Question

Find A⁻¹, if \[A = \begin{bmatrix}1 & 2 & 5 \\ 1 & - 1 & - 1 \\ 2 & 3 & - 1\end{bmatrix}\] . Hence solve the following system of linear equations:x + 2y + 5z = 10, x − y − z = −2, 2x + 3y − z = −11

Answer 1

Here,
\[A = \begin{bmatrix}1 & 2 & 5 \\ 1 & - 1 & - 1 \\ 2 & 3 & - 1\end{bmatrix}\]
\[\left| A \right| = \begin{vmatrix}1 & 2 & 5 \\ 1 & - 1 & - 1 \\ 2 & 3 & - 1\end{vmatrix}\]
\[ = 1\left( 1 + 3 \right) - 2\left( - 1 + 2 \right) + 5(3 + 2)\]
\[ = 4 - 2 + 25\]
\[ = 27\]
\[ {\text{ Let }C}_{ij} {\text{ be the co factors of the elements a }}_{ij} \text{ in }A\left[ a_{ij} \right]\text{. Then,}\]
\[ C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}- 1 & - 1 \\ 3 & - 1\end{vmatrix} = 4, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}1 & - 1 \\ 2 & - 1\end{vmatrix} = - 1, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}1 & - 1 \\ - 2 & 3\end{vmatrix} = 5\]
\[ C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}2 & 5 \\ 3 & - 1\end{vmatrix} = 17, C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}1 & 5 \\ 2 & - 1\end{vmatrix} = - 11, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}1 & 2 \\ 2 & 3\end{vmatrix} = 1\]
\[ C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}2 & 5 \\ - 1 & - 1\end{vmatrix} = 3, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}1 & 5 \\ 1 & - 1\end{vmatrix} = 6, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}1 & 2 \\ 1 & - 1\end{vmatrix} = - 3\]
\[adj A = \begin{bmatrix}4 & - 1 & 5 \\ 17 & - 11 & 1 \\ 3 & 6 & - 3\end{bmatrix}^T \]
\[ = \begin{bmatrix}4 & 17 & 3 \\ - 1 & - 11 & 6 \\ 5 & 1 & - 3\end{bmatrix}\]
\[ \Rightarrow A^{- 1} = \frac{1}{\left| A \right|}adj A\]
\[ = \frac{1}{27}\begin{bmatrix}4 & 17 & 3 \\ - 1 & - 11 & 6 \\ 5 & 1 & - 3\end{bmatrix}\]
The given system of equations can be written in matrix form as follows:
\[\begin{bmatrix}1 & 2 & 5 \\ 1 & - 1 & - 1 \\ 2 & 3 & - 1\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}10 \\ - 2 \\ - 11\end{bmatrix}\]
\[X = A^{- 1} B\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{27}\begin{bmatrix}4 & 17 & 3 \\ - 1 & - 11 & 6 \\ 5 & 1 & - 3\end{bmatrix}\begin{bmatrix}10 \\ - 2 \\ - 11\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{27}\begin{bmatrix}40 - 34 - 33 \\ - 10 + 22 - 66 \\ 50 - 2 + 33\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{27}\begin{bmatrix}- 27 \\ - 54 \\ 81\end{bmatrix}\]
\[ \therefore x = \frac{- 27}{27}, y = \frac{- 54}{27}\text{ and }z = \frac{81}{27}\]
\[ \Rightarrow x = - 1, y = - 2\text{ and }z = 3\]