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Question
Solve the following system of equations by matrix method:
x + y − z = 3
2x + 3y + z = 10
3x − y − 7z = 1
Solution
Here,
\[A = \begin{bmatrix}1 & 1 & - 1 \\ 2 & 3 & 1 \\ 3 & - 1 & - 7\end{bmatrix}\]
\[\left| A \right| = \begin{vmatrix}1 & 1 & - 1 \\ 2 & 3 & 1 \\ 3 & - 1 & - 7\end{vmatrix}\]
\[ = 1\left( - 21 + 1 \right) - 1\left( - 14 - 3 \right) - 1( - 2 - 9)\]
\[ = - 20 + 17 + 11\]
\[ = 8\]
\[ {\text{ Let }C}_{ij} {\text{ be the cofactors of the elements a }}_{ij}\text{ in }A\left[ a_{ij} \right].\text{ Then,}\]
\[ C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}3 & 1 \\ - 1 & - 7\end{vmatrix} = - 20 , C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}2 & 1 \\ 3 & - 7\end{vmatrix} = 17, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}2 & 3 \\ 3 & - 1\end{vmatrix} = - 11\]
\[ C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}1 & - 1 \\ - 1 & - 7\end{vmatrix} = 8 , C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}1 & - 1 \\ 3 & - 7\end{vmatrix} = - 4, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}1 & 1 \\ 3 & - 1\end{vmatrix} = 4\]
\[ C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}1 & - 1 \\ 3 & 1\end{vmatrix} = 4 , C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}1 & - 1 \\ 2 & 1\end{vmatrix} = - 3, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}1 & 1 \\ 2 & 3\end{vmatrix} = 1\]
\[adj A = \begin{bmatrix}- 20 & 17 & - 11 \\ 8 & - 4 & 4 \\ 4 & - 3 & 1\end{bmatrix}^T \]
\[ = \begin{bmatrix}- 20 & 8 & 4 \\ 17 & - 4 & - 3 \\ - 11 & 4 & 1\end{bmatrix}\]
\[ \Rightarrow A^{- 1} = \frac{1}{\left| A \right|}adj A\]
\[ = \frac{1}{8}\begin{bmatrix}- 20 & 8 & 4 \\ 17 & - 4 & - 3 \\ - 11 & 4 & 1\end{bmatrix}\]
\[X = A^{- 1} B\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{8}\begin{bmatrix}- 20 & 8 & 4 \\ 17 & - 4 & - 3 \\ - 11 & 4 & 1\end{bmatrix}\begin{bmatrix}3 \\ 10 \\ 1\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{8}\begin{bmatrix}- 60 + 80 + 4 \\ 51 - 40 - 3 \\ - 33 + 40 + 1\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{8}\begin{bmatrix}24 \\ 8 \\ 8\end{bmatrix}\]
\[ \Rightarrow x = \frac{24}{8}, y = \frac{8}{8}\text{ and }z = \frac{8}{8}\]
\[ \therefore x = 3, y = 1\text{ and }z = 1\]
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