Question

Solve the given question

Topic:Determinants .

Sub:Mathematics.​

Answer 1

Given :    

$\sf \: \begin{lgathered}A = \begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix}\end{lgathered} and</p><p>\begin{lgathered} \: \: B = \begin{bmatrix} 6 & 8 \\ 7& 9 \end{bmatrix}\end{lgathered} \:$

To Find :

verify that (AB)⁻¹= B⁻¹A⁻¹

Solution :

A⁻¹  =  AdjA / | A |

$: \implies \sf \: \begin{lgathered}A = \begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix}\end{lgathered} </p><p>\begin{lgathered} \: \: B = \begin{bmatrix} 6 & 8 \\ 7& 9 \end{bmatrix}\end{lgathered} \:$

$: \implies\sf \: \begin{lgathered}AB = \begin{bmatrix} 3\times6 + 7\times 7 & 3\times8 + 7\times 9 \\ 2\times6 + 5\times 7 & 2\times8 + 5\times 9\end{bmatrix}\end{lgathered} \:$

$: \implies\sf \: \begin{lgathered}AB = \begin{bmatrix} 67 & 87 \\ 47 & 61\end{bmatrix}\end{lgathered} \:$

| AB |  =  (67 * 61 ) - (47 * 87)  = -2

$: \implies\sf \: \begin{lgathered}Adj \: \: B= \begin{bmatrix} \: \: \: \: 61 & -87 \\ -47 & \: \: \: \: 67\end{bmatrix}\end{lgathered}</p><p>$

$: \implies\sf \: B ^{ - 1} =\begin{lgathered}\frac{-1}{2} \begin{bmatrix} 9 & -8 \\ -7 & 6 \end{bmatrix}\end{lgathered}$

$: \implies\sf \: \begin{lgathered}A = \begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix}\end{lgathered}$

| A | = 3 * 5 - 2 * 7 = 1

$: \implies \sf \: \begin{lgathered}AdjA = \begin{bmatrix} \: \: \: 5 & -7 \\ -2 & \: \: \: 3 \end{bmatrix}\end{lgathered}$

$: \implies\sf \: A ^{ - 1} = \begin{lgathered}\begin{bmatrix} \: \: \: 5 & -7 \\ -2 & \: \: \: 3 \end{bmatrix}\end{lgathered}$

$: \implies \sf \: B ^{ - 1} A ^{ - 1} = \begin{lgathered}\frac{-1 \: \: }{ \: \: 2} \begin{bmatrix} \: \: 9 & -8 \\ -7 & \: \: \: 6 \end{bmatrix}\end{lgathered} \begin{lgathered}\begin{bmatrix} \: \: \: 5 & -7 \\ -2 & \: \: \: 3 \end{bmatrix}\end{lgathered} \:$

$: \implies\sf \: B ^{ - 1} A ^{ - 1} = \begin{lgathered}\frac{-1}{2} \begin{bmatrix} 61 & -87 \\ -47 & 67\end{bmatrix}\end{lgathered}$

$: \implies\sf \: (AB) ^{ - 1} = \begin{lgathered}\frac{-1}{2} \begin{bmatrix} \: \: \: \: 61 & -87 \\ -47 & \: \: \: 67\end{bmatrix}\end{lgathered}$

$: \implies\sf \: (AB) ^{ - 1} = B ^{ - 1}A ^{ - 1}= </p><p>\begin{lgathered}\frac{-1}{2} \begin{bmatrix} 61 & -87 \\ -47 & 67\end{bmatrix}\end{lgathered}$

$\bf \: (AB) ^{ - 1} = B ^{ - 1}A ^{ - 1} \:$