × ( − 4 ) − 1 × ( − 3 ) + 3 × ( − 6 ) 2 × ( − 3 ) + 4 × ( − 6 ) ] = [ − 2 − 12 4 − 16 3 − 18 − 6 − 24 ] = [ − 14 − 12 − 15 − 30 ] BA = [ 2 − 4 − 3 − 6 ] [ − 1 2 3 4 ] = [ 2 × ( − 1 ) + ( − 3 ) × 2 ( − 4 ) × ( − 1 ) + ( − 6 ) × 2 2 × 3 + ( − 3 ) × 4 ( − 4 ) × 3 + ( − 6 ) × 4 ] = [ − 2 − 6 4 − 12 6 − 12 − 12 − 24 ] = [ − 8 − 8 − 6 − 36 ] Given, AB + BA = [ − 14 − 12 − 15 − 30 ] + [ − 8 − 8 − 6 − 36 ] = [ − 14 + ( − 8 ) − 12 + ( − 8 ) − 15 + ( − 6 ) − 30 + ( − 36 ) ] = [ − 22 − 20 − 21 − 66 ] .
Hence, the matrix AB + BA = [ − 22 − 21 − 20 − 66 ] . \begin{bmatrix*}[r] -22 & -21 \\ -20 & -66 \end{bmatrix*}. [ − 22 − 20 − 21 − 66 ] .
If A = [ 1 − 2 2 − 1 ] and B = [ 3 2 − 2 1 ] , \begin{bmatrix*}[r] 1 & -2 \\ 2 & -1 \end{bmatrix*} \text{and B } = \begin{bmatrix*}[r] 3 & 2 \\ -2 & 1 \end{bmatrix*}, [ 1 2 − 2 − 1 ] and B = [ 3 − 2 2 1 ] , 2 .
Answer
2 B = 2 [ 3 2 − 2 1 ] = [ 6 4 − 4 2 ] A 2 = [ 1 − 2 2 − 1 ] [ 1 − 2 2 − 1 ] = [ 1 × 1 + ( − 2 ) × 2 1 × ( − 2 ) + ( − 2 ) × ( − 1 ) 2 × 1 + ( − 1 ) × 2 2 × ( − 2 ) + ( − 1 ) × ( − 1 ) ] = [ 1 − 4 − 2 + 2 2 − 2 − 4 + 1 ] = [ − 3 0 0 − 3 ] ∴ 2 B − A 2 = [ 6 4 − 4 2 ] − [ − 3 0 0 − 3 ] = [ 6 − ( − 3 ) 4 − 0 − 4 − 0 2 − ( − 3 ) ] = [ 9 4 − 4 5 ] 2B = 2\begin{bmatrix*}[r] 3 & 2 \\ -2 & 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 6 & 4 \\ -4 & 2 \end{bmatrix*} \\[1em] A^2 = \begin{bmatrix*}[r] 1 & -2 \\ 2 & -1 \end{bmatrix*} \begin{bmatrix*}[r] 1 & -2 \\ 2 & -1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 \times 1 + (-2) \times 2 & 1 \times (-2) + (-2) \times (-1) \\ 2 \times 1 + (-1) \times 2 & 2 \times (-2) + (-1)\times (-1) \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 - 4 & -2 + 2 \\ 2 - 2 & -4 + 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -3 & 0 \\ 0 & -3 \end{bmatrix*} \\[1em] \therefore 2B - A^2 = \begin{bmatrix*}[r] 6 & 4 \\ -4 & 2 \end{bmatrix*} - \begin{bmatrix*}[r] -3 & 0 \\ 0 & -3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 6 - (-3) & 4 - 0 \\ -4 - 0 & 2 - (-3) \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 9 & 4 \\ -4 & 5 \end{bmatrix*} 2 B = 2 [ 3 − 2 2 1 ] = [ 6 − 4 4 2 ] A 2 = [ 1 2 − 2 − 1 ] [ 1 2 − 2 − 1 ] = [ 1 × 1 + ( − 2 ) × 2 2 × 1 + ( − 1 ) × 2 1 × ( − 2 ) + ( − 2 ) × ( − 1 ) 2 × ( − 2 ) + ( − 1 ) × ( − 1 ) ] = [ 1 − 4 2 − 2 − 2 + 2 − 4 + 1 ] = [ − 3 0 0 − 3 ] ∴ 2 B − A 2 = [ 6 − 4 4 2 ] − [ − 3 0 0 − 3 ] = [ 6 − ( − 3 ) − 4 − 0 4 − 0 2 − ( − 3 ) ] = [ 9 − 4 4 5 ]
Hence, the matrix 2B - A2 = [ 9 4 − 4 5 ] . \begin{bmatrix*}[r] 9 & 4 \\ -4 & 5 \end{bmatrix*}. [ 9 − 4 4 5 ] .
If A = [ 1 2 3 4 ] , B = [ 2 1 4 2 ] and C = [ 5 1 7 4 ] , \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}, \text{ B } = \begin{bmatrix} 2 & 1 \\ 4 & 2 \end{bmatrix} \text{ and C } = \begin{bmatrix} 5 & 1 \\ 7 & 4 \end{bmatrix}, [ 1 3 2 4 ] , B = [ 2 4 1 2 ] and C = [ 5 7 1 4 ] ,
(i) A(B + C)
(ii) (B + C)A
Answer
(i) A(B + C)
B + C = [ 2 1 4 2 ] + [ 5 1 7 4 ] = [ 2 + 5 1 + 1 4 + 7 2 + 4 ] = [ 7 2 11 6 ] ∴ A(B + C) = [ 1 2 3 4 ] [ 7 2 11 6 ] = [ 1 × 7 + 2 × 11 1 × 2 + 2 × 6 3 × 7 + 4 × 11 3 × 2 + 4 × 6 ] = [ 7 + 22 2 + 12 21 + 44 6 + 24 ] = [ 29 14 65 30 ] . \text{B + C } = \begin{bmatrix} 2 & 1 \\ 4 & 2 \end{bmatrix} + \begin{bmatrix} 5 & 1 \\ 7 & 4 \end{bmatrix} \\[1em] = \begin{bmatrix} 2 + 5 & 1 + 1 \\ 4 + 7 & 2 + 4 \end{bmatrix} \\[1em] = \begin{bmatrix} 7 & 2 \\ 11 & 6 \end{bmatrix} \\[1em] \therefore \text{ A(B + C) } = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 7 & 2 \\ 11 & 6 \end{bmatrix} \\[1em] = \begin{bmatrix} 1 \times 7 + 2 \times 11 & 1 \times 2 + 2 \times 6 \\ 3 \times 7 + 4 \times 11 & 3 \times 2 + 4 \times 6 \end{bmatrix} \\[1em] = \begin{bmatrix} 7 + 22 & 2 + 12 \\ 21 + 44 & 6 + 24 \end{bmatrix} \\[1em] = \begin{bmatrix} 29 & 14 \\ 65 & 30 \end{bmatrix}. \\[1em] B + C = [ 2 4 1 2 ] + [ 5 7 1 4 ] = [ 2 + 5 4 + 7 1 + 1 2 + 4 ] = [ 7 11 2 6 ] ∴ A(B + C) = [ 1 3 2 4 ] [ 7 11 2 6 ] = [ 1 × 7 + 2 × 11 3 × 7 + 4 × 11 1 × 2 + 2 × 6 3 × 2 + 4 × 6 ] = [ 7 + 22 21 + 44 2 + 12 6 + 24 ] = [ 29 65 14 30 ] .
Hence, the matrix A(B + C) = [ 29 14 65 30 ] . \begin{bmatrix} 29 & 14 \\ 65 & 30 \end{bmatrix}. [ 29 65 14 30 ] .
(ii) (B + C)A
B + C = [ 2 1 4 2 ] + [ 5 1 7 4 ] = [ 2 + 5 1 + 1 4 + 7 2 + 4 ] = [ 7 2 11 6 ] ∴ (B + C)A = [ 7 2 11 6 ] [ 1 2 3 4 ] = [ 7 × 1 + 2 × 3 7 × 2 + 2 × 4 11 × 1 + 6 × 3 11 × 2 + 6 × 4 ] = [ 7 + 6 14 + 8 11 + 18 22 + 24 ] = [ 13 22 29 46 ] . \text{B + C } = \begin{bmatrix} 2 & 1 \\ 4 & 2 \end{bmatrix} + \begin{bmatrix} 5 & 1 \\ 7 & 4 \end{bmatrix} \\[1em] = \begin{bmatrix} 2 + 5 & 1 + 1 \\ 4 + 7 & 2 + 4 \end{bmatrix} \\[1em] = \begin{bmatrix} 7 & 2 \\ 11 & 6 \end{bmatrix} \\[1em] \therefore \text{(B + C)A } = \begin{bmatrix} 7 & 2 \\ 11 & 6 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \\[1em] = \begin{bmatrix} 7 \times 1 + 2 \times 3 & 7 \times 2 + 2 \times 4 \\ 11 \times 1 + 6 \times 3 & 11 \times 2 + 6 \times 4 \end{bmatrix} \\[1em] = \begin{bmatrix} 7 + 6 & 14 + 8 \\ 11 + 18 & 22 + 24 \end{bmatrix} \\[1em] = \begin{bmatrix} 13 & 22 \\ 29 & 46 \end{bmatrix}. B + C = [ 2 4 1 2 ] + [ 5 7 1 4 ] = [ 2 + 5 4 + 7 1 + 1 2 + 4 ] = [ 7 11 2 6 ] ∴ (B + C)A = [ 7 11 2 6 ] [ 1 3 2 4 ] = [ 7 × 1 + 2 × 3 11 × 1 + 6 × 3 7 × 2 + 2 × 4 11 × 2 + 6 × 4 ] = [ 7 + 6 11 + 18 14 + 8 22 + 24 ] = [ 13 29 22 46 ] .
Hence, the matrix (B + C)A = [ 13 22 29 46 ] . \begin{bmatrix} 13 & 22 \\ 29 & 46 \end{bmatrix}. [ 13 29 22 46 ] .
If A = [ 1 2 2 3 ] , B = [ 2 1 3 2 ] and C = [ 1 3 3 1 ] , \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix}, \text{ B } = \begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix} \text{ and C } = \begin{bmatrix} 1 & 3 \\ 3 & 1 \end{bmatrix}, [ 1 2 2 3 ] , B = [ 2 3 1 2 ] and C = [ 1 3 3 1 ] ,
Answer
B − A = [ 2 1 3 2 ] − [ 1 2 2 3 ] = [ 2 − 1 1 − 2 3 − 2 2 − 3 ] = [ 1 − 1 1 − 1 ] ∴ C ( B − A ) = [ 1 3 3 1 ] [ 1 − 1 1 − 1 ] = [ 1 × 1 + 3 × 1 1 × ( − 1 ) + 3 × ( − 1 ) 3 × 1 + 1 × 1 3 × ( − 1 ) + 1 × ( − 1 ) ] = [ 1 + 3 − 1 − 3 3 + 1 − 3 − 1 ] = [ 4 − 4 4 − 4 ] . B - A = \begin{bmatrix*}[r] 2 & 1 \\ 3 & 2 \end{bmatrix*} - \begin{bmatrix*}[r] 1 & 2 \\ 2 & 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 - 1 & 1 - 2 \\ 3 - 2 & 2 - 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 & -1 \\ 1 & -1 \end{bmatrix*} \\[1.5em] \therefore C(B - A) = \begin{bmatrix*}[r] 1 & 3 \\ 3 & 1 \end{bmatrix*} \begin{bmatrix*}[r] 1 & -1 \\ 1 & -1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 \times 1 + 3 \times 1 & 1 \times (-1) + 3 \times (-1) \\ 3 \times 1 + 1 \times 1 & 3 \times (-1) + 1 \times (-1) \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 + 3 & -1 - 3 \\ 3 + 1 & -3 - 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 4 & -4 \\ 4 & -4 \end{bmatrix*}. B − A = [ 2 3 1 2 ] − [ 1 2 2 3 ] = [ 2 − 1 3 − 2 1 − 2 2 − 3 ] = [ 1 1 − 1 − 1 ] ∴ C ( B − A ) = [ 1 3 3 1 ] [ 1 1 − 1 − 1 ] = [ 1 × 1 + 3 × 1 3 × 1 + 1 × 1 1 × ( − 1 ) + 3 × ( − 1 ) 3 × ( − 1 ) + 1 × ( − 1 ) ] = [ 1 + 3 3 + 1 − 1 − 3 − 3 − 1 ] = [ 4 4 − 4 − 4 ] .
Hence, the matrix C(B - A) = [ 4 − 4 4 − 4 ] \begin{bmatrix} 4 & -4 \\ 4 & -4 \end{bmatrix} [ 4 4 − 4 − 4 ]
Let A = [ 1 0 2 1 ] and B = [ 2 3 − 1 0 ] , \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} \text{ and B } = \begin{bmatrix*}[r] 2 & 3 \\ -1 & 0 \end{bmatrix*}, [ 1 2 0 1 ] and B = [ 2 − 1 3 0 ] , 2 + AB + B2 .
