Question

Solve the following system of equations by matrix method:
x − y + 2z = 7
3x + 4y − 5z = −5
2x − y + 3z = 12

Answer 1

Here,
\[A = \begin{bmatrix}1 & - 1 & 2 \\ 3 & 4 & - 5 \\ 2 & - 1 & 3\end{bmatrix}\]
\[\left| A \right| = \begin{vmatrix}1 & - 1 & 2 \\ 3 & 4 & - 5 \\ 2 & - 1 & 3\end{vmatrix}\]
\[ = 1\left( 12 - 5 \right) + 1\left( 9 + 10 \right) + 2( - 3 - 8)\]
\[ = 7 + 19 - 22\]
\[ = 4\]
\[\text{ Let }C_{ij}\text{ be the cofactors of elements }a_{ij}\text{ in }A = \left[ a_{ij} \right] .\text{ Then, }\]
\[ C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}4 & - 5 \\ - 1 & 3\end{vmatrix} = 7, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}3 & - 5 \\ 2 & 3\end{vmatrix} = - 19, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}3 & 4 \\ 2 & - 1\end{vmatrix} = - 11\]
\[ C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}- 1 & 2 \\ - 1 & 3\end{vmatrix} = 1 , C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}1 & 2 \\ 2 & 3\end{vmatrix} = - 1 , C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}1 & - 1 \\ 2 & - 1\end{vmatrix} = - 1\]
\[ C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}- 1 & 2 \\ 4 & - 5\end{vmatrix} = - 3, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}1 & 2 \\ 3 & - 5\end{vmatrix} = 11, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}1 & - 1 \\ 3 & 4\end{vmatrix} = 7\]
\[adj A = \begin{bmatrix}7 & - 19 & - 11 \\ 1 & - 1 & - 1 \\ - 3 & 11 & 7\end{bmatrix}^T \]
\[ = \begin{bmatrix}7 & 1 & - 3 \\ - 19 & - 1 & 11 \\ - 11 & - 1 & 7\end{bmatrix}\]
\[ \Rightarrow A^{- 1} = \frac{1}{\left| A \right|}adj A\]
\[ = \frac{1}{4}\begin{bmatrix}7 & 1 & - 3 \\ - 19 & - 1 & 11 \\ - 11 & - 1 & 7\end{bmatrix}\]
\[X = A^{- 1} B\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{4}\begin{bmatrix}7 & 1 & - 3 \\ - 19 & - 1 & 11 \\ - 11 & - 1 & 7\end{bmatrix}\begin{bmatrix}7 \\ - 5 \\ 12\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{4}\begin{bmatrix}49 - 5 - 36 \\ - 133 + 5 + 132 \\ - 77 + 5 + 84\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{4}\begin{bmatrix}8 \\ 4 \\ 12\end{bmatrix}\]
\[ \Rightarrow x = \frac{8}{4}, y = \frac{4}{4}\text{ and }z = \frac{12}{4}\]
\[ \therefore x = 2, y = 1\text{ and }z = 3 .\]