Question

Solve the following system of equations by matrix method:
x + y + z = 3
2x − y + z = − 1
2x + y − 3z = − 9

Answer 1

Here,
\[A = \begin{bmatrix}1 & 1 & 1 \\ 2 & - 1 & 1 \\ 2 & 1 & - 3\end{bmatrix}\]
\[\left| A \right| = \begin{vmatrix}1 & 1 & 1 \\ 2 & - 1 & 1 \\ 2 & 1 & - 3\end{vmatrix}\]
\[ = 1\left( 3 - 1 \right) - 1\left( - 6 - 2 \right) + 1(2 + 2)\]
\[ = 2 + 8 + 4\]
\[ = 14\]
\[ {\text{ Let }C}_{ij} {\text{be the cofactors of the elements a}}_{ij}\text{ in }A\left[ a_{ij} \right]. \text{ Then,}\]
\[ C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}- 1 & 1 \\ 1 & - 3\end{vmatrix} = 2 , C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}2 & 1 \\ 2 & - 3\end{vmatrix} = 8, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}2 & - 1 \\ 2 & 1\end{vmatrix} = 4\]
\[ C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}1 & 1 \\ 1 & - 3\end{vmatrix} = 4, C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}1 & 1 \\ 2 & - 3\end{vmatrix} = - 5, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}1 & 1 \\ 2 & 1\end{vmatrix} = 1\]
\[ C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}1 & 1 \\ - 1 & 1\end{vmatrix} = 2 , C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}1 & 1 \\ 2 & 1\end{vmatrix} = 1, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}1 & 1 \\ 2 & - 1\end{vmatrix} = - 3\]
\[adj A = \begin{bmatrix}2 & 8 & 4 \\ 4 & - 5 & 1 \\ 2 & 1 & - 3\end{bmatrix}^T \]
\[ = \begin{bmatrix}2 & 4 & 2 \\ 8 & - 5 & 1 \\ 4 & 1 & - 3\end{bmatrix}\]
\[ \Rightarrow A^{- 1} = \frac{1}{\left| A \right|}adj A\]
\[ = \frac{1}{14}\begin{bmatrix}2 & 4 & 2 \\ 8 & - 5 & 1 \\ 4 & 1 & - 3\end{bmatrix}\]
\[X = A^{- 1} B\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{14}\begin{bmatrix}2 & 4 & 2 \\ 8 & - 5 & 1 \\ 4 & 1 & - 3\end{bmatrix}\begin{bmatrix}3 \\ - 1 \\ - 9\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{14}\begin{bmatrix}6 - 4 - 18 \\ 24 + 5 - 9 \\ 12 - 1 + 27\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{14}\begin{bmatrix}- 16 \\ 20 \\ 38\end{bmatrix}\]
\[ \Rightarrow x = \frac{- 16}{14}, y = \frac{20}{14}\text{ and }z = \frac{38}{14}\]
\[ \therefore x = \frac{- 8}{7}, y = \frac{10}{7}\text{ and }z = \frac{19}{7}\]