Answer
A 2 = [ 1 0 2 1 ] [ 1 0 2 1 ] = [ 1 × 1 + 0 × 2 1 × 0 + 0 × 1 2 × 1 + 1 × 2 2 × 0 + 1 × 1 ] = [ 1 + 0 0 + 0 2 + 2 0 + 1 ] = [ 1 0 4 1 ] A B = [ 1 0 2 1 ] [ 2 3 − 1 0 ] = [ 1 × 2 + 0 × ( − 1 ) 1 × 3 + 0 × 0 2 × 2 + 1 × ( − 1 ) 2 × 3 + 1 × 0 ] = [ 2 + 0 3 + 0 4 − 1 6 + 0 ] = [ 2 3 3 6 ] B 2 = [ 2 3 − 1 0 ] [ 2 3 − 1 0 ] = [ 2 × 2 + 3 × ( − 1 ) 2 × 3 + 3 × 0 ( − 1 ) × 2 + 0 × ( − 1 ) ( − 1 ) × 3 + 0 × 0 ] = [ 4 − 3 6 + 0 − 2 + 0 − 3 + 0 ] = [ 1 6 − 2 − 3 ] . ∴ A 2 + A B + B 2 = [ 1 0 4 1 ] + [ 2 3 3 6 ] + [ 1 6 − 2 − 3 ] = [ 1 + 2 + 1 0 + 3 + 6 4 + 3 + ( − 2 ) 1 + 6 + ( − 3 ) ] = [ 4 9 5 4 ] . A^2 = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} \\[1em] = \begin{bmatrix} 1 \times 1 + 0 \times 2 & 1 \times 0 + 0 \times 1 \\ 2 \times 1 + 1 \times 2 & 2 \times 0 + 1 \times 1 \end{bmatrix} \\[1em] = \begin{bmatrix} 1 + 0 & 0 + 0 \\ 2 + 2 & 0 + 1 \end{bmatrix} \\[1em] = \begin{bmatrix} 1 & 0 \\ 4 & 1 \end{bmatrix} \\[1em] AB = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} \begin{bmatrix*}[r] 2 & 3 \\ -1 & 0 \end{bmatrix*} \\[1em] = \begin{bmatrix} 1 \times 2 + 0 \times (-1) & 1 \times 3 + 0 \times 0 \\ 2 \times 2 + 1 \times (-1) & 2 \times 3 + 1 \times 0 \end{bmatrix} \\[1em] = \begin{bmatrix*}[r] 2 + 0 & 3 + 0 \\ 4 - 1 & 6 + 0 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 & 3 \\ 3 & 6 \end{bmatrix*} \\[1em] B^2 = \begin{bmatrix*}[r] 2 & 3 \\ -1 & 0 \end{bmatrix*} \begin{bmatrix*}[r] 2 & 3 \\ -1 & 0 \end{bmatrix*} \\[1em] = \begin{bmatrix} 2 \times 2 + 3 \times (-1) & 2 \times 3 + 3 \times 0 \\ (-1) \times 2 + 0 \times (-1) & (-1) \times 3 + 0 \times 0 \end{bmatrix} \\[1em] = \begin{bmatrix*}[r] 4 - 3 & 6 + 0 \\ -2 + 0 & -3 + 0 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 & 6 \\ -2 & -3 \end{bmatrix*}. \\[1em] \therefore A^2 + AB + B^2 = \begin{bmatrix} 1 & 0 \\ 4 & 1 \end{bmatrix} + \begin{bmatrix*}[r] 2 & 3 \\ 3 & 6 \end{bmatrix*} + \begin{bmatrix*}[r] 1 & 6 \\ -2 & -3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 + 2 + 1 & 0 + 3 + 6 \\ 4 + 3 + (-2) & 1 + 6 + (-3) \end{bmatrix*} \\[1em] = \begin{bmatrix} 4 & 9 \\ 5 & 4 \end{bmatrix}. A 2 = [ 1 2 0 1 ] [ 1 2 0 1 ] = [ 1 × 1 + 0 × 2 2 × 1 + 1 × 2 1 × 0 + 0 × 1 2 × 0 + 1 × 1 ] = [ 1 + 0 2 + 2 0 + 0 0 + 1 ] = [ 1 4 0 1 ] A B = [ 1 2 0 1 ] [ 2 − 1 3 0 ] = [ 1 × 2 + 0 × ( − 1 ) 2 × 2 + 1 × ( − 1 ) 1 × 3 + 0 × 0 2 × 3 + 1 × 0 ] = [ 2 + 0 4 − 1 3 + 0 6 + 0 ] = [ 2 3 3 6 ] B 2 = [ 2 − 1 3 0 ] [ 2 − 1 3 0 ] = [ 2 × 2 + 3 × ( − 1 ) ( − 1 ) × 2 + 0 × ( − 1 ) 2 × 3 + 3 × 0 ( − 1 ) × 3 + 0 × 0 ] = [ 4 − 3 − 2 + 0 6 + 0 − 3 + 0 ] = [ 1 − 2 6 − 3 ] . ∴ A 2 + A B + B 2 = [ 1 4 0 1 ] + [ 2 3 3 6 ] + [ 1 − 2 6 − 3 ] = [ 1 + 2 + 1 4 + 3 + ( − 2 ) 0 + 3 + 6 1 + 6 + ( − 3 ) ] = [ 4 5 9 4 ] .
Hence, the matrix A2 + AB + B2 = [ 4 9 5 4 ] \begin{bmatrix} 4 & 9 \\ 5 & 4 \end{bmatrix} [ 4 5 9 4 ]
If A = [ 3 0 5 1 ] and B [ − 4 2 1 0 ] \begin{bmatrix*}[r] 3 & 0 \\ 5 & 1 \end{bmatrix*} \text{ and B} \begin{bmatrix*}[r] -4 & 2 \\ 1 & 0 \end{bmatrix*} [ 3 5 0 1 ] and B [ − 4 1 2 0 ] 2 - 2AB + B2 .
Answer
Given, A = [ 3 0 5 1 ] and B = [ − 4 2 1 0 ] \begin{bmatrix*}[r] 3 & 0 \\ 5 & 1 \end{bmatrix*} \text{ and B} = \begin{bmatrix*}[r] -4 & 2 \\ 1 & 0 \end{bmatrix*} [ 3 5 0 1 ] and B = [ − 4 1 2 0 ]
⇒ A 2 = [ 3 0 5 1 ] × [ 3 0 5 1 ] = [ 3 × 3 + 0 × 5 3 × 0 + 0 × 1 5 × 3 + 1 × 5 5 × 0 + 1 × 1 ] = [ 9 + 0 0 + 0 15 + 5 0 + 1 ] = [ 9 0 20 1 ] ⇒ 2AB = 2 × [ 3 0 5 1 ] × [ − 4 2 1 0 ] = [ 6 0 10 2 ] × [ − 4 2 1 0 ] = [ 6 × ( − 4 ) + 0 × 1 6 × 2 + 0 × 0 10 × ( − 4 ) + 2 × 1 10 × 2 + 2 × 0 ] = [ − 24 + 0 12 + 0 − 40 + 2 20 + 0 ] = [ − 24 12 − 38 20 ] ⇒ B 2 = [ − 4 2 1 0 ] × [ − 4 2 1 0 ] = [ − 4 × ( − 4 ) + 2 × 1 − 4 × 2 + 2 × 0 1 × ( − 4 ) + 0 × 1 1 × 2 + 0 × 0 ] = [ 16 + 2 − 8 + 0 − 4 + 0 2 + 0 ] = [ 18 − 8 − 4 2 ] \Rightarrow A^2 =\begin{bmatrix*}[r] 3 & 0 \\ 5 & 1 \end{bmatrix*} \times \begin{bmatrix*}[r] 3 & 0 \\ 5 & 1 \end{bmatrix*}\\[1em] = \begin{bmatrix*}[r] 3 \times 3 + 0 \times 5 & 3 \times 0 + 0 \times 1 \\ 5 \times 3 + 1 \times 5 & 5 \times 0 + 1 \times 1 \end{bmatrix*}\\[1em] = \begin{bmatrix*}[r] 9 + 0 & 0 + 0 \\ 15 + 5 & 0 + 1 \end{bmatrix*}\\[1em] = \begin{bmatrix*}[r] 9 & 0 \\ 20 & 1 \end{bmatrix*}\\[1em] \Rightarrow \text{2AB} = 2 \times \begin{bmatrix*}[r] 3 & 0 \\ 5 & 1 \end{bmatrix*} \times \begin{bmatrix*}[r] -4 & 2 \\ 1 & 0 \end{bmatrix*}\\[1em] = \begin{bmatrix*}[r] 6 & 0 \\ 10 & 2 \end{bmatrix*} \times \begin{bmatrix*}[r] -4 & 2 \\ 1 & 0 \end{bmatrix*}\\[1em] = \begin{bmatrix*}[r] 6 \times (-4) + 0 \times 1 & 6 \times 2 + 0 \times 0 \\ 10 \times (-4) + 2 \times 1 & 10 \times 2 + 2 \times 0 \end{bmatrix*}\\[1em] = \begin{bmatrix*}[r] -24 + 0 & 12 + 0 \\ -40 + 2 & 20 + 0 \end{bmatrix*}\\[1em] = \begin{bmatrix*}[r] -24 & 12 \\ -38 & 20 \end{bmatrix*}\\[1em] \Rightarrow \text{B}^2 = \begin{bmatrix*}[r] -4 & 2 \\ 1 & 0 \end{bmatrix*} \times \begin{bmatrix*}[r] -4 & 2 \\ 1 & 0 \end{bmatrix*}\\[1em] = \begin{bmatrix*}[r] -4 \times (-4) + 2 \times 1 & -4 \times 2 + 2 \times 0 \\ 1 \times (-4) + 0 \times 1 & 1 \times 2 + 0 \times 0 \end{bmatrix*}\\[1em] = \begin{bmatrix*}[r] 16 + 2 & -8 + 0 \\ -4 + 0 & 2 + 0 \end{bmatrix*}\\[1em] = \begin{bmatrix*}[r] 18 & -8 \\ -4 & 2 \end{bmatrix*} ⇒ A 2 = [ 3 5 0 1 ] × [ 3 5 0 1 ] = [ 3 × 3 + 0 × 5 5 × 3 + 1 × 5 3 × 0 + 0 × 1 5 × 0 + 1 × 1 ] = [ 9 + 0 15 + 5 0 + 0 0 + 1 ] = [ 9 20 0 1 ] ⇒ 2AB = 2 × [ 3 5 0 1 ] × [ − 4 1 2 0 ] = [ 6 10 0 2 ] × [ − 4 1 2 0 ] = [ 6 × ( − 4 ) + 0 × 1 10 × ( − 4 ) + 2 × 1 6 × 2 + 0 × 0 10 × 2 + 2 × 0 ] = [ − 24 + 0 − 40 + 2 12 + 0 20 + 0 ] = [ − 24 − 38 12 20 ] ⇒ B 2 = [ − 4 1 2 0 ] × [ − 4 1 2 0 ] = [ − 4 × ( − 4 ) + 2 × 1 1 × ( − 4 ) + 0 × 1 − 4 × 2 + 2 × 0 1 × 2 + 0 × 0 ] = [ 16 + 2 − 4 + 0 − 8 + 0 2 + 0 ] = [ 18 − 4 − 8 2 ]
Substituting values in A2 - 2AB + B2 , we get :
A 2 − 2 A B + B 2 = [ 9 0 20 1 ] − [ − 24 12 − 38 20 ] + [ 18 − 8 − 4 2 ] = [ 9 − ( − 24 ) + 18 0 − 12 + ( − 8 ) 20 − ( − 38 ) + ( − 4 ) 1 − 20 + 2 ] = [ 51 − 20 54 − 17 ] A^2 - 2AB + B^2 = \begin{bmatrix*}[r] 9 & 0 \\ 20 & 1 \end{bmatrix*} - \begin{bmatrix*}[r] -24 & 12 \\ -38 & 20 \end{bmatrix*} + \begin{bmatrix*}[r] 18 & -8 \\ -4 & 2 \end{bmatrix*}\\[1em] = \begin{bmatrix*}[r] 9 - (-24) + 18 & 0 - 12 + (-8) \\ 20 - (-38) + (-4) & 1 - 20 + 2 \end{bmatrix*}\\[1em] = \begin{bmatrix*}[r] 51 & -20 \\ 54 & -17 \end{bmatrix*} A 2 − 2 A B + B 2 = [ 9 20 0 1 ] − [ − 24 − 38 12 20 ] + [ 18 − 4 − 8 2 ] = [ 9 − ( − 24 ) + 18 20 − ( − 38 ) + ( − 4 ) 0 − 12 + ( − 8 ) 1 − 20 + 2 ] = [ 51 54 − 20 − 17 ]
Hence, the value of A2 - 2AB + B2 = [ 51 − 20 54 − 17 ] \begin{bmatrix*}[r] 51 & -20 \\ 54 & -17 \end{bmatrix*} [ 51 54 − 20 − 17 ]
Let A = [ 2 1 0 − 2 ] , B = [ 4 1 − 3 − 2 ] and C = [ − 3 2 − 1 4 ] , \begin{bmatrix*}[r] 2 & 1 \\ 0 & -2 \end{bmatrix*}, \text{ B } = \begin{bmatrix*}[r] 4 & 1 \\ -3 & -2 \end{bmatrix*} \text{ and C } = \begin{bmatrix*}[r] -3 & 2 \\ -1 & 4 \end{bmatrix*}, [ 2 0 1 − 2 ] , B = [ 4 − 3 1 − 2 ] and C = [ − 3 − 1 2 4 ] , 2 + AC - 5B.
Answer
A 2 = [ 2 1 0 − 2 ] [ 2 1 0 − 2 ] = [ 2 × 2 + 1 × 0 2 × 1 + 1 × ( − 2 ) 0 × 2 + ( − 2 ) × 0 0 × 1 + ( − 2 ) × ( − 2 ) ] = [ 4 + 0 2 − 2 0 + 0 0 + 4 ] = [ 4 0 0 4 ] . A C = [ 2 1 0 − 2 ] [ − 3 2 − 1 4 ] = [ 2 × ( − 3 ) + 1 × ( − 1 ) 2 × 2 + 1 × 4 0 × ( − 3 ) + ( − 2 ) × ( − 1 ) 0 × 2 + ( − 2 ) × 4 ] = [ − 6 + ( − 1 ) 4 + 4 0 + 2 0 + ( − 8 ) ] = [ − 7 8 2 − 8 ] . 5 B = 5 [ 4 1 − 3 − 2 ] = [ 20 5 − 15 − 10 ] ∴ A 2 + A C − 5 B = [ 4 0 0 4 ] + [ − 7 8 2 − 8 ] − [ 20 5 − 15 − 10 ] = [ 4 + ( − 7 ) − 20 0 + 8 − 5 0 + 2 − ( − 15 ) 4 + ( − 8 ) − ( − 10 ) ] = [ − 23 3 17 6 ] . A^2 = \begin{bmatrix*}[r] 2 & 1 \\ 0 & -2 \end{bmatrix*} \begin{bmatrix*}[r] 2 & 1 \\ 0 & -2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 \times 2 + 1 \times 0 & 2 \times 1 + 1 \times (-2) \\ 0 \times 2 + (-2) \times 0 & 0 \times 1 + (-2) \times (-2) \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 4 + 0 & 2 - 2 \\ 0 + 0 & 0 + 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 4 & 0 \\ 0 & 4 \end{bmatrix*}. \\[1.5em] AC = \begin{bmatrix*}[r] 2 & 1 \\ 0 & -2 \end{bmatrix*} \begin{bmatrix*}[r] -3 & 2 \\ -1 & 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 \times (-3) + 1 \times (-1) & 2 \times 2 + 1 \times 4 \\ 0 \times (-3) + (-2) \times (-1) & 0 \times 2 + (-2) \times 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -6 + (-1) & 4 + 4 \\ 0 + 2 & 0 + (-8) \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -7 & 8 \\ 2 & -8 \end{bmatrix*}. \\[1.5em] 5B = 5 \begin{bmatrix*}[r] 4 & 1 \\ -3 & -2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 20 & 5 \\ -15 & -10 \end{bmatrix*} \\[1.5em] \therefore A^2 + AC - 5B = \begin{bmatrix*}[r] 4 & 0 \\ 0 & 4 \end{bmatrix*} + \begin{bmatrix*}[r] -7 & 8 \\ 2 & -8 \end{bmatrix*} - \begin{bmatrix*}[r] 20 & 5 \\ -15 & -10 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 4 + (-7) - 20 & 0 + 8 - 5 \\ 0 + 2 - (-15) & 4 + (-8) - (-10) \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -23 & 3 \\ 17 & 6 \end{bmatrix*} . A 2 = [ 2 0 1 − 2 ] [ 2 0 1 − 2 ] = [ 2 × 2 + 1 × 0 0 × 2 + ( − 2 ) × 0 2 × 1 + 1 × ( − 2 ) 0 × 1 + ( − 2 ) × ( − 2 ) ] = [ 4 + 0 0 + 0 2 − 2 0 + 4 ] = [ 4 0 0 4 ] . A C = [ 2 0 1 − 2 ] [ − 3 − 1 2 4 ] = [ 2 × ( − 3 ) + 1 × ( − 1 ) 0 × ( − 3 ) + ( − 2 ) × ( − 1 ) 2 × 2 + 1 × 4 0 × 2 + ( − 2 ) × 4 ] = [ − 6 + ( − 1 ) 0 + 2 4 + 4 0 + ( − 8 ) ] = [ − 7 2 8 − 8 ] . 5 B = 5 [ 4 − 3 1 − 2 ] = [ 20 − 15 5 − 10 ] ∴ A 2 + A C − 5 B = [ 4 0 0 4 ] + [ − 7 2 8 − 8 ] − [ 20 − 15 5 − 10 ] = [ 4 + ( − 7 ) − 20 0 + 2 − ( − 15 ) 0 + 8 − 5 4 + ( − 8 ) − ( − 10 ) ] = [ − 23 17 3 6 ] .
