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Question
2x − 3y − 4z = 29
− 2x + 5y − z = − 15
3x − y + 5z = − 11
Solution
Given: 2x − 3y − 4z = 29
− 2x + 5y − z = − 15
3x − y + 5z = − 11
\[D = \begin{vmatrix}2 & - 3 & - 4 \\ - 2 & 5 & - 1 \\ 3 & - 1 & 5\end{vmatrix}\]
\[ = 2(25 - 1) + 3( - 10 + 3) - 4(2 - 15)\]
\[ = 2(24) + 3( - 7) - 4( - 13)\]
\[ = 79\]
\[ D_1 = \begin{vmatrix}29 & - 3 & - 4 \\ - 15 & 5 & - 1 \\ - 11 & - 1 & 5\end{vmatrix}\]
\[ = 29(25 - 1) + 3( - 75 - 11) - 4(15 + 55)\]
\[ = 29(24) + 3( - 86) - 4(70)\]
\[ = 158\]
\[ D_2 = \begin{vmatrix}2 & 29 & - 4 \\ - 2 & - 15 & - 1 \\ 3 & - 11 & 5\end{vmatrix}\]
\[ = 2( - 75 - 11) - 29( - 10 + 3) - 4(22 + 45)\]
\[ = 2( - 86) - 29( - 7) - 4(67)\]
\[ = - 237\]
\[ D_3 = \begin{vmatrix}2 & - 3 & 29 \\ - 2 & 5 & - 15 \\ 3 & - 1 & - 11\end{vmatrix}\]
\[ = 2( - 55 - 15) + 3(22 + 45) + 29(2 - 15)\]
\[ = 2( - 70) + 3(67) + 29( - 13)\]
\[ = - 316\]
Now,
\[x = \frac{D_1}{D} = \frac{158}{79} = 2\]
\[y = \frac{D_2}{D} = \frac{- 237}{79} = - 3\]
\[z = \frac{D_3}{D} = \frac{- 316}{79} = - 4\]
\[ \therefore x = 2, y = - 3\text{ and }z = - 4\]
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