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प्रश्न
Solve the following system of equations by matrix method:
6x − 12y + 25z = 4
4x + 15y − 20z = 3
2x + 18y + 15z = 10
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उत्तर
Here,
\[A = \begin{bmatrix}6 & - 12 & 25 \\ 4 & 15 & - 20 \\ 2 & 18 & 15\end{bmatrix}\]
\[\left| A \right| = \begin{vmatrix}6 & - 12 & 25 \\ 4 & 15 & - 20 \\ 2 & 18 & 15\end{vmatrix}\]
\[ = 6\left( 225 + 360 \right) + 12\left( 60 + 40 \right) + 25(72 - 30)\]
\[ = 3510 + 1200 + 1050\]
\[ = 5760\]
\[ {\text{ Let }C}_{ij} {\text{ be the cofactors of the elements a }}_{ij}\text{ in }A\left[ a_{ij} \right]. \text{ Then,}\]
\[ C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}15 & - 20 \\ 18 & 15\end{vmatrix} = 585, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}4 & - 20 \\ 2 & 15\end{vmatrix} = - 100 , C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}4 & 15 \\ 2 & 18\end{vmatrix} = 42\]
\[ C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}- 12 & 25 \\ 18 & 15\end{vmatrix} = 630, C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}6 & 25 \\ 2 & 15\end{vmatrix} = 40, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}6 & - 12 \\ 2 & 18\end{vmatrix} = - 132\]
\[ C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}- 12 & 25 \\ 15 & - 20\end{vmatrix} = - 135, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}6 & 25 \\ 4 & - 20\end{vmatrix} = 220, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}6 & - 12 \\ 4 & 15\end{vmatrix} = 138\]
\[adj A = \begin{bmatrix}585 & - 100 & 42 \\ 630 & 40 & - 132 \\ - 135 & 220 & 138\end{bmatrix}^T \]
\[ = \begin{bmatrix}585 & 630 & - 135 \\ - 100 & 40 & 220 \\ 42 & - 132 & 138\end{bmatrix}\]
\[ \Rightarrow A^{- 1} = \frac{1}{\left| A \right|}adj A\]
\[ = \frac{1}{5760}\begin{bmatrix}585 & 630 & - 135 \\ - 100 & 40 & 220 \\ 42 & - 132 & 138\end{bmatrix}\]
\[X = A^{- 1} B\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{5760}\begin{bmatrix}585 & 630 & - 135 \\ - 100 & 40 & 220 \\ 42 & - 132 & 138\end{bmatrix}\begin{bmatrix}4 \\ 3 \\ 10\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{5760}\begin{bmatrix}2340 + 1890 - 1350 \\ - 400 + 120 + 2200 \\ 168 - 396 + 1380\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{5760}\begin{bmatrix}2880 \\ 1920 \\ 1152\end{bmatrix}\]
\[ \Rightarrow x = \frac{2880}{5760}, y = \frac{1920}{5760}\text{ and }z = \frac{1152}{5760}\]
\[ \therefore x = \frac{1}{2}, y = \frac{1}{3}\text{ and }z = \frac{1}{5}\]
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