Hence, the matrix A2 + AC - 5B = [ − 23 3 17 6 ] . \begin{bmatrix*}[r] -23 & 3 \\ 17 & 6 \end{bmatrix*} . [ − 23 17 3 6 ] .
If A = [ 2 3 5 7 ] , B = [ 0 4 − 1 7 ] and C = [ 1 0 − 1 4 ] , \begin{bmatrix*}[r] 2 & 3 \\ 5 & 7 \end{bmatrix*}, \text{ B } = \begin{bmatrix*}[r] 0 & 4 \\ -1 & 7 \end{bmatrix*} \text{ and C } = \begin{bmatrix*}[r] 1 & 0 \\ -1 & 4 \end{bmatrix*}, [ 2 5 3 7 ] , B = [ 0 − 1 4 7 ] and C = [ 1 − 1 0 4 ] , 2 - 10C.
Answer
A C = [ 2 3 5 7 ] [ 1 0 − 1 4 ] = [ 2 × 1 + 3 × ( − 1 ) 2 × 0 + 3 × 4 5 × 1 + 7 × ( − 1 ) 5 × 0 + 7 × 4 ] = [ 2 − 3 0 + 12 5 − 7 0 + 28 ] = [ − 1 12 − 2 28 ] . AC = \begin{bmatrix*}[r] 2 & 3 \\ 5 & 7 \end{bmatrix*} \begin{bmatrix*}[r] 1 & 0 \\ -1 & 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 \times 1 + 3 \times (-1) & 2 \times 0 + 3 \times 4 \\ 5 \times 1 + 7 \times (-1) & 5 \times 0 + 7 \times 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 - 3 & 0 + 12 \\ 5 - 7 & 0 + 28 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -1 & 12 \\ -2 & 28 \end{bmatrix*}. A C = [ 2 5 3 7 ] [ 1 − 1 0 4 ] = [ 2 × 1 + 3 × ( − 1 ) 5 × 1 + 7 × ( − 1 ) 2 × 0 + 3 × 4 5 × 0 + 7 × 4 ] = [ 2 − 3 5 − 7 0 + 12 0 + 28 ] = [ − 1 − 2 12 28 ] .
B 2 = [ 0 4 − 1 7 ] [ 0 4 − 1 7 ] = [ 0 × 0 + 4 × ( − 1 ) 0 × 4 + 4 × 7 ( − 1 ) × 0 + 7 × ( − 1 ) ( − 1 ) × 4 + 7 × 7 ] = [ 0 − 4 0 + 28 0 − 7 − 4 + 49 ] = [ − 4 28 − 7 45 ] . B^2 = \begin{bmatrix*}[r] 0 & 4 \\ -1 & 7 \end{bmatrix*} \begin{bmatrix*}[r] 0 & 4 \\ -1 & 7 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 0 \times 0 + 4 \times (-1) & 0 \times 4 + 4 \times 7 \\ (-1) \times 0 + 7 \times (-1) & (-1) \times 4 + 7 \times 7 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 0 - 4 & 0 + 28 \\ 0 - 7 & -4 + 49 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -4 & 28 \\ -7 & 45 \end{bmatrix*}. B 2 = [ 0 − 1 4 7 ] [ 0 − 1 4 7 ] = [ 0 × 0 + 4 × ( − 1 ) ( − 1 ) × 0 + 7 × ( − 1 ) 0 × 4 + 4 × 7 ( − 1 ) × 4 + 7 × 7 ] = [ 0 − 4 0 − 7 0 + 28 − 4 + 49 ] = [ − 4 − 7 28 45 ] .
10 C = 10 [ 1 0 − 1 4 ] = [ 10 0 − 10 40 ] . ∴ A C + B 2 − 10 C = [ − 1 12 − 2 28 ] + [ − 4 28 − 7 45 ] − [ 10 0 − 10 40 ] = [ − 1 + ( − 4 ) − 10 12 + 28 − 0 − 2 + ( − 7 ) − ( − 10 ) 28 + 45 − 40 ] = [ − 15 40 1 33 ] 10C = 10 \begin{bmatrix*}[r] 1 & 0 \\ -1 & 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 10 & 0 \\ -10 & 40 \end{bmatrix*}. \\[1em] \therefore AC + B^2 - 10C = \begin{bmatrix*}[r] -1 & 12 \\ -2 & 28 \end{bmatrix*} + \begin{bmatrix*}[r] -4 & 28 \\ -7 & 45 \end{bmatrix*} - \begin{bmatrix*}[r] 10 & 0 \\ -10 & 40 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -1 + (-4) - 10 & 12 + 28 - 0 \\ -2 + (-7) - (-10) & 28 + 45 - 40 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -15 & 40 \\ 1 & 33 \end{bmatrix*} 10 C = 10 [ 1 − 1 0 4 ] = [ 10 − 10 0 40 ] . ∴ A C + B 2 − 10 C = [ − 1 − 2 12 28 ] + [ − 4 − 7 28 45 ] − [ 10 − 10 0 40 ] = [ − 1 + ( − 4 ) − 10 − 2 + ( − 7 ) − ( − 10 ) 12 + 28 − 0 28 + 45 − 40 ] = [ − 15 1 40 33 ]
Hence, the matrix AC + B2 - 10C = [ − 15 40 1 33 ] \begin{bmatrix*}[r] -15 & 40 \\ 1 & 33 \end{bmatrix*} [ − 15 1 40 33 ]
If A = [ 1 0 0 − 1 ] , \begin{bmatrix*}[r] 1 & 0 \\ 0 & -1 \end{bmatrix*}, [ 1 0 0 − 1 ] , 2 and A3 . Also state which of these is equal to A.
Answer
A 2 = [ 1 0 0 − 1 ] [ 1 0 0 − 1 ] = [ 1 × 1 + 0 × 0 1 × 0 + 0 × ( − 1 ) 0 × 1 + ( − 1 ) × 0 0 × 0 + ( − 1 ) × ( − 1 ) ] = [ 1 0 0 1 ] A^2 = \begin{bmatrix*}[r] 1 & 0 \\ 0 & -1 \end{bmatrix*} \begin{bmatrix*}[r] 1 & 0 \\ 0 & -1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 \times 1 + 0 \times 0 & 1 \times 0 + 0 \times (-1) \\ 0 \times 1 + (-1) \times 0 & 0 \times 0 + (-1) \times (-1) \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} A 2 = [ 1 0 0 − 1 ] [ 1 0 0 − 1 ] = [ 1 × 1 + 0 × 0 0 × 1 + ( − 1 ) × 0 1 × 0 + 0 × ( − 1 ) 0 × 0 + ( − 1 ) × ( − 1 ) ] = [ 1 0 0 1 ]
A 3 = A 2 × A = [ 1 0 0 1 ] [ 1 0 0 − 1 ] = [ 1 × 1 + 0 × 0 1 × 0 + 0 × ( − 1 ) 0 × 1 + 1 × 0 0 × 0 + 1 × ( − 1 ) ] [ 1 + 0 0 + 0 0 + 0 0 − 1 ] = [ 1 0 0 − 1 ] A^3 = A^2 \times A \\[1em] = \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \begin{bmatrix*}[r] 1 & 0 \\ 0 & -1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 \times 1 + 0 \times 0 & 1 \times 0 + 0 \times (-1) \\ 0 \times 1 + 1 \times 0 & 0 \times 0 + 1 \times (-1) \end{bmatrix*} \\[1em] \begin{bmatrix*}[r] 1 + 0 & 0 + 0 \\ 0 + 0 & 0 -1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 & 0 \\ 0 & -1 \end{bmatrix*} A 3 = A 2 × A = [ 1 0 0 1 ] [ 1 0 0 − 1 ] = [ 1 × 1 + 0 × 0 0 × 1 + 1 × 0 1 × 0 + 0 × ( − 1 ) 0 × 0 + 1 × ( − 1 ) ] [ 1 + 0 0 + 0 0 + 0 0 − 1 ] = [ 1 0 0 − 1 ]
Hence, the matrix A 2 = [ 1 0 0 1 ] and A 3 = [ 1 0 0 − 1 ] . A^2 = \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \text{ and } A^3 = \begin{bmatrix*}[r] 1 & 0 \\ 0 & -1 \end{bmatrix*}. A 2 = [ 1 0 0 1 ] and A 3 = [ 1 0 0 − 1 ] . 3 = A.
If X = [ 4 1 − 1 2 ] , \begin{bmatrix*}[r] 4 & 1 \\ -1 & 2 \end{bmatrix*}, [ 4 − 1 1 2 ] , 2 = 9I where I is the unit matrix.
Answer
We have to prove 6X - X2 = 9I,
L.H.S. = 6 X − X 2 6 X − X 2 = 6 [ 4 1 − 1 2 ] − [ 4 1 − 1 2 ] [ 4 1 − 1 2 ] = [ 24 6 − 6 12 ] − [ 4 × 4 + 1 × ( − 1 ) 4 × 1 + 1 × 2 ( − 1 ) × 4 + 2 × ( − 1 ) ( − 1 ) × 1 + 2 × 2 ] = [ 24 6 − 6 12 ] − [ 16 − 1 4 + 2 − 4 − 2 − 1 + 4 ] = [ 24 6 − 6 12 ] − [ 15 6 − 6 3 ] = [ 24 − 15 6 − 6 − 6 + 6 12 − 3 ] = [ 9 0 0 9 ] \text{L.H.S. } = 6X - X^2 \\[1em] 6X - X^2 = 6\begin{bmatrix*}[r] 4 & 1 \\ -1 & 2 \end{bmatrix*} - \begin{bmatrix*}[r] 4 & 1 \\ -1 & 2 \end{bmatrix*}\begin{bmatrix*}[r] 4 & 1 \\ -1 & 2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 24 & 6 \\ -6 & 12 \end{bmatrix*} - \begin{bmatrix*}[r] 4 \times 4 + 1 \times (-1) & 4 \times 1 + 1 \times 2 \\ (-1) \times 4 + 2 \times (-1) & (-1) \times 1 + 2 \times 2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 24 & 6 \\ -6 & 12 \end{bmatrix*} - \begin{bmatrix*}[r] 16 - 1 & 4 + 2 \\ -4 - 2 & -1 + 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 24 & 6 \\ -6 & 12 \end{bmatrix*} - \begin{bmatrix*}[r] 15 & 6 \\ -6 & 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 24 - 15 & 6 - 6 \\ -6 + 6 & 12 - 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 9 & 0 \\ 0 & 9 \end{bmatrix*} L.H.S. = 6 X − X 2 6 X − X 2 = 6 [ 4 − 1 1 2 ] − [ 4 − 1 1 2 ] [ 4 − 1 1 2 ] = [ 24 − 6 6 12 ] − [ 4 × 4 + 1 × ( − 1 ) ( − 1 ) × 4 + 2 × ( − 1 ) 4 × 1 + 1 × 2 ( − 1 ) × 1 + 2 × 2 ] = [ 24 − 6 6 12 ] − [ 16 − 1 − 4 − 2 4 + 2 − 1 + 4 ] = [ 24 − 6 6 12 ] − [ 15 − 6 6 3 ] = [ 24 − 15 − 6 + 6 6 − 6 12 − 3 ] = [ 9 0 0 9 ]
R.H.S. = 9 I 9 I = 9 [ 1 0 0 1 ] = [ 9 0 0 9 ] . \text{R.H.S. } = 9I \\[1em] 9I = 9\begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 9 & 0 \\ 0 & 9 \end{bmatrix*}. R.H.S. = 9 I 9 I = 9 [ 1 0 0 1 ] = [ 9 0 0 9 ] .
Since, L.H.S. = [ 9 0 0 9 ] \begin{bmatrix*}[r] 9 & 0 \\ 0 & 9 \end{bmatrix*} [ 9 0 0 9 ] 2 = 9I.
Show that [ 1 2 2 1 ] \begin{bmatrix*}[r] 1 & 2 \\ 2 & 1 \end{bmatrix*} [ 1 2 2 1 ] 2 - 2X - 3I = 0 where I is the unit matrix of order 2.
