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Question
x − 4y − z = 11
2x − 5y + 2z = 39
− 3x + 2y + z = 1
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Solution
Given: x − 4y − z = 11
2x − 5y + 2z = 39
− 3x + 2y + z = 1
\[D = \begin{vmatrix}1 & - 4 & - 1 \\ 2 & - 5 & 2 \\ - 3 & 2 & 1\end{vmatrix}\]
\[ = 1\left( - 5 - 4 \right) - ( - 4)(2 + 6) + ( - 1)(4 - 15)\]
\[ = 1( - 9) - ( - 4)(8) + ( - 1)( - 11) = 34\]
\[ D_1 = \begin{vmatrix}11 & - 4 & - 1 \\ 39 & - 5 & 2 \\ 1 & 2 & 1\end{vmatrix}\]
\[ = 11( - 5 - 4) - ( - 4)(39 - 2) + ( - 1)(78 + 5)\]
\[ = 11( - 9) - ( - 4)(37) + ( - 1)(83) = - 34\]
\[ D_2 = \begin{vmatrix}1 & 11 & - 1 \\ 2 & 39 & 2 \\ - 3 & 1 & 1\end{vmatrix}\]
\[ = 1(39 - 2) - 11(2 + 6) + ( - 1)(2 + 117)\]
\[ = 1(37) - 11(8) + ( - 1)(119) = - 170\]
\[ D_3 = \begin{vmatrix}1 & - 4 & 11 \\ 2 & - 5 & 39 \\ - 3 & 2 & 1\end{vmatrix}\]
\[ = 1( - 5 - 78) - ( - 4)(2 + 117) + 11(4 - 15)\]
\[ = 1( - 83) - ( - 4)(119) + 11( - 11) = 272\]
Now,
\[x = \frac{D_1}{D} = \frac{- 34}{34} = - 1\]
\[y = \frac{D_2}{D} = \frac{- 170}{34} = - 5\]
\[z = \frac{D_3}{D} = \frac{272}{34} = 8\]
\[ \therefore x = -1, y = -5\text{ and }z = 8\]
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