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Question
Let \[A = \begin{bmatrix}3 & 2 \\ 7 & 5\end{bmatrix}\text{ and }B = \begin{bmatrix}6 & 7 \\ 8 & 9\end{bmatrix} .\text{ Find }\left( AB \right)^{- 1}\]
Solution
Given:
\[A = \begin{bmatrix}3 & 2 \\ 7 & 5\end{bmatrix}\]
\[B = \begin{bmatrix}6 & 7 \\ 8 & 9\end{bmatrix}\]
\[AB = \begin{bmatrix}34 & 39 \\ 82 & 94\end{bmatrix}\]
Now,
\[\left| AB \right| = - 2\]
\[\text{ Since, }\left| AB \right| \neq 0\]
\[\text{ Hence, AB is invertible . Let }C_{ij} \text{ be the cofactor of }a_{in}\text{ in AB = }\left[ a_{ij} \right]\]
\[ C_{11} = 94 , C_{12} = - 82, C_{21} = - 39\text{ and }C_{22} = 34\]
\[adj(AB) = \begin{bmatrix}94 & - 82 \\ - 39 & 34\end{bmatrix}^T = \begin{bmatrix}94 & - 39 \\ - 82 & 34\end{bmatrix}\]
\[ \therefore \left( AB \right)^{- 1} = - \frac{1}{2}\begin{bmatrix}94 & - 39 \\ - 82 & 34\end{bmatrix} = \begin{bmatrix}- 47 & \frac{39}{2} \\ 41 & - 17\end{bmatrix}\]
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