Answer
I = [ 1 0 0 1 ] , X = [ 1 2 2 1 ] I = \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*}, X = \begin{bmatrix*}[r] 1 & 2 \\ 2 & 1 \end{bmatrix*} \\[1em] I = [ 1 0 0 1 ] , X = [ 1 2 2 1 ]
Given,
X2 - 2X - 3I = 0
Putting value of X and I in above equation we get,
L.H.S. = [ 1 2 2 1 ] [ 1 2 2 1 ] − 2 [ 1 2 2 1 ] − 3 [ 1 0 0 1 ] = [ 1 × 1 + 2 × 2 1 × 2 + 2 × 1 2 × 1 + 1 × 2 2 × 2 + 1 × 1 ] − [ 2 4 4 2 ] − [ 3 0 0 3 ] [ 1 + 4 2 + 2 2 + 2 4 + 1 ] − [ 2 4 4 2 ] − [ 3 0 0 3 ] [ 5 4 4 5 ] − [ 2 4 4 2 ] − [ 3 0 0 3 ] = [ 5 − 2 − 3 4 − 4 − 0 4 − 4 − 0 5 − 2 − 3 ] = [ 0 0 0 0 ] \text{L.H.S. } = \begin{bmatrix*}[r] 1 & 2 \\ 2 & 1 \end{bmatrix*} \begin{bmatrix*}[r] 1 & 2 \\ 2 & 1 \end{bmatrix*} - 2 \begin{bmatrix*}[r] 1 & 2 \\ 2 & 1 \end{bmatrix*} - 3 \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 \times 1 + 2 \times 2 & 1 \times 2 + 2 \times 1 \\ 2 \times 1 + 1 \times 2 & 2 \times 2 + 1 \times 1 \end{bmatrix*} - \begin{bmatrix*}[r] 2 & 4 \\ 4 & 2 \end{bmatrix*} - \begin{bmatrix*}[r] 3 & 0 \\ 0 & 3 \end{bmatrix*} \\[1em] \begin{bmatrix*}[r] 1 + 4 & 2 + 2 \\ 2 + 2 & 4 + 1 \end{bmatrix*} - \begin{bmatrix*}[r] 2 & 4 \\ 4 & 2 \end{bmatrix*} - \begin{bmatrix*}[r] 3 & 0 \\ 0 & 3 \end{bmatrix*} \\[1em] \begin{bmatrix*}[r] 5 & 4 \\ 4 & 5 \end{bmatrix*} - \begin{bmatrix*}[r] 2 & 4 \\ 4 & 2 \end{bmatrix*} - \begin{bmatrix*}[r] 3 & 0 \\ 0 & 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 5 - 2 - 3 & 4 - 4 - 0 \\ 4 - 4 - 0 & 5 - 2 - 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*} L.H.S. = [ 1 2 2 1 ] [ 1 2 2 1 ] − 2 [ 1 2 2 1 ] − 3 [ 1 0 0 1 ] = [ 1 × 1 + 2 × 2 2 × 1 + 1 × 2 1 × 2 + 2 × 1 2 × 2 + 1 × 1 ] − [ 2 4 4 2 ] − [ 3 0 0 3 ] [ 1 + 4 2 + 2 2 + 2 4 + 1 ] − [ 2 4 4 2 ] − [ 3 0 0 3 ] [ 5 4 4 5 ] − [ 2 4 4 2 ] − [ 3 0 0 3 ] = [ 5 − 2 − 3 4 − 4 − 0 4 − 4 − 0 5 − 2 − 3 ] = [ 0 0 0 0 ]
Since, L.H.S. = [ 0 0 0 0 ] \begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*} [ 0 0 0 0 ] 2 - 2X -3I = 0.
∴[ 1 2 2 1 ] \begin{bmatrix*}[r] 1 & 2 \\ 2 & 1 \end{bmatrix*} [ 1 2 2 1 ] is a solution of the matrix equation X2 - 2X - 3I = 0
Find the matrix X of order 2 x 2 which satisfies the equation :
[ 3 7 2 4 ] [ 0 2 5 3 ] + 2 X = [ 1 − 5 − 4 6 ] . \begin{bmatrix*}[r] 3 & 7 \\ 2 & 4 \end{bmatrix*} \begin{bmatrix*}[r] 0 & 2 \\ 5 & 3 \end{bmatrix*} + 2X = \begin{bmatrix*}[r] 1 & -5 \\ -4 & 6 \end{bmatrix*}. [ 3 2 7 4 ] [ 0 5 2 3 ] + 2 X = [ 1 − 4 − 5 6 ] .
Answer
Given,
[ 3 7 2 4 ] [ 0 2 5 3 ] + 2 X = [ 1 − 5 − 4 6 ] , ⇒ [ 3 × 0 + 7 × 5 3 × 2 + 7 × 3 2 × 0 + 4 × 5 2 × 2 + 4 × 3 ] + 2 X = [ 1 − 5 − 4 6 ] ⇒ [ 0 + 35 6 + 21 0 + 20 4 + 12 ] + 2 X = [ 1 − 5 − 4 6 ] ⇒ [ 35 27 20 16 ] + 2 X = [ 1 − 5 − 4 6 ] ⇒ 2 X = [ 1 − 5 − 4 6 ] − [ 35 27 20 16 ] ⇒ 2 X = [ 1 − 35 − 5 − 27 − 4 − 20 6 − 16 ] ⇒ 2 X = [ − 34 − 32 − 24 − 10 ] ⇒ X = 1 2 [ − 34 − 32 − 24 − 10 ] X = [ − 17 − 16 − 12 − 5 ] . \begin{bmatrix*}[r] 3 & 7 \\ 2 & 4 \end{bmatrix*} \begin{bmatrix*}[r] 0 & 2 \\ 5 & 3 \end{bmatrix*} + 2X = \begin{bmatrix*}[r] 1 & -5 \\ -4 & 6 \end{bmatrix*}, \\[1em] \Rightarrow \begin{bmatrix*}[r] 3 \times 0 + 7 \times 5 & 3 \times 2 + 7 \times 3 \\ 2 \times 0 + 4 \times 5 & 2 \times 2 + 4 \times 3 \end{bmatrix*} + 2X = \begin{bmatrix*}[r] 1 & -5 \\ -4 & 6 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 0 + 35 & 6 + 21 \\ 0 + 20 & 4 + 12 \end{bmatrix*} + 2X = \begin{bmatrix*}[r] 1 & -5 \\ -4 & 6 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 35 & 27 \\ 20 & 16 \end{bmatrix*} + 2X = \begin{bmatrix*}[r] 1 & -5 \\ -4 & 6 \end{bmatrix*} \\[1em] \Rightarrow 2X = \begin{bmatrix*}[r] 1 & -5 \\ -4 & 6 \end{bmatrix*} - \begin{bmatrix*}[r] 35 & 27 \\ 20 & 16 \end{bmatrix*} \\[1em] \Rightarrow 2X = \begin{bmatrix*}[r] 1 - 35 & -5 - 27 \\ -4 - 20 & 6 - 16 \end{bmatrix*} \\[1em] \Rightarrow 2X = \begin{bmatrix*}[r] -34 & -32 \\ -24 & -10 \end{bmatrix*} \\[1em] \Rightarrow X = \dfrac{1}{2} \begin{bmatrix*}[r] -34 & -32 \\ -24 & -10 \end{bmatrix*} \\[1em] X = \begin{bmatrix*}[r] -17 & -16 \\ -12 & -5 \end{bmatrix*}. [ 3 2 7 4 ] [ 0 5 2 3 ] + 2 X = [ 1 − 4 − 5 6 ] , ⇒ [ 3 × 0 + 7 × 5 2 × 0 + 4 × 5 3 × 2 + 7 × 3 2 × 2 + 4 × 3 ] + 2 X = [ 1 − 4 − 5 6 ] ⇒ [ 0 + 35 0 + 20 6 + 21 4 + 12 ] + 2 X = [ 1 − 4 − 5 6 ] ⇒ [ 35 20 27 16 ] + 2 X = [ 1 − 4 − 5 6 ] ⇒ 2 X = [ 1 − 4 − 5 6 ] − [ 35 20 27 16 ] ⇒ 2 X = [ 1 − 35 − 4 − 20 − 5 − 27 6 − 16 ] ⇒ 2 X = [ − 34 − 24 − 32 − 10 ] ⇒ X = 2 1 [ − 34 − 24 − 32 − 10 ] X = [ − 17 − 12 − 16 − 5 ] .
Hence, the matrix X = [ − 17 − 16 − 12 − 5 ] \begin{bmatrix*}[r] -17 & -16 \\ -12 & -5 \end{bmatrix*} [ − 17 − 12 − 16 − 5 ]
If A = [ 1 1 x x ] , \begin{bmatrix*}[r] 1 & 1 \\ x & x \end{bmatrix*}, [ 1 x 1 x ] , 2 = O.
Answer
Given, A2 = O.
or , [ 1 1 x x ] [ 1 1 x x ] = [ 0 0 0 0 ] L.H.S. = [ 1 1 x x ] [ 1 1 x x ] = [ 1 × 1 + 1 × x 1 × 1 + 1 × x x × 1 + x × x x × 1 + x × x ] = [ 1 + x 1 + x x + x 2 x + x 2 ] \text{or}, \begin{bmatrix*}[r] 1 & 1 \\ x & x \end{bmatrix*} \begin{bmatrix*}[r] 1 & 1 \\ x & x \end{bmatrix*} = \begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*} \\[1em] \text{L.H.S.} = \begin{bmatrix*}[r] 1 & 1 \\ x & x \end{bmatrix*} \begin{bmatrix*}[r] 1 & 1 \\ x & x \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 \times 1 + 1 \times x & 1 \times 1 + 1 \times x \\ x \times 1 + x \times x & x \times 1 + x \times x \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 + x & 1 + x \\ x + x^2 & x + x^2 \end{bmatrix*} or , [ 1 x 1 x ] [ 1 x 1 x ] = [ 0 0 0 0 ] L.H.S. = [ 1 x 1 x ] [ 1 x 1 x ] = [ 1 × 1 + 1 × x x × 1 + x × x 1 × 1 + 1 × x x × 1 + x × x ] = [ 1 + x x + x 2 1 + x x + x 2 ]
Comparing with R.H.S. we get,
⇒ 1 + x = 0 or x = -1 (...Eq 1)
⇒ x + x2 = 0 (...Eq 2)
Putting the value x = -1 in equation 2,
⇒ -1 + (-1)2 = -1 + 1 = 0.
Since, x = -1 satisfies the equation,
Hence, the required value of x = -1.
Find x and y, if [ 2 x x y 3 y ] [ 3 2 ] = [ 16 9 ] . \begin{bmatrix*}[r] 2x & x \\ y & 3y \end{bmatrix*} \begin{bmatrix*}[r] 3 \\ 2 \end{bmatrix*} = \begin{bmatrix*}[r] 16 \\ 9 \end{bmatrix*}. [ 2 x y x 3 y ] [ 3 2 ] = [ 16 9 ] .
Answer
Given,
[ 2 x x y 3 y ] [ 3 2 ] = [ 16 9 ] L.H.S. = [ 2 x x y 3 y ] [ 3 2 ] = [ 2 x × 3 + x × 2 y × 3 + 3 y × 2 ] = [ 6 x + 2 x 3 y + 6 y ] = [ 8 x 9 y ] \begin{bmatrix*}[r] 2x & x \\ y & 3y \end{bmatrix*} \begin{bmatrix*}[r] 3 \\ 2 \end{bmatrix*} = \begin{bmatrix*}[r] 16 \\ 9 \end{bmatrix*} \\[1em] \text{L.H.S.} = \begin{bmatrix*}[r] 2x & x \\ y & 3y \end{bmatrix*} \begin{bmatrix*}[r] 3 \\ 2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2x \times 3 + x \times 2 \\ y \times 3 + 3y \times 2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 6x + 2x \\ 3y + 6y \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 8x \\ 9y \end{bmatrix*} [ 2 x y x 3 y ] [ 3 2 ] = [ 16 9 ] L.H.S. = [ 2 x y x 3 y ] [ 3 2 ] = [ 2 x × 3 + x × 2 y × 3 + 3 y × 2 ] = [ 6 x + 2 x 3 y + 6 y ] = [ 8 x 9 y ]
Comparing L.H.S. with R.H.S. we get,
⇒ [ 8 x 9 y ] = [ 16 9 ] ⇒ 8 x = 16 and 9 y = 9 ∴ x = 2 and y = 1. \Rightarrow \begin{bmatrix*}[r] 8x \\ 9y \end{bmatrix*} = \begin{bmatrix*}[r] 16 \\ 9 \end{bmatrix*} \\[1em] \Rightarrow 8x = 16 \text{ and } 9y = 9 \\[1em] \therefore x = 2 \text{ and } y = 1. ⇒ [ 8 x 9 y ] = [ 16 9 ] ⇒ 8 x = 16 and 9 y = 9 ∴ x = 2 and y = 1.
Hence, the value of x = 2 and y = 1.
Find the values of x and y if [ x + y y 2 x x − y ] [ 2 − 1 ] = [ 3 2 ] . \begin{bmatrix*}[r] x + y & y \\ 2x & x - y \end{bmatrix*} \begin{bmatrix*}[r] 2 \\ -1 \end{bmatrix*} = \begin{bmatrix*}[r] 3 \\ 2 \end{bmatrix*}. [ x + y 2 x y x − y ] [ 2 − 1 ] = [ 3 2 ] .
Answer
Given,
[ x + y y 2 x x − y ] [ 2 − 1 ] = [ 3 2 ] ⇒ [ ( x + y ) × 2 + y × ( − 1 ) 2 x × 2 + ( x − y ) × ( − 1 ) ] = [ 3 2 ] ⇒ [ 2 x + 2 y − y 4 x − x + y ] = [ 3 2 ] ⇒ [ 2 x + y 3 x + y ] = [ 3 2 ] \begin{bmatrix*}[r] x + y & y \\ 2x & x - y \end{bmatrix*} \begin{bmatrix*}[r] 2 \\ -1 \end{bmatrix*} = \begin{bmatrix*}[r] 3 \\ 2 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] (x + y) \times 2 + y \times (-1) \\ 2x \times 2 + (x - y) \times (-1) \end{bmatrix*} = \begin{bmatrix*}[r] 3 \\ 2 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2x + 2y - y \\ 4x - x + y \end{bmatrix*} = \begin{bmatrix*}[r] 3 \\ 2 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2x + y \\ 3x + y \end{bmatrix*} = \begin{bmatrix*}[r] 3 \\ 2 \end{bmatrix*} \\[1em] [ x + y 2 x y x − y ] [ 2 − 1 ] = [ 3 2 ] ⇒ [ ( x + y ) × 2 + y × ( − 1 ) 2 x × 2 + ( x − y ) × ( − 1 ) ] = [ 3 2 ] ⇒ [ 2 x + 2 y − y 4 x − x + y ] = [ 3 2 ] ⇒ [ 2 x + y 3 x + y ] = [ 3 2 ]
By definition of equality of matrices we have,
⇒ 2x + y = 3 or y = 3 - 2x (...Eq 1)
Putting value of y from equation 1 in equation 2,
⇒ 3x + 3 - 2x = 2
Now finding value of y,
⇒ y = 3 - 2x
Hence, the value of x = -1 and y = 5.
