Question

For the following pair of matrix verify that \[\left( AB \right)^{- 1} = B^{- 1} A^{- 1} :\]

\[A = \begin{bmatrix}3 & 2 \\ 7 & 5\end{bmatrix}\text{ and }B \begin{bmatrix}4 & 6 \\ 3 & 2\end{bmatrix}\]

Answer 1

\[\text{ We have, }A = \begin{bmatrix}3 & 2 \\ 7 & 5\end{bmatrix}\text{ and }B = \begin{bmatrix}4 & 6 \\ 3 & 2\end{bmatrix}\]
\[ \therefore AB = \begin{bmatrix}18 & 22 \\ 43 & 52\end{bmatrix}\]
\[\left| AB \right| = - 10\]
\[\text{Since, }\left| AB \right| \neq 0\]
\[\text{Hence, AB is invertible . Let } C_{ij}\text{ be the cofactor of }a_{in}\text{ in AB = }\left[ a_{ij} \right]\]
\[ C_{11} = 52 , C_{12} = - 43, C_{21} = - 22\text{ and }C_{22} = 18\]
\[adj\left( AB \right) = \begin{bmatrix}52 & - 43 \\ - 22 & 18\end{bmatrix}^T = \begin{bmatrix}52 & - 22 \\ - 43 & 18\end{bmatrix}\]
\[ \therefore \left( AB \right)^{- 1} = - \frac{1}{10}\begin{bmatrix}52 & - 22 \\ - 43 & 18\end{bmatrix} . . . \left( 1 \right)\]
\[\text{ Now, }B = \begin{bmatrix}4 & 6 \\ 3 & 2\end{bmatrix}\]
\[\left| B \right| = - 10\]
\[\text{ Since, }\left| B \right| \neq 0\]
\[\text{Hence, B is invertible . Let }C_{ij}\text{ be the cofactor of } a_{in}\text{ in B = }\left[ a_{ij} \right]\]
\[ C_{11} = 2 , C_{12} = - 3, C_{21} = - 6\text{ and }C_{22} = 4\]
\[adjB = \begin{bmatrix}2 & - 3 \\ - 6 & 4\end{bmatrix}^T = \begin{bmatrix}2 & - 6 \\ - 3 & 4\end{bmatrix}\]
\[ \therefore B^{- 1} = - \frac{1}{10}\begin{bmatrix}2 & - 6 \\ - 3 & 4\end{bmatrix}\]
\[\left| A \right| = 1\]
\[\text{Since, }\left| A \right| \neq 0\]
\[\text{ Hence, A is invertible . Let }C_{ij}\text{ be the cofactor of } a_{in}\text{ in A = }\left[ a_{ij} \right]\]
\[ C_{11} = 5 , C_{12} = - 7, C_{21} = - 2\text{ and }C_{22} = 3\]
\[adjA = \begin{bmatrix}5 & - 7 \\ - 2 & 3\end{bmatrix}^T = \begin{bmatrix}5 & - 2 \\ - 7 & 3\end{bmatrix}\]
\[ \therefore A^{- 1} = \begin{bmatrix}5 & - 2 \\ - 7 & 3\end{bmatrix}\]
\[\text{ Now, }B^{- 1} A^{- 1} = - \frac{1}{10}\begin{bmatrix}52 & - 22 \\ - 43 & 18\end{bmatrix} . . . \left( 2 \right)\]
\[\text{From eq . }\left( 1 \right)\text{ and }\left( 2 \right),\text{ we have}\]
\[ \left( AB \right)^{- 1} = B^{- 1} A^{- 1} \]
Hence verified .