Question

\[\text{ If }A^{- 1} = \begin{bmatrix}3 & - 1 & 1 \\ - 15 & 6 & - 5 \\ 5 & - 2 & 2\end{bmatrix}\text{ and }B = \begin{bmatrix}1 & 2 & - 2 \\ - 1 & 3 & 0 \\ 0 & - 2 & 1\end{bmatrix},\text{ find }\left( AB \right)^{- 1} .\]

Answer 1

We know that (AB)⁻¹ = B⁻¹ A⁻¹. 
\[B = \begin{bmatrix}1 & 2 & - 2 \\ - 1 & 3 & 0 \\ 0 & - 2 & 1\end{bmatrix}\]
\[ B^{- 1} = \frac{1}{\left| B \right|}Adj . B\]
Now, 
\[\left| B \right| = \begin{vmatrix}1 & 2 & - 2 \\ - 1 & 3 & 0 \\ 0 & - 2 & 1\end{vmatrix}\]
\[ = 1\left( 3 + 0 \right) + 1\left( 2 - 4 \right)\]
\[ = 1\]
\[\text{ Now, to find Adj . B}\]
\[B_{11} = \left( - 1 \right)^{1 + 1} \left( 3 \right) = 3\]
\[ B_{12} = \left( - 1 \right)^{1 + 2} \left( - 1 \right) = 1\]
\[ B_{13} = \left( - 1 \right)^{1 + 3} \left( 2 \right) = 2\]
\[B_{21} = \left( - 1 \right)^{2 + 1} \left( 2 - 4 \right) = 2\]
\[ B_{22} = \left( - 1 \right)^{2 + 2} \left( 1 \right) = 1 \]
\[ B_{23} = \left( - 1 \right)^{2 + 3} \left( - 2 \right) = 2\]
\[B_{31} = \left( - 1 \right)^{3 + 1} \left( 6 \right) = 6\]
\[ B_{32} = \left( - 1 \right)^{3 + 2} \left( - 2 \right) = 2\]
\[ B_{33} = \left( - 1 \right)^{3 + 3} \left( 3 + 2 \right) = 5\]
Therefore, 
\[Adj . B = \begin{bmatrix}3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5\end{bmatrix}\]
Thus, 
\[ B^{- 1} = \begin{bmatrix}3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5\end{bmatrix} . \]
\[ \left( AB \right)^{- 1} = B^{- 1} A^{- 1} \]
\[ = \begin{bmatrix}3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5\end{bmatrix}\begin{bmatrix}3 & - 1 & 1 \\ - 15 & 6 & - 5 \\ 5 & - 2 & 2\end{bmatrix}\]
\[ = \begin{bmatrix}9 - 30 + 30 & - 3 + 12 - 12 & 3 - 10 + 12 \\ 3 - 15 + 10 & - 1 + 6 - 4 & 1 - 5 + 4 \\ 6 - 30 + 25 & - 2 + 12 - 10 & 2 - 10 + 10\end{bmatrix}\]
\[ = \begin{bmatrix}9 & - 3 & 5 \\ - 2 & 1 & 0 \\ 1 & 0 & 2\end{bmatrix}\]
\[\text{ Hence,} \left( AB \right)^{- 1} = \begin{bmatrix}9 & - 3 & 5 \\ - 2 & 1 & 0 \\ 1 & 0 & 2\end{bmatrix}\]