If [ 1 2 3 3 ] [ x 0 0 y ] = [ x 0 9 0 ] , \begin{bmatrix*}[r] 1 & 2 \\ 3 & 3 \end{bmatrix*} \begin{bmatrix*}[r] x & 0 \\ 0 & y \end{bmatrix*} = \begin{bmatrix*}[r] x & 0 \\ 9 & 0 \end{bmatrix*}, [ 1 3 2 3 ] [ x 0 0 y ] = [ x 9 0 0 ] ,
Answer
Given,
[ 1 2 3 3 ] [ x 0 0 y ] = [ x 0 9 0 ] ⇒ [ 1 × x + 2 × 0 1 × 0 + 2 × y 3 × x + 3 × 0 3 × 0 + 3 × y ] = [ x 0 9 0 ] ⇒ [ x + 0 0 + 2 y 3 x + 0 0 + 3 y ] = [ x 0 9 0 ] ⇒ [ x 2 y 3 x 3 y ] = [ x 0 9 0 ] \begin{bmatrix*}[r] 1 & 2 \\ 3 & 3 \end{bmatrix*} \begin{bmatrix*}[r] x & 0 \\ 0 & y \end{bmatrix*} = \begin{bmatrix*}[r] x & 0 \\ 9 & 0 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 1 \times x + 2 \times 0 & 1 \times 0 + 2 \times y \\ 3 \times x + 3 \times 0 & 3 \times 0 + 3 \times y \end{bmatrix*} = \begin{bmatrix*}[r] x & 0 \\ 9 & 0 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] x + 0 & 0 + 2y \\ 3x + 0 & 0 + 3y \end{bmatrix*} = \begin{bmatrix*}[r] x & 0 \\ 9 & 0 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] x & 2y \\ 3x & 3y \end{bmatrix*} = \begin{bmatrix*}[r] x & 0 \\ 9 & 0 \end{bmatrix*} \\[1em] [ 1 3 2 3 ] [ x 0 0 y ] = [ x 9 0 0 ] ⇒ [ 1 × x + 2 × 0 3 × x + 3 × 0 1 × 0 + 2 × y 3 × 0 + 3 × y ] = [ x 9 0 0 ] ⇒ [ x + 0 3 x + 0 0 + 2 y 0 + 3 y ] = [ x 9 0 0 ] ⇒ [ x 3 x 2 y 3 y ] = [ x 9 0 0 ]
By definition of equality of matrices,
⇒ 2y = 0, 3x = 9 and 3y = 0
Hence, the values are x = 3 and y = 0.
If [ 3 4 2 5 ] = [ a b c d ] [ 1 0 0 1 ] , \begin{bmatrix*}[r] 3 & 4 \\ 2 & 5 \end{bmatrix*} = \begin{bmatrix*}[r] a & b \\ c & d \end{bmatrix*} \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*}, [ 3 2 4 5 ] = [ a c b d ] [ 1 0 0 1 ] ,
Answer
Given,
[ 3 4 2 5 ] = [ a b c d ] [ 1 0 0 1 ] ⇒ [ 3 4 2 5 ] = [ a × 1 + b × 0 a × 0 + b × 1 c × 1 + d × 0 c × 0 + d × 1 ] ⇒ [ 3 4 2 5 ] = [ a b c d ] \begin{bmatrix*}[r] 3 & 4 \\ 2 & 5 \end{bmatrix*} = \begin{bmatrix*}[r] a & b \\ c & d \end{bmatrix*} \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 3 & 4 \\ 2 & 5 \end{bmatrix*} = \begin{bmatrix*}[r] a \times 1 + b \times 0 & a \times 0 + b \times 1 \\ c \times 1 + d \times 0 & c \times 0 + d \times 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 3 & 4 \\ 2 & 5 \end{bmatrix*} = \begin{bmatrix*}[r] a & b \\ c & d \end{bmatrix*} [ 3 2 4 5 ] = [ a c b d ] [ 1 0 0 1 ] ⇒ [ 3 2 4 5 ] = [ a × 1 + b × 0 c × 1 + d × 0 a × 0 + b × 1 c × 0 + d × 1 ] ⇒ [ 3 2 4 5 ] = [ a c b d ]
By definition of equality of matrices,
⇒ a = 3, b = 4, c = 2 and d = 5
Hence, the value of a = 3, b = 4, c = 2 and d = 5.
Find the value of x given that A2 = B where,
A = [ 2 12 0 1 ] and B = [ 4 x 0 1 ] . A = \begin{bmatrix*}[r] 2 & 12 \\ 0 & 1 \end{bmatrix*} \text{ and } B = \begin{bmatrix*}[r] 4 & x \\ 0 & 1 \end{bmatrix*}. A = [ 2 0 12 1 ] and B = [ 4 0 x 1 ] .
Answer
Given A 2 = B , ⇒ [ 2 12 0 1 ] [ 2 12 0 1 ] = [ 4 x 0 1 ] ⇒ [ 2 × 2 + 12 × 0 2 × 12 + 12 × 1 0 × 2 + 1 × 0 0 × 12 + 1 × 1 ] = [ 4 x 0 1 ] ⇒ [ 4 + 0 24 + 12 0 + 0 0 + 1 ] = [ 4 x 0 1 ] ⇒ [ 4 36 0 1 ] = [ 4 x 0 1 ] \text{Given } A^2 = B, \\[1em] \Rightarrow \begin{bmatrix*}[r] 2 & 12 \\ 0 & 1 \end{bmatrix*} \begin{bmatrix*}[r] 2 & 12 \\ 0 & 1 \end{bmatrix*} = \begin{bmatrix*}[r] 4 & x \\ 0 & 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2 \times 2 + 12 \times 0 & 2 \times 12 + 12 \times 1 \\ 0 \times 2 + 1 \times 0 & 0 \times 12 + 1 \times 1 \end{bmatrix*} = \begin{bmatrix*}[r] 4 & x \\ 0 & 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 + 0 & 24 + 12 \\ 0 + 0 & 0 + 1 \end{bmatrix*} = \begin{bmatrix*}[r] 4 & x \\ 0 & 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 & 36 \\ 0 & 1 \end{bmatrix*} = \begin{bmatrix*}[r] 4 & x \\ 0 & 1 \end{bmatrix*} \\[1em] Given A 2 = B , ⇒ [ 2 0 12 1 ] [ 2 0 12 1 ] = [ 4 0 x 1 ] ⇒ [ 2 × 2 + 12 × 0 0 × 2 + 1 × 0 2 × 12 + 12 × 1 0 × 12 + 1 × 1 ] = [ 4 0 x 1 ] ⇒ [ 4 + 0 0 + 0 24 + 12 0 + 1 ] = [ 4 0 x 1 ] ⇒ [ 4 0 36 1 ] = [ 4 0 x 1 ]
By definition of equality of matrices,
⇒ x = 36.
Hence, the value of x = 36.
If A = [ 2 x 0 1 ] and B = [ 4 36 0 1 ] , \begin{bmatrix*}[r] 2 & x \\ 0 & 1 \end{bmatrix*} \text{and B } = \begin{bmatrix*}[r] 4 & 36 \\ 0 & 1 \end{bmatrix*}, [ 2 0 x 1 ] and B = [ 4 0 36 1 ] , 2 = B.
Answer
Given,
A2 = B
⇒ [ 2 x 0 1 ] [ 2 x 0 1 ] = [ 4 36 0 1 ] ⇒ [ 2 × 2 + x × 0 2 × x + x × 1 0 × 2 + 1 × 0 0 × x + 1 × 1 ] = [ 4 36 0 1 ] ⇒ [ 4 + 0 2 x + x 0 + 0 0 + 1 ] = [ 4 36 0 1 ] ⇒ [ 4 3 x 0 1 ] = [ 4 36 0 1 ] \Rightarrow \begin{bmatrix*}[r] 2 & x \\ 0 & 1 \end{bmatrix*} \begin{bmatrix*}[r] 2 & x \\ 0 & 1 \end{bmatrix*} = \begin{bmatrix*}[r] 4 & 36 \\ 0 & 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2 \times 2 + x \times 0 & 2 \times x + x \times 1 \\ 0 \times 2 + 1 \times 0 & 0 \times x + 1 \times 1 \end{bmatrix*} = \begin{bmatrix*}[r] 4 & 36 \\ 0 & 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 + 0 & 2x + x \\ 0 + 0 & 0 + 1 \end{bmatrix*} = \begin{bmatrix*}[r] 4 & 36 \\ 0 & 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 & 3x \\ 0 & 1 \end{bmatrix*} = \begin{bmatrix*}[r] 4 & 36 \\ 0 & 1 \end{bmatrix*} \\[1em] ⇒ [ 2 0 x 1 ] [ 2 0 x 1 ] = [ 4 0 36 1 ] ⇒ [ 2 × 2 + x × 0 0 × 2 + 1 × 0 2 × x + x × 1 0 × x + 1 × 1 ] = [ 4 0 36 1 ] ⇒ [ 4 + 0 0 + 0 2 x + x 0 + 1 ] = [ 4 0 36 1 ] ⇒ [ 4 0 3 x 1 ] = [ 4 0 36 1 ]
By definition of equality of matrices we get,
⇒ 3x = 36
Hence, the value of x = 12.
If A = [ 3 x 0 1 ] and B = [ 9 16 0 − y ] , \begin{bmatrix*}[r] 3 & x \\ 0 & 1 \end{bmatrix*} \text{ and B} = \begin{bmatrix*}[r] 9 & 16 \\ 0 & -y \end{bmatrix*}, [ 3 0 x 1 ] and B = [ 9 0 16 − y ] , 2 = B.
Answer
Given,
A2 = B
⇒ [ 3 x 0 1 ] [ 3 x 0 1 ] = [ 9 16 0 − y ] ⇒ [ 3 × 3 + x × 0 3 × x + x × 1 0 × 3 + 1 × 0 0 × x + 1 × 1 ] = [ 9 16 0 − y ] ⇒ [ 9 + 0 3 x + x 0 + 0 0 + 1 ] = [ 9 16 0 − y ] ⇒ [ 9 4 x 0 1 ] = [ 9 16 0 − y ] \Rightarrow \begin{bmatrix*}[r] 3 & x \\ 0 & 1 \end{bmatrix*} \begin{bmatrix*}[r] 3 & x \\ 0 & 1 \end{bmatrix*} = \begin{bmatrix*}[r] 9 & 16 \\ 0 & -y \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 3 \times 3 + x \times 0 & 3 \times x + x \times 1 \\ 0 \times 3 + 1 \times 0 & 0 \times x + 1 \times 1 \end{bmatrix*} = \begin{bmatrix*}[r] 9 & 16 \\ 0 & -y \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 9 + 0 & 3x + x \\ 0 + 0 & 0 + 1 \end{bmatrix*} = \begin{bmatrix*}[r] 9 & 16 \\ 0 & -y \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 9 & 4x \\ 0 & 1 \end{bmatrix*} = \begin{bmatrix*}[r] 9 & 16 \\ 0 & -y \end{bmatrix*} \\[1em] ⇒ [ 3 0 x 1 ] [ 3 0 x 1 ] = [ 9 0 16 − y ] ⇒ [ 3 × 3 + x × 0 0 × 3 + 1 × 0 3 × x + x × 1 0 × x + 1 × 1 ] = [ 9 0 16 − y ] ⇒ [ 9 + 0 0 + 0 3 x + x 0 + 1 ] = [ 9 0 16 − y ] ⇒ [ 9 0 4 x 1 ] = [ 9 0 16 − y ]
By definition of equality of matrices we get,
⇒ 4x = 16 and -y = 1
Hence, the values are x = 4 and y = -1.
Find x, y if [ − 2 0 3 1 ] [ − 1 2 x ] + 3 [ − 2 1 ] = 2 [ y 3 ] . \begin{bmatrix*}[r] -2 & 0 \\ 3 & 1 \end{bmatrix*} \begin{bmatrix*}[r] -1 \\ 2x \end{bmatrix*} + 3\begin{bmatrix*}[r] -2 \\ 1 \end{bmatrix*} = 2\begin{bmatrix*}[r] y \\ 3 \end{bmatrix*}. [ − 2 3 0 1 ] [ − 1 2 x ] + 3 [ − 2 1 ] = 2 [ y 3 ] .
Answer
Given,
[ − 2 0 3 1 ] [ − 1 2 x ] + 3 [ − 2 1 ] = 2 [ y 3 ] ⇒ [ ( − 2 ) × ( − 1 ) + 0 × 2 x 3 × ( − 1 ) + 1 × 2 x ] + [ − 6 3 ] = [ 2 y 6 ] ⇒ [ 2 − 3 + 2 x ] + [ − 6 3 ] = [ 2 y 6 ] ⇒ [ 2 + ( − 6 ) − 3 + 2 x + 3 ] = [ 2 y 6 ] ⇒ [ − 4 2 x ] = [ 2 y 6 ] \begin{bmatrix*}[r] -2 & 0 \\ 3 & 1 \end{bmatrix*} \begin{bmatrix*}[r] -1 \\ 2x \end{bmatrix*} + 3\begin{bmatrix*}[r] -2 \\ 1 \end{bmatrix*} = 2\begin{bmatrix*}[r] y \\ 3 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] (-2) \times (-1) + 0 \times 2x \\ 3 \times (-1) + 1 \times 2x \end{bmatrix*} + \begin{bmatrix*}[r] -6 \\ 3 \end{bmatrix*} = \begin{bmatrix*}[r] 2y \\ 6 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2 \\ -3 + 2x \end{bmatrix*} + \begin{bmatrix*}[r] -6 \\ 3 \end{bmatrix*} = \begin{bmatrix*}[r] 2y \\ 6 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2 + (-6) \\ -3 + 2x + 3 \end{bmatrix*} = \begin{bmatrix*}[r] 2y \\ 6 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] -4 \\ 2x \end{bmatrix*} = \begin{bmatrix*}[r] 2y \\ 6 \end{bmatrix*} \\[1em] [ − 2 3 0 1 ] [ − 1 2 x ] + 3 [ − 2 1 ] = 2 [ y 3 ] ⇒ [ ( − 2 ) × ( − 1 ) + 0 × 2 x 3 × ( − 1 ) + 1 × 2 x ] + [ − 6 3 ] = [ 2 y 6 ] ⇒ [ 2 − 3 + 2 x ] + [ − 6 3 ] = [ 2 y 6 ] ⇒ [ 2 + ( − 6 ) − 3 + 2 x + 3 ] = [ 2 y 6 ] ⇒ [ − 4 2 x ] = [ 2 y 6 ]
By definition of equality of matrices we get,
2y = -4 and 2x = 6
Hence, the values are x = 3 and y = -2.
