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Question
Solve the following system of equations by matrix method:
2x + 6y = 2
3x − z = −8
2x − y + z = −3
Solution
Here,
\[A = \begin{bmatrix}2 & 6 & 0 \\ 3 & 0 & - 1 \\ 2 & - 1 & 1\end{bmatrix}\]
\[\left| A \right| = \begin{vmatrix}2 & 6 & 0 \\ 3 & 0 & - 1 \\ 2 & - 1 & 1\end{vmatrix}\]
\[ = 2\left( 0 - 1 \right) - 6\left( 3 + 2 \right) + 0( - 3 + 0)\]
\[ = - 2 - 30\]
\[ = - 32\]
\[ {\text{ Let }C}_{ij} {\text{ be the cofactors of the elements a }}_{ij}\text{ in }A\left[ a_{ij} \right].\text{ Then,}\]
\[ C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}0 & - 1 \\ - 1 & 1\end{vmatrix} = - 1 , C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}3 & - 1 \\ 2 & 1\end{vmatrix} = - 5 , C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}3 & 0 \\ 2 & - 1\end{vmatrix} = - 3\]
\[ C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}6 & 0 \\ - 1 & 1\end{vmatrix} = - 6 , C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}2 & 0 \\ 2 & 1\end{vmatrix} = 2 , C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}2 & 6 \\ 2 & - 1\end{vmatrix} = 14\]
\[ C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}6 & 0 \\ 0 & - 1\end{vmatrix} = - 6 , C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}2 & 0 \\ 3 & - 1\end{vmatrix} = 2 , C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}2 & 6 \\ 3 & 0\end{vmatrix} = - 18\]
\[adj A = \begin{bmatrix}- 1 & - 5 & - 3 \\ - 6 & 2 & 14 \\ - 6 & 2 & - 18\end{bmatrix}^T \]
\[ = \begin{bmatrix}- 1 & - 6 & - 6 \\ - 5 & 2 & 2 \\ - 3 & 14 & - 18\end{bmatrix}\]
\[ \Rightarrow A^{- 1} = \frac{1}{\left| A \right|}adj A\]
\[ = \frac{1}{- 32}\begin{bmatrix}- 1 & - 6 & - 6 \\ - 5 & 2 & 2 \\ - 3 & 14 & - 18\end{bmatrix}\]
\[X = A^{- 1} B\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{- 32}\begin{bmatrix}- 1 & - 6 & - 6 \\ - 5 & 2 & 2 \\ - 3 & 14 & - 18\end{bmatrix}\begin{bmatrix}2 \\ - 8 \\ - 3\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{- 32}\begin{bmatrix}- 2 + 48 + 18 \\ - 10 - 16 - 6 \\ - 6 - 112 + 54\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{- 32}\begin{bmatrix}64 \\ - 32 \\ - 64\end{bmatrix}\]
\[ \Rightarrow x = \frac{64}{- 32}, y = \frac{- 32}{- 32}\text{ and }z = \frac{- 64}{- 32}\]
\[ \therefore x = - 2, y = 1\text{ and }z = 2\]
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