Question

If \[A = \begin{bmatrix}3 & - 4 & 2 \\ 2 & 3 & 5 \\ 1 & 0 & 1\end{bmatrix}\] , find A⁻¹ and hence solve the following system of equations: 

Answer 1

Here,
\[ A = \begin{bmatrix}3 & - 4 & 2 \\ 2 & 3 & 5 \\ 1 & 0 & 1\end{bmatrix}\]
\[\left| A \right|=3 \left( 3 - 0 \right) + 4\left( 2 - 5 \right) + 2\left( 0 - 3 \right)\]
\[ = 9 - 12 - 6\]
\[ = - 9\]
\[ {\text{ Let }C}_{ij} {\text{ be the cofactors of the elements a }}_{ij}\text{ in }A=\left[ a_{ij} \right].\text{ Then,}\]
\[ C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}3 & 5 \\ 0 & 1\end{vmatrix} = 3, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}2 & 5 \\ 1 & 1\end{vmatrix} = 3, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}2 & 3 \\ 1 & 0\end{vmatrix} = - 3\]
\[ C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}- 4 & 2 \\ 0 & 1\end{vmatrix} = 4, C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}3 & 2 \\ 1 & 1\end{vmatrix} = 1, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}3 & - 4 \\ 1 & 0\end{vmatrix} = - 4\]
\[ C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}- 4 & 2 \\ 3 & 5\end{vmatrix} = - 26, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}3 & 2 \\ 2 & 5\end{vmatrix} = - 11, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}3 & - 4 \\ 2 & 3\end{vmatrix} = 17\]
\[adj A = \begin{bmatrix}3 & 3 & - 3 \\ 4 & 1 & - 4 \\ - 26 & - 11 & 17\end{bmatrix}^T \]
\[ = \begin{bmatrix}3 & 4 & - 26 \\ 3 & 1 & - 11 \\ - 3 & - 4 & 17\end{bmatrix}\]
\[ \Rightarrow A^{- 1} = \frac{1}{\left| A \right|}adj A\]
\[ = \frac{1}{- 9}\begin{bmatrix}3 & 4 & - 26 \\ 3 & 1 & - 11 \\ - 3 & - 4 & 17\end{bmatrix}\]
\[AX = B\]
Here, 
\[A = \begin{bmatrix}3 & - 4 & 2 \\ 2 & 3 & 5 \\ 1 & 0 & 1\end{bmatrix}, X = \begin{bmatrix}x \\ y \\ z\end{bmatrix}\text{ and }B = \begin{bmatrix}- 1 \\ 7 \\ 2\end{bmatrix}\]
\[X = A^{- 1} B\]
\[ \Rightarrow X = \frac{1}{- 9}\begin{bmatrix}3 & 4 & - 26 \\ 3 & 1 & - 11 \\ - 3 & - 4 & 17\end{bmatrix}\begin{bmatrix}- 1 \\ 7 \\ 2\end{bmatrix}\]
\[ \Rightarrow X = \frac{1}{- 9}\begin{bmatrix}- 3 + 28 - 52 \\ - 3 + 7 - 22 \\ 3 - 28 + 34\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{- 9}\begin{bmatrix}- 27 \\ - 18 \\ 9\end{bmatrix}\]
\[ \therefore x = 3, y = 2\text{ and }z = - 1\]