If [ a 1 1 0 ] [ 4 3 − 3 2 ] = [ b 11 4 c ] , \begin{bmatrix*}[r] a & 1 \\ 1 & 0 \end{bmatrix*} \begin{bmatrix*}[r] 4 & 3 \\ -3 & 2 \end{bmatrix*} = \begin{bmatrix*}[r] b & 11 \\ 4 & c \end{bmatrix*}, [ a 1 1 0 ] [ 4 − 3 3 2 ] = [ b 4 11 c ] ,
Answer
Given,
[ a 1 1 0 ] [ 4 3 − 3 2 ] = [ b 11 4 c ] ⇒ [ a × 4 + 1 × ( − 3 ) a × 3 + 1 × 2 1 × 4 + 0 × ( − 3 ) 1 × 3 + 0 × 2 ] = [ b 11 4 c ] ⇒ [ 4 a − 3 3 a + 2 4 + 0 3 + 0 ] = [ b 11 4 c ] ⇒ [ 4 a − 3 3 a + 2 4 3 ] = [ b 11 4 c ] \begin{bmatrix*}[r] a & 1 \\ 1 & 0 \end{bmatrix*} \begin{bmatrix*}[r] 4 & 3 \\ -3 & 2 \end{bmatrix*} = \begin{bmatrix*}[r] b & 11 \\ 4 & c \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] a \times 4 + 1 \times (-3) & a \times 3 + 1 \times 2 \\ 1 \times 4 + 0 \times (-3) & 1 \times 3 + 0 \times 2 \end{bmatrix*} = \begin{bmatrix*}[r] b & 11 \\ 4 & c \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 4a - 3 & 3a + 2 \\ 4 + 0 & 3 + 0 \end{bmatrix*} = \begin{bmatrix*}[r] b & 11 \\ 4 & c \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 4a - 3 & 3a + 2 \\ 4 & 3 \end{bmatrix*} = \begin{bmatrix*}[r] b & 11 \\ 4 & c \end{bmatrix*} \\[1em] [ a 1 1 0 ] [ 4 − 3 3 2 ] = [ b 4 11 c ] ⇒ [ a × 4 + 1 × ( − 3 ) 1 × 4 + 0 × ( − 3 ) a × 3 + 1 × 2 1 × 3 + 0 × 2 ] = [ b 4 11 c ] ⇒ [ 4 a − 3 4 + 0 3 a + 2 3 + 0 ] = [ b 4 11 c ] ⇒ [ 4 a − 3 4 3 a + 2 3 ] = [ b 4 11 c ]
By definition of equality of matrices we get,
4a - 3 = b (...Eq 1)
Solving (Eq 2) first,
⇒ 3a + 2 = 11
Putting value of a in Eq 1,
⇒ 4a - 3 = b
∴ a = 3, b = 9 and c = 3.
Hence, the value of a = 3, b = 9 and c = 3.
If A = [ 2 3 1 2 ] , \begin{bmatrix*}[r] 2 & 3 \\ 1 & 2 \end{bmatrix*}, [ 2 1 3 2 ] , 2 = xA + yI.
Answer
Given,
A = [ 2 3 1 2 ] , I = [ 1 0 0 1 ] A 2 = x A + y I ⇒ [ 2 3 1 2 ] [ 2 3 1 2 ] = x [ 2 3 1 2 ] + y [ 1 0 0 1 ] ⇒ [ 2 × 2 + 3 × 1 2 × 3 + 3 × 2 1 × 2 + 2 × 1 1 × 3 + 2 × 2 ] = [ 2 x 3 x x 2 x ] + [ y 0 0 y ] ⇒ [ 4 + 3 6 + 6 2 + 2 3 + 4 ] = [ 2 x + y 3 x + 0 x + 0 2 x + y ] ⇒ [ 7 12 4 7 ] = [ 2 x + y 3 x + 0 x + 0 2 x + y ] ⇒ [ 7 12 4 7 ] = [ 2 x + y 3 x x 2 x + y ] \text{A } = \begin{bmatrix*}[r] 2 & 3 \\ 1 & 2 \end{bmatrix*}, \text{ I } = \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em] \text{A}^2 = x\text{A} + y\text{I } \\[1em] \Rightarrow \begin{bmatrix*}[r] 2 & 3 \\ 1 & 2 \end{bmatrix*} \begin{bmatrix*}[r] 2 & 3 \\ 1 & 2 \end{bmatrix*} = x \begin{bmatrix*}[r] 2 & 3 \\ 1 & 2 \end{bmatrix*} + y \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2 \times 2 + 3 \times 1 & 2 \times 3 + 3 \times 2 \\ 1 \times 2 + 2 \times 1 & 1 \times 3 + 2 \times 2 \end{bmatrix*} = \begin{bmatrix*}[r] 2x & 3x \\ x & 2x \end{bmatrix*} + \begin{bmatrix*}[r] y & 0 \\ 0 & y \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 + 3 & 6 + 6 \\ 2 + 2 & 3 + 4 \end{bmatrix*} = \begin{bmatrix*}[r] 2x + y & 3x + 0 \\ x + 0 & 2x + y \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 7 & 12 \\ 4 & 7 \end{bmatrix*} = \begin{bmatrix*}[r] 2x + y & 3x + 0 \\ x + 0 & 2x + y \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 7 & 12 \\ 4 & 7 \end{bmatrix*} = \begin{bmatrix*}[r] 2x + y & 3x \\ x & 2x + y \end{bmatrix*} \\[1em] A = [ 2 1 3 2 ] , I = [ 1 0 0 1 ] A 2 = x A + y I ⇒ [ 2 1 3 2 ] [ 2 1 3 2 ] = x [ 2 1 3 2 ] + y [ 1 0 0 1 ] ⇒ [ 2 × 2 + 3 × 1 1 × 2 + 2 × 1 2 × 3 + 3 × 2 1 × 3 + 2 × 2 ] = [ 2 x x 3 x 2 x ] + [ y 0 0 y ] ⇒ [ 4 + 3 2 + 2 6 + 6 3 + 4 ] = [ 2 x + y x + 0 3 x + 0 2 x + y ] ⇒ [ 7 4 12 7 ] = [ 2 x + y x + 0 3 x + 0 2 x + y ] ⇒ [ 7 4 12 7 ] = [ 2 x + y x 3 x 2 x + y ]
By definition of equality of matrices we get,
3x = 12 or x = 4 (...Eq 1)
2x + y = 7 (...Eq 2)
Putting value of x from Eq1 in Eq 2,
⇒ 2(4) + y = 7
Hence, the value of x = 4 and y = -1.
If P = [ 2 6 3 9 ] and Q = [ 3 x y 2 ] , \begin{bmatrix*}[r] 2 & 6 \\ 3 & 9 \end{bmatrix*} \text{ and Q } = \begin{bmatrix*}[r] 3 & x \\ y & 2 \end{bmatrix*}, [ 2 3 6 9 ] and Q = [ 3 y x 2 ] ,
Answer
Given,
PQ = 0
⇒ [ 2 6 3 9 ] [ 3 x y 2 ] = [ 0 0 0 0 ] ⇒ [ 2 × 3 + 6 × y 2 × x + 6 × 2 3 × 3 + 9 × y 3 × x + 9 × 2 ] = [ 0 0 0 0 ] ⇒ [ 6 + 6 y 2 x + 12 9 + 9 y 3 x + 18 ] = [ 0 0 0 0 ] \Rightarrow \begin{bmatrix*}[r] 2 & 6 \\ 3 & 9 \end{bmatrix*} \begin{bmatrix*}[r] 3 & x \\ y & 2 \end{bmatrix*} = \begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2 \times 3 + 6 \times y & 2 \times x + 6 \times 2 \\ 3 \times 3 + 9 \times y & 3 \times x + 9 \times 2 \end{bmatrix*} = \begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 6 + 6y & 2x + 12 \\ 9 + 9y & 3x + 18 \end{bmatrix*} = \begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*} \\[1em] ⇒ [ 2 3 6 9 ] [ 3 y x 2 ] = [ 0 0 0 0 ] ⇒ [ 2 × 3 + 6 × y 3 × 3 + 9 × y 2 × x + 6 × 2 3 × x + 9 × 2 ] = [ 0 0 0 0 ] ⇒ [ 6 + 6 y 9 + 9 y 2 x + 12 3 x + 18 ] = [ 0 0 0 0 ]
By definition of equality of matrices we get,
⇒ 6 + 6y = 0 and 2x + 12 = 0
Checking whether x = -6 and y = -1, satisfies other equations 9 + 9y = 0 and 3x + 18 = 0,
⇒ 9 + 9y = 0
⇒ 3x + 18 = 0
∴ x = -6 and y = -1.
Hence, the value of x = -6 and y = -1.
Let M × [ 1 1 0 2 ] = [ 1 2 ] \times \begin{bmatrix*}[r] 1 & 1 \\ 0 & 2 \end{bmatrix*} = \begin{bmatrix*}[r] 1 & 2 \end{bmatrix*} × [ 1 0 1 2 ] = [ 1 2 ]
(i) State the order of the matrix M.
(ii) Find the matrix M.
Answer
(i)
Since,
M × [ 1 1 0 2 ] = [ 1 2 ] ⇒ M × [ 1 1 0 2 ] is a 1 × 2 matrix, but [ 1 1 0 2 ] is a 2 × 2 matrix . ⇒ M is a 1 × 2 matrix . \text{M} \times \begin{bmatrix*}[r] 1 & 1 \\ 0 & 2 \end{bmatrix*} = \begin{bmatrix*}[r] 1 & 2 \end{bmatrix*} \\[1em] \Rightarrow \text{M} \times \begin{bmatrix*}[r] 1 & 1 \\ 0 & 2 \end{bmatrix*} \text{is a } 1 \times 2 \text{ matrix, but} \begin{bmatrix*}[r] 1 & 1 \\ 0 & 2 \end{bmatrix*} \text{ is a } 2 \times 2 \text{ matrix}. \\[1em] \Rightarrow \text{M is a } 1 \times 2 \text{ matrix}. M × [ 1 0 1 2 ] = [ 1 2 ] ⇒ M × [ 1 0 1 2 ] is a 1 × 2 matrix, but [ 1 0 1 2 ] is a 2 × 2 matrix . ⇒ M is a 1 × 2 matrix .
The order of matrix M is 1 × 2.
(ii)
Let M = [ x y ] Given, M × [ 1 1 0 2 ] = [ 1 2 ] ⇒ [ x y ] × [ 1 1 0 2 ] = [ 1 2 ] ⇒ [ x × 1 + y × 0 x × 1 + y × 2 ] = [ 1 2 ] ⇒ [ x x + 2 y ] = [ 1 2 ] \text{Let M =} \begin{bmatrix*}[r] x & y \end{bmatrix*} \\[1em] \text{Given, } \text{M} \times \begin{bmatrix*}[r] 1 & 1 \\ 0 & 2 \end{bmatrix*} = \begin{bmatrix*}[r] 1 & 2 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] x & y \end{bmatrix*} \times \begin{bmatrix*}[r] 1 & 1 \\ 0 & 2 \end{bmatrix*} = \begin{bmatrix*}[r] 1 & 2 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] x \times 1 + y \times 0 & x \times 1 + y \times 2 \\ \end{bmatrix*} = \begin{bmatrix*}[r] 1 & 2 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] x & x + 2y \\ \end{bmatrix*} = \begin{bmatrix*}[r] 1 & 2 \end{bmatrix*} \\[1em] Let M = [ x y ] Given, M × [ 1 0 1 2 ] = [ 1 2 ] ⇒ [ x y ] × [ 1 0 1 2 ] = [ 1 2 ] ⇒ [ x × 1 + y × 0 x × 1 + y × 2 ] = [ 1 2 ] ⇒ [ x x + 2 y ] = [ 1 2 ]
By definition of equality of matrices we get,
x = 1 (...Eq 1)
Putting value of x from Eq 1 in Eq 2,
⇒ x + 2y = 21 2 \dfrac{1}{2} 2 1
Since, M = [ x y ] ∴ M = [ 1 1 2 ] . \text{Since, M }= \begin{bmatrix*}[r] x & y \\ \end{bmatrix*} \\[1em] \therefore \text{M} = \begin{bmatrix*}[r] 1 & \dfrac{1}{2} \\ \end{bmatrix*}. Since, M = [ x y ] ∴ M = [ 1 2 1 ] .
Hence, the matrix M = [ 1 1 2 ] \begin{bmatrix*}[r] 1 & \dfrac{1}{2} \end{bmatrix*} [ 1 2 1 ]
Given [ 2 1 − 3 4 ] X = [ 7 6 ] , \begin{bmatrix*}[r] 2 & 1 \\ -3 & 4 \end{bmatrix*}X = \begin{bmatrix*}[r] 7 \\ 6 \end{bmatrix*}, [ 2 − 3 1 4 ] X = [ 7 6 ] ,
(i) the order of the matrix X
(ii) the matrix X.
Answer
(i) Since,
[ 2 1 − 3 4 ] X = [ 7 6 ] ⇒ [ 2 1 − 3 4 ] X is a 2 × 1 matrix, but [ 2 1 − 3 4 ] is a 2 × 2 matrix . ⇒ X is a 2 × 1 matrix . \begin{bmatrix*}[r] 2 & 1 \\ -3 & 4 \end{bmatrix*}X = \begin{bmatrix*}[r] 7 \\ 6 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2 & 1 \\ -3 & 4 \end{bmatrix*}X \text{ is a } 2 \times 1 \text{ matrix, but} \begin{bmatrix*}[r] 2 & 1 \\ -3 & 4 \end{bmatrix*} \text{ is a } 2 \times 2 \text{ matrix}. \\[1em] \Rightarrow \text{X is a } 2 \times 1 \text{ matrix}. [ 2 − 3 1 4 ] X = [ 7 6 ] ⇒ [ 2 − 3 1 4 ] X is a 2 × 1 matrix, but [ 2 − 3 1 4 ] is a 2 × 2 matrix . ⇒ X is a 2 × 1 matrix .
The order of the matrix is 2 × 1.
(ii) Let X = [ x y ] \text{X} = \begin{bmatrix*}[r] x \\ y \end{bmatrix*} X = [ x y ]
Given,
[ 2 1 − 3 4 ] X = [ 7 6 ] ⇒ [ 2 1 − 3 4 ] [ x y ] = [ 7 6 ] ⇒ [ 2 × x + 1 × y − 3 × x + 4 × y ] = [ 7 6 ] ⇒ [ 2 x + y − 3 x + 4 y ] = [ 7 6 ] \begin{bmatrix*}[r] 2 & 1 \\ -3 & 4 \end{bmatrix*}X = \begin{bmatrix*}[r] 7 \\ 6 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2 & 1 \\ -3 & 4 \end{bmatrix*} \begin{bmatrix*}[r] x \\ y \end{bmatrix*} = \begin{bmatrix*}[r] 7 \\ 6 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2 \times x + 1 \times y \\ -3 \times x + 4 \times y \end{bmatrix*} = \begin{bmatrix*}[r] 7 \\ 6 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2x + y \\ -3x + 4y \end{bmatrix*} = \begin{bmatrix*}[r] 7 \\ 6 \end{bmatrix*} \\[1em] [ 2 − 3 1 4 ] X = [ 7 6 ] ⇒ [ 2 − 3 1 4 ] [ x y ] = [ 7 6 ] ⇒ [ 2 × x + 1 × y − 3 × x + 4 × y ] = [ 7 6 ] ⇒ [ 2 x + y − 3 x + 4 y ] = [ 7 6 ]
By definition of equality of matrices we get,
2x + y = 7 or y = 7 - 2x (...Eq 1)
-3x + 4y = 6 (...Eq 2)
Putting value of y from Eq 1 in Eq 2
⇒ -3x + 4y = 6
∴ x = 2 and y = 7 - 2x = 7 - 2(2) = 7 - 4 = 3.
