Question

If \[A = \begin{bmatrix}1 & - 2 & 3 \\ 0 & - 1 & 4 \\ - 2 & 2 & 1\end{bmatrix},\text{ find }\left( A^T \right)^{- 1} .\]

Answer 1

We know that (A^T)⁻¹ = (A⁻¹)^T.
\[A = \begin{bmatrix}1 & - 2 & 3 \\ 0 & - 1 & 4 \\ - 2 & 2 & 1\end{bmatrix}\]
\[ A^{- 1} = \frac{1}{\left| A \right|}Adj . A\]
Now,
\[\left| A \right| = \begin{vmatrix}1 & - 2 & 3 \\ 0 & - 1 & 4 \\ - 2 & 2 & 1\end{vmatrix}\]
\[ = 1\left( - 1 - 8 \right) - 2\left( - 8 + 3 \right)\]
\[ = - 9 + 10\]
\[ = 1\]
\[\text{ Now, to find Adj . A}\]
\[A_{11} = \left( - 1 \right)^{1 + 1} \left( - 9 \right) = - 9\]
\[ A_{12} = \left( - 1 \right)^{1 + 2} \left( 8 \right) = - 8\]
\[ A_{13} = \left( - 1 \right)^{1 + 3} \left( - 2 \right) = - 2\]
\[A_{21} = \left( - 1 \right)^{2 + 1} \left( - 8 \right) = 8\]
\[ A_{22} = \left( - 1 \right)^{2 + 2} \left( 7 \right) = 7 \]
\[ A_{23} = \left( - 1 \right)^{2 + 3} \left( - 2 \right) = 2\]
\[ A_{31} = \left( - 1 \right)^{3 + 1} \left( - 5 \right) = - 5\]
\[ A_{32} = \left( - 1 \right)^{3 + 2} \left( 4 \right) = - 4\]
\[ A_{33} = \left( - 1 \right)^{3 + 3} \left( - 1 \right) = - 1\]
Therefore, 
\[Adj . A = \begin{bmatrix}- 9 & 8 & - 5 \\ - 8 & 7 & - 4 \\ - 2 & 2 & - 1\end{bmatrix}\]
Thus, 
\[ A^{- 1} = \begin{bmatrix}- 9 & 8 & - 5 \\ - 8 & 7 & - 4 \\ - 2 & 2 & - 1\end{bmatrix} . \]
\[ \left( A^T \right)^{- 1} = \left( A^{- 1} \right)^T \]
\[ = \begin{bmatrix}- 9 & 8 & - 5 \\ - 8 & 7 & - 4 \\ - 2 & 2 & - 1\end{bmatrix}^T \]
\[ = \begin{bmatrix}- 9 & - 8 & - 2 \\ 8 & 7 & 2 \\ - 5 & - 4 & - 1\end{bmatrix}\]
\[\text{ Hence, }\left( A^T \right)^{- 1} = \begin{bmatrix}- 9 & - 8 & - 2 \\ 8 & 7 & 2 \\ - 5 & - 4 & - 1\end{bmatrix} .\]