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Question
Solve the following system of equations by matrix method:
8x + 4y + 3z = 18
2x + y +z = 5
x + 2y + z = 5
Solution
Here,
\[A = \begin{bmatrix}8 & 4 & 3 \\ 2 & 1 & 1 \\ 1 & 2 & 1\end{bmatrix}\]
\[\left| A \right| = \begin{vmatrix}8 & 4 & 3 \\ 2 & 1 & 1 \\ 1 & 2 & 1\end{vmatrix}\]
\[ = 8\left( 1 - 2 \right) - 4\left( 2 - 1 \right) + 3(4 - 1)\]
\[ = - 8 - 4 + 9\]
\[ = - 3\]
\[ {\text{ Let }C}_{ij} {\text{ be the cofactors of the elements a }}_{ij}\text{ in }A\left[ a_{ij} \right].\text{ Then, }\]
\[ C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}1 & 1 \\ 2 & 1\end{vmatrix} = - 1, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}2 & 1 \\ 1 & 1\end{vmatrix} = - 1, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}2 & 1 \\ 1 & 2\end{vmatrix} = 3\]
\[ C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}4 & 3 \\ 2 & 1\end{vmatrix} = 2, C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}8 & 3 \\ 1 & 1\end{vmatrix} = 5, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}8 & 4 \\ 1 & 2\end{vmatrix} = - 12\]
\[ C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}4 & 3 \\ 1 & 1\end{vmatrix} = 1, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}8 & 3 \\ 2 & 1\end{vmatrix} = - 2 , C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}8 & 4 \\ 2 & 1\end{vmatrix} = 0\]
\[adj A = \begin{bmatrix}- 1 & - 1 & 3 \\ 2 & 5 & - 12 \\ 1 & - 2 & 0\end{bmatrix}^T \]
\[ = \begin{bmatrix}- 1 & 2 & 1 \\ - 1 & 5 & - 2 \\ 3 & - 12 & 0\end{bmatrix}\]
\[ \Rightarrow A^{- 1} = \frac{1}{\left| A \right|}adj A\]
\[ = \frac{1}{- 3}\begin{bmatrix}- 1 & 2 & 1 \\ - 1 & 5 & - 2 \\ 3 & - 12 & 0\end{bmatrix}\]
\[X = A^{- 1} B\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{- 3}\begin{bmatrix}- 1 & 2 & 1 \\ - 1 & 5 & - 2 \\ 3 & - 12 & 0\end{bmatrix}\begin{bmatrix}18 \\ 5 \\ 5\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{- 3}\begin{bmatrix}- 18 + 10 + 5 \\ - 18 + 25 - 10 \\ 54 - 60\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{- 3}\begin{bmatrix}- 3 \\ - 3 \\ - 6\end{bmatrix}\]
\[ \Rightarrow x = \frac{- 3}{- 3}, y = \frac{- 3}{- 3}\text{ and }z = \frac{- 6}{- 3}\]
\[ \therefore x = 1, y = 1\text{ and }z = 2\]
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