Question

If \[A = \begin{bmatrix}2 & - 3 & 5 \\ 3 & 2 & - 4 \\ 1 & 1 & - 2\end{bmatrix}\], find A⁻¹ and hence solve the system of linear equations 2x − 3y + 5z = 11, 3x + 2y − 4z = −5, x + y + 2z = −3

Answer 1

Here,
\[A = \begin{bmatrix}2 & - 3 & 5 \\ 3 & 2 & - 4 \\ 1 & 1 & - 2\end{bmatrix}\]
\[\left| A \right| = \begin{vmatrix}2 & - 3 & 5 \\ 3 & 2 & - 4 \\ 1 & 1 & - 2\end{vmatrix}\]
\[ = 2\left( - 4 + 4 \right) + 3\left( - 6 + 4 \right) + 5(3 - 2)\]
\[ = 0 - 6 + 5\]
\[ = - 1\]
\[ {\text{ Let }C}_{ij} {\text{ be the co factors of the elements a }}_{ij}\text{ in }A\left[ a_{ij} \right].\text{ Then,}\]
\[ C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}2 & - 4 \\ 1 & - 2\end{vmatrix} = 0, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}3 & - 4 \\ 1 & - 2\end{vmatrix} = 2, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}3 & 2 \\ 1 & 1\end{vmatrix} = 1\]
\[ C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}- 3 & 5 \\ 1 & - 2\end{vmatrix} = - 1, C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}2 & 5 \\ 1 & - 2\end{vmatrix} = - 9, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}2 & - 3 \\ 1 & 1\end{vmatrix} = - 5\]
\[ C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}- 3 & 5 \\ 2 & - 4\end{vmatrix} = 2 , C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}2 & 5 \\ 3 & - 4\end{vmatrix} = 23, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}2 & - 3 \\ 3 & 2\end{vmatrix} = 13\]
\[adj A = \begin{bmatrix}0 & 2 & 1 \\ - 1 & - 9 & - 5 \\ 2 & 23 & 13\end{bmatrix}^T \]
\[ = \begin{bmatrix}0 & - 1 & 2 \\ 2 & - 9 & 23 \\ 1 & - 5 & 13\end{bmatrix}\]
\[ \Rightarrow A^{- 1} = \frac{1}{\left| A \right|}adj A\]
\[ = \frac{1}{- 1}\begin{bmatrix}0 & - 1 & 2 \\ 2 & - 9 & 23 \\ 1 & - 5 & 13\end{bmatrix}\]
The given system of equations can be written in matrix form as follows:
\[\begin{bmatrix}2 & - 3 & 5 \\ 3 & 2 & - 4 \\ 1 & 1 & - 2\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}11 \\ - 5 \\ - 3\end{bmatrix}\]
\[X = A^{- 1} B\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{- 1}\begin{bmatrix}0 & - 1 & 2 \\ 2 & - 9 & 23 \\ 1 & - 5 & 13\end{bmatrix}\begin{bmatrix}11 \\ - 5 \\ - 3\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{- 1}\begin{bmatrix}0 + 5 - 6 \\ 22 + 45 - 69 \\ 11 + 25 - 39\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{- 1}\begin{bmatrix}- 1 \\ - 2 \\ - 3\end{bmatrix}\]
\[ \Rightarrow x = \frac{- 1}{- 1}, y = \frac{- 2}{- 1}\text{ and }z = \frac{- 3}{- 1}\]
\[ \therefore x = 1, y = 2\text{ and }z = 3\]