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प्रश्न
Solve the following system of equations by matrix method:
5x + 3y + z = 16
2x + y + 3z = 19
x + 2y + 4z = 25
उत्तर
Here,
\[A = \begin{bmatrix}5 & 3 & 1 \\ 2 & 1 & 3 \\ 1 & 2 & 4\end{bmatrix}\]
\[\left| A \right| = \begin{vmatrix}5 & 3 & 1 \\ 2 & 1 & 3 \\ 1 & 2 & 4\end{vmatrix} = 5\left( 4 - 6 \right) - 3\left( 8 - 3 \right) + 1(4 - 1)\]
\[ = - 10 - 15 + 3\]
\[ = - 22\]
\[ {\text{ Let }C}_{ij} {\text{ be the cofactors of the elements a }}_{ij}\text{ in }A\left[ a_{ij} \right].\text{ Then,}\]
\[ C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}1 & 3 \\ 2 & 4\end{vmatrix} = - 2, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}2 & 3 \\ 1 & 4\end{vmatrix} = - 5 , C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}2 & 1 \\ 1 & 2\end{vmatrix} = 3\]
\[ C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}3 & 1 \\ 2 & 4\end{vmatrix} = - 10 , C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}5 & 1 \\ 1 & 4\end{vmatrix} = 19, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}5 & 3 \\ 1 & 2\end{vmatrix} = - 7\]
\[ C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}3 & 1 \\ 1 & 3\end{vmatrix} = 8 , C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}5 & 1 \\ 2 & 3\end{vmatrix} = - 13 , C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}5 & 3 \\ 2 & 1\end{vmatrix} = - 1\]
\[adj A = \begin{bmatrix}- 2 & - 5 & 3 \\ - 10 & 19 & - 7 \\ 8 & - 13 & - 1\end{bmatrix}^T \]
\[ = \begin{bmatrix}- 2 & - 10 & 8 \\ - 5 & 19 & - 13 \\ 3 & - 7 & - 1\end{bmatrix}\]
\[ \Rightarrow A^{- 1} = \frac{1}{\left| A \right|}adj A\]
\[ = \frac{1}{- 22}\begin{bmatrix}- 2 & - 10 & 8 \\ - 5 & 19 & - 13 \\ 3 & - 7 & - 1\end{bmatrix}\]
\[X = A^{- 1} B\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{- 22}\begin{bmatrix}- 2 & - 10 & 8 \\ - 5 & 19 & - 13 \\ 3 & - 7 & - 1\end{bmatrix}\begin{bmatrix}16 \\ 19 \\ 25\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{- 22}\begin{bmatrix}- 32 - 190 + 200 \\ - 80 + 361 - 325 \\ 48 - 133 - 25\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{- 22}\begin{bmatrix}- 22 \\ - 44 \\ - 110\end{bmatrix}\]
\[ \Rightarrow x = \frac{- 22}{- 22}, y = \frac{- 44}{- 22}\text{ and }z = \frac{- 110}{- 22}\]
\[ \therefore x = 1, y = 2\text{ and }z = 5\]
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