Since, X = [ x y ] ∴ X = [ 2 3 ] \text{X} = \begin{bmatrix*}[r] x \\ y \end{bmatrix*} \\[1em] \therefore \text{X} = \begin{bmatrix*}[r] 2 \\ 3 \end{bmatrix*} X = [ x y ] ∴ X = [ 2 3 ]
Hence, the matrix X = [ 2 3 ] . \text{X} = \begin{bmatrix*}[r] 2 \\ 3 \end{bmatrix*} . X = [ 2 3 ] .
Solve the matrix equation [ 4 1 ] X = [ − 4 8 − 1 2 ] . \begin{bmatrix*}[r] 4 \\ 1 \end{bmatrix*} X = \begin{bmatrix*}[r] -4 & 8 \\ -1 & 2 \end{bmatrix*} . [ 4 1 ] X = [ − 4 − 1 8 2 ] .
Answer
Since,
[ 4 1 ] X = [ − 4 8 − 1 2 ] ⇒ [ 4 1 ] X is a 2 × 2 matrix, but [ 4 1 ] is a 2 × 1 matrix . ⇒ X is a 1 × 2 matrix . \begin{bmatrix*}[r] 4 \\ 1 \end{bmatrix*} X = \begin{bmatrix*}[r] -4 & 8 \\ -1 & 2 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 \\ 1 \end{bmatrix*} X \text{ is a } 2 \times 2 \text{ matrix, but} \begin{bmatrix*}[r] 4 \\ 1 \end{bmatrix*} \text{ is a } 2 \times 1 \text{ matrix}. \\[1em] \Rightarrow \text{X is a } 1 \times 2 \text{ matrix}. [ 4 1 ] X = [ − 4 − 1 8 2 ] ⇒ [ 4 1 ] X is a 2 × 2 matrix, but [ 4 1 ] is a 2 × 1 matrix . ⇒ X is a 1 × 2 matrix .
We know that X matrix will be of order 1 × 2. So, let matrix X be [ x y ] . \begin{bmatrix*}[r] x & y \end{bmatrix*}. [ x y ] .
Given,
[ 4 1 ] X = [ − 4 8 − 1 2 ] ⇒ [ 4 1 ] [ x y ] = [ − 4 8 − 1 2 ] ⇒ [ 4 x 4 y x y ] = [ − 4 8 − 1 2 ] \begin{bmatrix*}[r] 4 \\ 1 \end{bmatrix*} X = \begin{bmatrix*}[r] -4 & 8 \\ -1 & 2 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 \\ 1 \end{bmatrix*} \begin{bmatrix*}[r] x & y \end{bmatrix*} = \begin{bmatrix*}[r] -4 & 8 \\ -1 & 2 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 4x & 4y \\ x & y \end{bmatrix*} = \begin{bmatrix*}[r] -4 & 8 \\ -1 & 2 \end{bmatrix*} \\[1em] [ 4 1 ] X = [ − 4 − 1 8 2 ] ⇒ [ 4 1 ] [ x y ] = [ − 4 − 1 8 2 ] ⇒ [ 4 x x 4 y y ] = [ − 4 − 1 8 2 ]
From definition of equality of matrices we get,
⇒ x = -1 and y = 2.
Since, X = [ x y ] ∴ X = [ − 1 2 ] \text{Since, X }= \begin{bmatrix*}[r] x & y \\ \end{bmatrix*} \\[1em] \therefore \text{X } = \begin{bmatrix*}[r] -1 & 2 \\ \end{bmatrix*} Since, X = [ x y ] ∴ X = [ − 1 2 ]
Hence, the matrix X = [ − 1 2 ] . \text{X} = \begin{bmatrix*}[r] -1 & 2 \end{bmatrix*}. X = [ − 1 2 ] .
If A = [ 2 − 1 − 4 5 ] and B = [ − 3 2 ] , \begin{bmatrix*}[r] 2 & -1 \\ -4 & 5 \end{bmatrix*} \text{and B } = \begin{bmatrix*}[r] -3 \\ 2 \end{bmatrix*}, [ 2 − 4 − 1 5 ] and B = [ − 3 2 ] ,
Answer
Given,
AC = B
⇒ [ 2 − 1 − 4 5 ] C = [ − 3 2 ] ⇒ [ 2 − 1 − 4 5 ] C is a 2 × 1 matrix, but [ 2 − 1 − 4 5 ] is a 2 × 2 matrix . ∴ C is a 1 × 2 matrix . Let matrix C = [ x y ] ⇒ [ 2 − 1 − 4 5 ] [ x y ] = [ − 3 2 ] ⇒ [ 2 × x + ( − 1 ) × y − 4 × x + 5 × y ] = [ − 3 2 ] ⇒ [ 2 x − y − 4 x + 5 y ] = [ − 3 2 ] \Rightarrow \begin{bmatrix*}[r] 2 & -1 \\ -4 & 5 \end{bmatrix*} C = \begin{bmatrix*}[r] -3 \\ 2 \end{bmatrix*} \\[0.5em] \Rightarrow \begin{bmatrix*}[r] 2 & -1 \\ -4 & 5 \end{bmatrix*} C \text{ is a } 2 \times 1 \text{ matrix, but} \begin{bmatrix*}[r] 2 & -1 \\ -4 & 5 \end{bmatrix*} \text{ is a } 2 \times 2 \text{ matrix}. \\[0.5em] \therefore \text{C is a } 1 \times 2 \text{ matrix}. \\[0.5em] \text{Let matrix C } = \begin{bmatrix*}[r] x \\ y \end{bmatrix*} \\[0.5em] \Rightarrow \begin{bmatrix*}[r] 2 & -1 \\ -4 & 5 \end{bmatrix*} \begin{bmatrix*}[r] x \\ y \end{bmatrix*} = \begin{bmatrix*}[r] -3 \\ 2 \end{bmatrix*} \\[0.5em] \Rightarrow \begin{bmatrix*}[r] 2 \times x + (-1) \times y \\ -4 \times x + 5 \times y \end{bmatrix*} = \begin{bmatrix*}[r] -3 \\ 2 \end{bmatrix*} \\[0.5em] \Rightarrow \begin{bmatrix*}[r] 2x - y \\ -4x + 5y \end{bmatrix*} = \begin{bmatrix*}[r] -3 \\ 2 \end{bmatrix*} \\[0.5em] ⇒ [ 2 − 4 − 1 5 ] C = [ − 3 2 ] ⇒ [ 2 − 4 − 1 5 ] C is a 2 × 1 matrix, but [ 2 − 4 − 1 5 ] is a 2 × 2 matrix . ∴ C is a 1 × 2 matrix . Let matrix C = [ x y ] ⇒ [ 2 − 4 − 1 5 ] [ x y ] = [ − 3 2 ] ⇒ [ 2 × x + ( − 1 ) × y − 4 × x + 5 × y ] = [ − 3 2 ] ⇒ [ 2 x − y − 4 x + 5 y ] = [ − 3 2 ]
By definition of equality of matrices we get,
⇒ 2x - y = -3 or y = 2x + 3 (...Eq 1)
⇒ -4x + 5y = 2 (...Eq 2)
Putting value of y from Eq 1 in Eq 2,
⇒ -4x + 5y = 2− 13 6 -\dfrac{13}{6} − 6 13
Now finding value of y,
y = 2 x + 3 = 2 ( − 13 6 ) + 3 = − 26 6 + 3 = − 26 + 18 6 = − 8 6 = − 4 3 . y = 2x + 3 \\[1em] = 2\Big(-\dfrac{13}{6}\Big) + 3 \\[1em] = -\dfrac{26}{6} + 3 \\[1em] = \dfrac{-26 + 18}{6} \\[1em] = -\dfrac{8}{6} \\[1em] = -\dfrac{4}{3}. y = 2 x + 3 = 2 ( − 6 13 ) + 3 = − 6 26 + 3 = 6 − 26 + 18 = − 6 8 = − 3 4 .
∴ x = − 13 6 and y = − 4 3 . -\dfrac{13}{6} \text{ and y } = -\dfrac{4}{3}. − 6 13 and y = − 3 4 .
Since,
C = [ x y ] ∴ C = [ − 13 6 − 4 3 ] \text{C }= \begin{bmatrix*}[r] x \\ y \end{bmatrix*} \\[1em] \therefore C = \begin{bmatrix*}[r] -\dfrac{13}{6} \\ -\dfrac{4}{3} \end{bmatrix*} C = [ x y ] ∴ C = − 6 13 − 3 4
Hence, the matrix C = [ − 13 6 − 4 3 ] . \begin{bmatrix*}[r] -\dfrac{13}{6} \\ -\dfrac{4}{3} \end{bmatrix*}. − 6 13 − 3 4 .
If A = [ 2 − 1 − 4 5 ] and B = [ 0 − 3 ] , \begin{bmatrix*}[r] 2 & -1 \\ -4 & 5 \end{bmatrix*} \text{and B } = \begin{bmatrix*}[r] 0 & -3 \\ \end{bmatrix*}, [ 2 − 4 − 1 5 ] and B = [ 0 − 3 ] ,
Answer
Given,
CA = B.
C [ 2 − 1 − 4 5 ] = [ 0 − 3 ] ⇒ C [ 2 − 1 − 4 5 ] is a 1 × 2 matrix, but [ 2 − 1 − 4 5 ] is a 2 × 2 matrix . ∴ C is a 1 × 2 matrix . Let matrix C = [ x y ] ⇒ [ x y ] [ 2 − 1 − 4 5 ] = [ 0 − 3 ] ⇒ [ x × 2 + y × ( − 4 ) x × ( − 1 ) + y × 5 ] = [ 0 − 3 ] ⇒ [ 2 x − 4 y − x + 5 y ] = [ 0 − 3 ] C\begin{bmatrix*}[r] 2 & -1 \\ -4 & 5 \end{bmatrix*} = \begin{bmatrix*}[r] 0 & -3 \\ \end{bmatrix*} \\[1em] \Rightarrow C\begin{bmatrix*}[r] 2 & -1 \\ -4 & 5 \end{bmatrix*} \text{ is a } 1 \times 2 \text{ matrix, but} \begin{bmatrix*}[r] 2 & -1 \\ -4 & 5 \end{bmatrix*} \text{ is a } 2 \times 2 \text{ matrix}. \\[1em] \therefore \text{C is a } 1 \times 2 \text{ matrix}. \\[1em] \text{Let matrix C } = \begin{bmatrix*}[r] x & y \\ \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] x & y \\ \end{bmatrix*} \begin{bmatrix*}[r] 2 & -1 \\ -4 & 5 \end{bmatrix*} = \begin{bmatrix*}[r] 0 & -3 \\ \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] x \times 2 + y \times (-4) & x \times (-1) + y \times 5 \end{bmatrix*} = \begin{bmatrix*}[r] 0 & -3 \\ \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2x - 4y & -x + 5y \\ \end{bmatrix*} = \begin{bmatrix*}[r] 0 & -3 \\ \end{bmatrix*} C [ 2 − 4 − 1 5 ] = [ 0 − 3 ] ⇒ C [ 2 − 4 − 1 5 ] is a 1 × 2 matrix, but [ 2 − 4 − 1 5 ] is a 2 × 2 matrix . ∴ C is a 1 × 2 matrix . Let matrix C = [ x y ] ⇒ [ x y ] [ 2 − 4 − 1 5 ] = [ 0 − 3 ] ⇒ [ x × 2 + y × ( − 4 ) x × ( − 1 ) + y × 5 ] = [ 0 − 3 ] ⇒ [ 2 x − 4 y − x + 5 y ] = [ 0 − 3 ]
By definition of equality of matrices we get,
2x - 4y = 0 (...Eq 1)
-x + 5y = -3 or x = 5y + 3 (...Eq 2)
Putting value of x from Eq 2 in Eq 1,
⇒ 2x - 4y = 0
Now finding value of x,
⇒ x = 5y + 3 = 5(-1) + 3 = -5 + 3 = -2.
∴ x = -2, y = -1.
