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प्रश्न
Solve the following system of equations by matrix method:
3x + 4y + 2z = 8
2y − 3z = 3
x − 2y + 6z = −2
उत्तर
Here,
\[A = \begin{bmatrix}3 & 4 & 2 \\ 0 & 2 & - 3 \\ 1 & - 2 & 6\end{bmatrix}\]
\[\left| A \right| = \begin{vmatrix}3 & 4 & 2 \\ 0 & 2 & - 3 \\ 1 & - 2 & 6\end{vmatrix}\]
\[ = 3\left( 12 - 6 \right) - 4\left( 0 + 3 \right) + 2(0 - 2)\]
\[ = 18 - 12 - 4\]
\[ = 2\]
\[ {\text{ Let }C}_{ij} {\text{ be the cofactors of the elements a }}_{ij}\text{ in }A\left[ a_{ij} \right].\text{ Then,}\]
\[ C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}2 & - 3 \\ - 2 & 6\end{vmatrix} = 6, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}0 & - 3 \\ 1 & 6\end{vmatrix} = - 3 , C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}0 & 2 \\ 1 & - 2\end{vmatrix} = - 2\]
\[ C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}4 & 2 \\ - 2 & 6\end{vmatrix} = - 28 , C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}3 & 2 \\ 1 & 6\end{vmatrix} = 16 , C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}3 & 4 \\ 1 & - 2\end{vmatrix} = 10\]
\[ C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}4 & 2 \\ 2 & - 3\end{vmatrix} = - 16, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}3 & 2 \\ 0 & - 3\end{vmatrix} = 9, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}3 & 4 \\ 0 & 2\end{vmatrix} = 6\]
\[adj A = \begin{bmatrix}6 & - 3 & - 2 \\ - 28 & 16 & 10 \\ - 16 & 9 & 6\end{bmatrix}^T \]
\[ = \begin{bmatrix}6 & - 28 & - 16 \\ - 3 & 16 & 9 \\ - 2 & 10 & 6\end{bmatrix}\]
\[ \Rightarrow A^{- 1} = \frac{1}{\left| A \right|}adj A\]
\[ = \frac{1}{2}\begin{bmatrix}6 & - 28 & - 16 \\ - 3 & 16 & 9 \\ - 2 & 10 & 6\end{bmatrix}\]
\[X = A^{- 1} B\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{2}\begin{bmatrix}6 & - 28 & - 16 \\ - 3 & 16 & 9 \\ - 2 & 10 & 6\end{bmatrix}\begin{bmatrix}8 \\ 3 \\ - 2\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{2}\begin{bmatrix}48 - 84 + 32 \\ - 24 + 48 - 18 \\ - 16 + 30 - 12\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{2}\begin{bmatrix}- 4 \\ 6 \\ 2\end{bmatrix}\]
\[ \Rightarrow x = \frac{- 4}{2}, y = \frac{6}{2}\text{ and }z = \frac{2}{2}\]
\[ \therefore x = - 2, y = 3\text{ and }z = 1\]
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