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Question
3x + y + z = 2
2x − 4y + 3z = − 1
4x + y − 3z = − 11
Solution
Given: 3x + y + z = 2
2x − 4y + 3z = − 1
4x + y − 3z = − 11
\[D = \begin{vmatrix}3 & 1 & 1 \\ 2 & - 4 & 3 \\ 4 & 1 & - 3\end{vmatrix}\]
\[ = 3\left( 12 - 3 \right) - 2\left( - 3 - 1 \right) + 4\left( 3 + 4 \right)\]
\[ = 27 + 8 + 28\]
\[ = 63\]
\[ D_1 = \begin{vmatrix}2 & 1 & 1 \\ - 1 & - 4 & 3 \\ - 11 & 1 & - 3\end{vmatrix}\]
\[ = 2\left( 12 - 3 \right) + 1\left( - 3 - 1 \right) - 11\left( 3 + 4 \right)\]
\[ = 18 - 4 - 77\]
\[ = - 63\]
\[ D_2 = \begin{vmatrix}3 & 2 & 1 \\ 2 & - 1 & 3 \\ 4 & - 11 & - 3\end{vmatrix}\]
\[ = 3\left( 3 + 33 \right) - 2\left( - 6 + 11 \right) + 4\left( 6 + 1 \right)\]
\[ = 108 - 10 + 28\]
\[ = 126\]
\[ D_3 = \begin{vmatrix}3 & 1 & 2 \\ 2 & - 4 & - 1 \\ 4 & 1 & - 11\end{vmatrix}\]
\[ = 3\left( 44 + 1 \right) - 2\left( - 11 - 2 \right) + 4\left( - 1 + 8 \right)\]
\[ = 135 + 26 + 28\]
\[ = 189\]
\[Now, \]
\[x = \frac{D_1}{D} = \frac{- 63}{63} = - 1\]
\[y = \frac{D_2}{D} = \frac{126}{63} = 2\]
\[z = \frac{D_3}{D} = \frac{189}{63} = 3\]
\[ \therefore x = - 1, y = 2\text{ and }z = 3\]
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