Since, C = [ x y ] ∴ C = [ − 2 − 1 ] \text{Since, C }= \begin{bmatrix*}[r] x & y \\ \end{bmatrix*} \\[0.5em] \therefore C = \begin{bmatrix*}[r] -2 & -1 \\ \end{bmatrix*} Since, C = [ x y ] ∴ C = [ − 2 − 1 ]
Hence, the matrix C = [ − 2 − 1 ] \begin{bmatrix*}[r] -2 & -1 \\ \end{bmatrix*} [ − 2 − 1 ]
If A = [ 3 − 4 − 1 2 ] \begin{bmatrix*}[r] 3 & -4 \\ -1 & 2 \end{bmatrix*} [ 3 − 1 − 4 2 ]
Answer
Given,
BA = I
B [ 3 − 4 − 1 2 ] = [ 1 0 0 1 ] ⇒ B [ 3 − 4 − 1 2 ] is a 2 × 2 matrix, and [ 3 − 4 − 1 2 ] is a 2 × 2 matrix . ∴ B is a 2 × 2 matrix . I = [ 1 0 0 1 ] \text{B}\begin{bmatrix*}[r] 3 & -4 \\ -1 & 2 \end{bmatrix*} = \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em] \Rightarrow \text{B}\begin{bmatrix*}[r] 3 & -4 \\ -1 & 2 \end{bmatrix*} \text{ is a } 2 \times 2 \text{ matrix, and} \begin{bmatrix*}[r] 3 & -4 \\ -1 & 2 \end{bmatrix*} \text{ is a } 2 \times 2 \text{ matrix}. \\[1em] \therefore \text{B is a } 2 \times 2 \text{ matrix}. \\[1em] \text{I } = \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} B [ 3 − 1 − 4 2 ] = [ 1 0 0 1 ] ⇒ B [ 3 − 1 − 4 2 ] is a 2 × 2 matrix, and [ 3 − 1 − 4 2 ] is a 2 × 2 matrix . ∴ B is a 2 × 2 matrix . I = [ 1 0 0 1 ]
We know that B will be of order 2 × 2. So, let
B = [ a b c d ] ⇒ [ a b c d ] [ 3 − 4 − 1 2 ] = [ 1 0 0 1 ] ⇒ [ a × 3 + b × ( − 1 ) a × ( − 4 ) + b × 2 c × 3 + d × ( − 1 ) c × ( − 4 ) + d × 2 ] = [ 1 0 0 1 ] ⇒ [ 3 a − b − 4 a + 2 b 3 c − d − 4 c + 2 d ] = [ 1 0 0 1 ] \text{B} = \begin{bmatrix*}[r] a & b \\ c & d \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] a & b \\ c & d \end{bmatrix*} \begin{bmatrix*}[r] 3 & -4 \\ -1 & 2 \end{bmatrix*} = \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] a \times 3 + b \times (-1) & a \times (-4) + b \times 2 \\ c \times 3 + d \times (-1) & c \times (-4) + d \times 2 \end{bmatrix*} = \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 3a - b & -4a + 2b \\ 3c - d & -4c + 2d \end{bmatrix*} = \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em] B = [ a c b d ] ⇒ [ a c b d ] [ 3 − 1 − 4 2 ] = [ 1 0 0 1 ] ⇒ [ a × 3 + b × ( − 1 ) c × 3 + d × ( − 1 ) a × ( − 4 ) + b × 2 c × ( − 4 ) + d × 2 ] = [ 1 0 0 1 ] ⇒ [ 3 a − b 3 c − d − 4 a + 2 b − 4 c + 2 d ] = [ 1 0 0 1 ]
By definition of equality of matrices we get,
3a - b = 1 (...Eq 1)
-4a + 2b = 0
3c - d = 0
-4c + 2d = 1 (...Eq 4)
Putting value of b from Eq 2 in Eq 1
⇒ 3a - b = 1
∴ a = 1, b = 2a = 2.
Putting value of d from Eq 3 in Eq 4
⇒ -4c + 2d = 11 2 \dfrac{1}{2} 2 1
∴ c = 1 2 \dfrac{1}{2} 2 1 3 2 \dfrac{3}{2} 2 3
Since, B = [ a b c d ] ∴ B = [ 1 2 1 2 3 2 ] \text{Since, B }= \begin{bmatrix*}[r] a & b \\ c & d \end{bmatrix*} \\[1em] \therefore \text{B} = \begin{bmatrix*}[r] 1 & 2 \\ \dfrac{1}{2} & \dfrac{3}{2} \end{bmatrix*} Since, B = [ a c b d ] ∴ B = [ 1 2 1 2 2 3 ]
Hence, the matrix B = [ 1 2 1 2 3 2 ] \begin{bmatrix*}[r] 1 & 2 \\ \dfrac{1}{2} & \dfrac{3}{2} \end{bmatrix*} [ 1 2 1 2 2 3 ]
Given [ 4 2 − 1 1 ] \begin{bmatrix*}[r] 4 & 2 \\ -1 & 1 \end{bmatrix*} [ 4 − 1 2 1 ]
(i) State the order of matrix M.
(ii) Find the matrix M.
Answer
(i) Given,
[ 4 2 − 1 1 ] M = 6 I ⇒ [ 4 2 − 1 1 ] M is a 2 × 2 matrix, and [ 4 2 − 1 1 ] is a 2 × 2 matrix . ∴ M is a 2 × 2 matrix . \begin{bmatrix*}[r] 4 & 2 \\ -1 & 1 \end{bmatrix*} M = 6I \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 & 2 \\ -1 & 1 \end{bmatrix*} M \text{ is a } 2 \times 2 \text{ matrix, and} \begin{bmatrix*}[r] 4 & 2 \\ -1 & 1 \end{bmatrix*} \text{ is a } 2 \times 2 \text{ matrix}. \\[1em] \therefore \text{M is a } 2 \times 2 \text{ matrix}. \\[1em] [ 4 − 1 2 1 ] M = 6 I ⇒ [ 4 − 1 2 1 ] M is a 2 × 2 matrix, and [ 4 − 1 2 1 ] is a 2 × 2 matrix . ∴ M is a 2 × 2 matrix .
Hence, the matrix M is of order 2 × 2.
(ii) Let matrix M be [ a b c d ] . \begin{bmatrix*}[r] a & b \\ c & d \end{bmatrix*}. [ a c b d ] .
Given,
[ 4 2 − 1 1 ] M = 6 I ⇒ [ 4 2 − 1 1 ] [ a b c d ] = 6 [ 1 0 0 1 ] ⇒ [ 4 × a + 2 × c 4 × b + 2 × d ( − 1 ) × a + 1 × c ( − 1 ) × b + 1 × d ] = [ 6 0 0 6 ] ⇒ [ 4 a + 2 c 4 b + 2 d − a + c − b + d ] = [ 6 0 0 6 ] \begin{bmatrix*}[r] 4 & 2 \\ -1 & 1 \end{bmatrix*} M = 6I \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 & 2 \\ -1 & 1 \end{bmatrix*} \begin{bmatrix*}[r] a & b \\ c & d \end{bmatrix*} = 6\begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 \times a + 2 \times c & 4 \times b + 2 \times d \\ (-1) \times a + 1 \times c & (-1) \times b + 1 \times d \end{bmatrix*} = \begin{bmatrix*}[r] 6 & 0 \\ 0 & 6 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 4a + 2c & 4b + 2d \\ -a + c & -b + d \end{bmatrix*} = \begin{bmatrix*}[r] 6 & 0 \\ 0 & 6 \end{bmatrix*} \\[1em] [ 4 − 1 2 1 ] M = 6 I ⇒ [ 4 − 1 2 1 ] [ a c b d ] = 6 [ 1 0 0 1 ] ⇒ [ 4 × a + 2 × c ( − 1 ) × a + 1 × c 4 × b + 2 × d ( − 1 ) × b + 1 × d ] = [ 6 0 0 6 ] ⇒ [ 4 a + 2 c − a + c 4 b + 2 d − b + d ] = [ 6 0 0 6 ]
By definition of equality of matrices we get,
4a + 2c = 6 (...Eq 1)
4b + 2d = 0
-a + c = 0
-b + d = 6 (...Eq 4)
Putting value of a from Eq 3 in Eq 1
⇒ 4a + 2c = 6
∴ c = 1 and a = c = 1.
Putting value of d from Eq 2 in Eq 4
⇒ -b + d = 6
∴ b = -2 and d = -2b = 4.
Since,
M = [ a b c d ] ∴ M = [ 1 − 2 1 4 ] \text{M} = \begin{bmatrix*}[r] a & b \\ c & d \end{bmatrix*} \\[1em] \therefore \text{M} = \begin{bmatrix*}[r] 1 & -2 \\ 1 & 4 \end{bmatrix*} M = [ a c b d ] ∴ M = [ 1 1 − 2 4 ]
Hence, the matrix M = [ 1 − 2 1 4 ] \text{M} = \begin{bmatrix*}[r] 1 & -2 \\ 1 & 4 \end{bmatrix*} M = [ 1 1 − 2 4 ]
If B = [ − 4 2 5 − 1 ] and C = [ 17 − 1 47 − 13 ] , \begin{bmatrix*}[r] -4 & 2 \\ 5 & -1 \end{bmatrix*} \text{ and C} = \begin{bmatrix*}[r] 17 & -1 \\ 47 & -13 \end{bmatrix*}, [ − 4 5 2 − 1 ] and C = [ 17 47 − 1 − 13 ] ,
Answer
Given,
AB = C
⇒ A [ − 4 2 5 − 1 ] is a 2 × 2 matrix, and [ 4 2 5 − 1 ] is a 2 × 2 matrix . ∴ A is a 2 × 2 matrix . \Rightarrow A\begin{bmatrix*}[r] -4 & 2 \\ 5 & -1 \end{bmatrix*} \text{ is a } 2 \times 2 \text{ matrix, and} \begin{bmatrix*}[r] 4 & 2 \\ 5 & -1 \end{bmatrix*} \text{ is a } 2 \times 2 \text{ matrix}. \\[1em] \therefore \text{A is a } 2 \times 2 \text{ matrix}. ⇒ A [ − 4 5 2 − 1 ] is a 2 × 2 matrix, and [ 4 5 2 − 1 ] is a 2 × 2 matrix . ∴ A is a 2 × 2 matrix .
We know that matrix A will be of order 2 × 2. Let matrix be
A = [ a b c d ] ⇒ [ a b c d ] [ − 4 2 5 − 1 ] = [ 17 − 1 47 − 13 ] ⇒ [ a × ( − 4 ) + b × 5 a × 2 + b × ( − 1 ) c × ( − 4 ) + d × 5 c × 2 + d × ( − 1 ) ] = [ 17 − 1 47 − 13 ] ⇒ [ − 4 a + 5 b 2 a − b − 4 c + 5 d 2 c − d ] = [ 17 − 1 47 − 13 ] \text{A} = \begin{bmatrix*}[r] a & b \\ c & d \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] a & b \\ c & d \end{bmatrix*} \begin{bmatrix*}[r] -4 & 2 \\ 5 & -1 \end{bmatrix*} = \begin{bmatrix*}[r] 17 & -1 \\ 47 & -13 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] a \times (-4) + b \times 5 & a \times 2 + b \times (-1) \\ c \times (-4) + d \times 5 & c \times 2 + d \times (-1) \end{bmatrix*} = \begin{bmatrix*}[r] 17 & -1 \\ 47 & -13 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] -4a + 5b & 2a - b \\ -4c + 5d & 2c - d \end{bmatrix*} = \begin{bmatrix*}[r] 17 & -1 \\ 47 & -13 \end{bmatrix*} \\[1em] A = [ a c b d ] ⇒ [ a c b d ] [ − 4 5 2 − 1 ] = [ 17 47 − 1 − 13 ] ⇒ [ a × ( − 4 ) + b × 5 c × ( − 4 ) + d × 5 a × 2 + b × ( − 1 ) c × 2 + d × ( − 1 ) ] = [ 17 47 − 1 − 13 ] ⇒ [ − 4 a + 5 b − 4 c + 5 d 2 a − b 2 c − d ] = [ 17 47 − 1 − 13 ]
By definition of equality of matrices we get,
-4a + 5b = 17 (...Eq 1)
2a - b = -1
-4c + 5d = 47 (...Eq 3)
2c - d = -13
Putting value of b from Eq 2 in Eq 1
⇒ -4a + 5b = 17
∴ a = 2, b = 2a + 1 = 2(2) + 1 = 5.
Putting value of d from Eq 4 in Eq 3
⇒ -4c + 5d = 47
∴ c = -3, d = 2c + 13 = 2(-3) + 13 = 7.
Since,
A = [ a b c d ] ∴ A = [ 2 5 − 3 7 ] \text{A} = \begin{bmatrix*}[r] a & b \\ c & d \end{bmatrix*} \\[1em] \therefore \text{A} = \begin{bmatrix*}[r] 2 & 5 \\ -3 & 7 \end{bmatrix*} A = [ a c b d ] ∴ A = [ 2 − 3 5 7 ]
Hence, the matrix A = [ 2 5 − 3 7 ] \begin{bmatrix*}[r] 2 & 5 \\ -3 & 7 \end{bmatrix*} [ 2 − 3 5 7 ]
If A = [ 4 − 4 − 4 4 ] \begin{bmatrix*}[r] 4 & -4 \\ -4 & 4 \end{bmatrix*} [ 4 − 4 − 4 4 ] 2 . If A2 = pA, then find the value of p.
Answer
Given,
⇒ A2 = pA
⇒ [ 4 − 4 − 4 4 ] [ 4 − 4 − 4 4 ] = p [ 4 − 4 − 4 4 ] ⇒ [ 4 × 4 + ( − 4 ) × ( − 4 ) 4 × ( − 4 ) + ( − 4 ) × 4 − 4 × 4 + 4 × ( − 4 ) ( − 4 ) × ( − 4 ) + 4 × 4 ] = p [ 4 − 4 − 4 4 ] ⇒ [ 16 + 16 − 16 − 16 − 16 − 16 16 + 16 ] = p [ 4 − 4 − 4 4 ] ⇒ [ 32 − 32 − 32 32 ] = p [ 4 − 4 − 4 4 ] ⇒ 32 [ 1 − 1 − 1 1 ] = 4 p [ 1 − 1 − 1 1 ] ⇒ 32 = 4 p ⇒ p = 32 4 = 8. \Rightarrow \begin{bmatrix*}[r] 4 & -4 \\ -4 & 4 \end{bmatrix*}\begin{bmatrix*}[r] 4 & -4 \\ -4 & 4 \end{bmatrix*} = p\begin{bmatrix*}[r] 4 & -4 \\ -4 & 4 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 \times 4 + (-4) \times (-4) & 4 \times (-4) + (-4) \times 4 \\ -4 \times 4 + 4 \times (-4) & (-4) \times (-4) + 4 \times 4 \end{bmatrix*} = p\begin{bmatrix*}[r] 4 & -4 \\ -4 & 4 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 16 + 16 & -16 - 16 \\ -16 - 16 & 16 + 16 \end{bmatrix*} = p\begin{bmatrix*}[r] 4 & -4 \\ -4 & 4 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 32 & -32 \\ -32 & 32 \end{bmatrix*} = p\begin{bmatrix*}[r] 4 & -4 \\ -4 & 4 \end{bmatrix*} \\[1em] \Rightarrow 32\begin{bmatrix*}[r] 1 & -1 \\ -1 & 1 \end{bmatrix*} = 4p\begin{bmatrix*}[r] 1 & -1 \\ -1 & 1 \end{bmatrix*} \\[1em] \Rightarrow 32 = 4p \\[1em] \Rightarrow p = \dfrac{32}{4} = 8. ⇒ [ 4 − 4 − 4 4 ] [ 4 − 4 − 4 4 ] = p [ 4 − 4 − 4 4 ] ⇒ [ 4 × 4 + ( − 4 ) × ( − 4 ) − 4 × 4 + 4 × ( − 4 ) 4 × ( − 4 ) + ( − 4 ) × 4 ( − 4 ) × ( − 4 ) + 4 × 4 ] = p [ 4 − 4 − 4 4 ] ⇒ [ 16 + 16 − 16 − 16 − 16 − 16 16 + 16 ] = p [ 4 − 4 − 4 4 ] ⇒ [ 32 − 32 − 32 32 ] = p [ 4 − 4 − 4 4 ] ⇒ 32 [ 1 − 1 − 1 1 ] = 4 p [ 1 − 1 − 1 1 ] ⇒ 32 = 4 p ⇒ p = 4 32 = 8.
Hence, p = 8.
