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Question
Solve the following system of equations by matrix method:
2x + y + z = 2
x + 3y − z = 5
3x + y − 2z = 6
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Solution
Here,
\[A = \begin{bmatrix}2 & 1 & 1 \\ 1 & 3 & - 1 \\ 3 & 1 & - 2\end{bmatrix}\]
\[\left| A \right| = \begin{vmatrix}2 & 1 & 1 \\ 1 & 3 & - 1 \\ 3 & 1 & - 2\end{vmatrix}\]
\[ = 2\left( - 6 + 1 \right) - 1\left( - 2 + 3 \right) + 1(1 - 9)\]
\[ = - 10 - 1 - 8\]
\[ = - 19\]
\[ {\text{ Let }C}_{ij} {\text{ be the cofactors of the elements a }}_{ij}\text{ in }A\left[ a_{ij} \right].\text{ Then,}\]
\[ C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}3 & - 1 \\ 1 & - 2\end{vmatrix} = - 5, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}1 & - 1 \\ 3 & - 2\end{vmatrix} = - 1, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}1 & 3 \\ 3 & 1\end{vmatrix} = - 8\]
\[ C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}1 & 1 \\ 1 & - 2\end{vmatrix} = 3, C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}2 & - 1 \\ 3 & - 2\end{vmatrix} = - 7, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}2 & 1 \\ 3 & 1\end{vmatrix} = 1\]
\[ C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}1 & 1 \\ 3 & - 1\end{vmatrix} = - 4, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}2 & 1 \\ 1 & - 1\end{vmatrix} = 3, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}2 & 1 \\ 1 & 3\end{vmatrix} = 5\]
\[adj A = \begin{bmatrix}- 5 & - 1 & - 8 \\ 3 & - 7 & 1 \\ - 4 & 3 & 5\end{bmatrix}^T \]
\[ = \begin{bmatrix}- 5 & 3 & - 4 \\ - 1 & - 7 & 3 \\ - 8 & 1 & 5\end{bmatrix}\]
\[ \Rightarrow A^{- 1} = \frac{1}{\left| A \right|}adj A\]
\[ = \frac{1}{- 19}\begin{bmatrix}- 5 & 3 & - 4 \\ - 1 & - 7 & 3 \\ - 8 & 1 & 5\end{bmatrix}\]
\[X = A^{- 1} B\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{- 19}\begin{bmatrix}- 5 & 3 & - 4 \\ - 1 & - 7 & 3 \\ - 8 & 1 & 5\end{bmatrix}\begin{bmatrix}2 \\ 5 \\ 6\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{- 19}\begin{bmatrix}- 10 + 15 - 24 \\ - 2 - 35 + 18 \\ - 16 + 5 + 30\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{- 19}\begin{bmatrix}- 19 \\ 19 \\ 19\end{bmatrix}\]
\[ \Rightarrow x = \frac{- 19}{- 19}, y = \frac{19}{- 19}\text{ and }z = \frac{19}{- 19}\]
\[ \therefore x = 1, y = 3\text{ and }z = - 1\]
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