Question

If \[A = \begin{bmatrix}1 & 2 & 0 \\ - 2 & - 1 & - 2 \\ 0 & - 1 & 1\end{bmatrix}\] , find A⁻¹. Using A⁻¹, solve the system of linear equations   x − 2y = 10, 2x − y − z = 8, −2y + z = 7

Answer 1

 Here, 
\[ A = \begin{bmatrix}1 & 2 & 0 \\ - 2 & - 1 & - 2 \\ 0 & - 1 & 1\end{bmatrix}\]
\[\left| A \right| = 1\left( - 1 - 2 \right) + 2\left( 2 \right)\]
\[ = - 3 + 4\]
\[ = 1\]
\[\text{ Let }C_{ij}\text{ be the cofactors of the elements }a_{ij}\text{ in }A = \left[ a_{ij} \right] .\text{ Then, }\]
\[ C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}- 1 & - 2 \\ - 1 & 1\end{vmatrix} = - 3, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}- 2 & - 2 \\ 0 & 1\end{vmatrix} = 2, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}- 2 & - 1 \\ 0 & - 1\end{vmatrix} = 2\]
\[ C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}2 & 0 \\ - 1 & 1\end{vmatrix} = - 2, C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}1 & 0 \\ 0 & 1\end{vmatrix} = 1, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}1 & 2 \\ 0 & - 1\end{vmatrix} = 1\]
\[ C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}2 & 0 \\ - 1 & - 2\end{vmatrix} = - 4, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}1 & 0 \\ - 2 & - 2\end{vmatrix} = 2, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}1 & 2 \\ - 2 & - 1\end{vmatrix} = 3\]
\[ \therefore adj A = \begin{bmatrix}- 3 & 2 & 2 \\ - 2 & 1 & 1 \\ - 4 & 2 & 3\end{bmatrix}^T \]
\[ = \begin{bmatrix}- 3 & - 2 & - 4 \\ 2 & 1 & 2 \\ 2 & 1 & 3\end{bmatrix}\]
\[ \Rightarrow A^{- 1} = \frac{1}{\left| A \right|}adj A\]
\[ = \frac{1}{1}\begin{bmatrix}- 3 & - 2 & - 4 \\ 2 & 1 & 2 \\ 2 & 1 & 3\end{bmatrix}\]
\[ = \begin{bmatrix}- 3 & - 2 & - 4 \\ 2 & 1 & 2 \\ 2 & 1 & 3\end{bmatrix}\]
\[\text{ We know that, }\left( A^T \right)^{- 1} = \left( A^{- 1} \right)^T . \]
\[\text{ Here, }C = A^T \]
\[i . e . , C = \begin{bmatrix}1 & - 2 & 0 \\ 2 & - 1 & - 1 \\ 0 & - 2 & 1\end{bmatrix}\]
\[ \therefore C^{- 1} = \begin{bmatrix}- 3 & 2 & 2 \\ - 2 & 1 & 1 \\ - 4 & 2 & 3\end{bmatrix}\]
\[\text{ or, }CX = B\]
\[\text{ where, }C = \begin{bmatrix}1 & - 2 & 0 \\ 2 & - 1 & - 1 \\ 0 & - 2 & 1\end{bmatrix}, X = \begin{bmatrix}x \\ y \\ z\end{bmatrix}\text{ and }B = \begin{bmatrix}10 \\ 8 \\ 7\end{bmatrix}\]
Now,
\[ \therefore X = C^{- 1} B\]
\[ \Rightarrow X = \begin{bmatrix}- 3 & 2 & 2 \\ - 2 & 1 & 1 \\ - 4 & 2 & 3\end{bmatrix}\begin{bmatrix}10 \\ 8 \\ 7\end{bmatrix}\]
\[ \Rightarrow X = \begin{bmatrix}- 30 + 16 + 14 \\ - 20 + 8 + 7 \\ - 40 + 16 + 21\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}0 \\ - 5 \\ - 3\end{bmatrix}\]
\[ \therefore x = 0, y = - 5\text{ and }z = - 3 .\]