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Question
3x − y + 2z = 6
2x − y + z = 2
3x + 6y + 5z = 20.
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Solution
Given: 3x − y + 2z = 6
2x − y + z = 2
3x + 6y + 5z = 20
\[D = \begin{vmatrix}3 & - 1 & 2 \\ 2 & - 1 & 1 \\ 3 & 6 & 5\end{vmatrix}\]
\[3\left( - 5 - 6 \right) + 1\left( 10 - 3 \right) + 2\left( 12 + 3 \right) = 4\]
Since D is non-zero, the system of linear equations is consistent and has a unique solution.
\[ D_1 = \begin{vmatrix}6 & - 1 & 2 \\ 2 & - 1 & 1 \\ 20 & 6 & 5\end{vmatrix}\]
\[ = 6\left( - 5 - 6 \right) + 1\left( 10 - 20 \right) + 2\left( 12 + 20 \right)\]
\[ = - 66 - 10 + 64\]
\[ = - 12\]
\[ D_2 = \begin{vmatrix}3 & 6 & 2 \\ 2 & 2 & 1 \\ 3 & 20 & 5\end{vmatrix}\]
\[ = 3\left( 10 - 20 \right) - 6\left( 10 - 3 \right) + 2\left( 40 - 6 \right)\]
\[ = - 30 - 42 + 68\]
\[ = - 4\]
\[ D_3 = \begin{vmatrix}3 & - 1 & 6 \\ 2 & - 1 & 2 \\ 3 & 6 & 20\end{vmatrix}\]
\[ = 3\left( - 20 - 12 \right) + 1\left( 40 - 6 \right) + 6\left( 12 + 3 \right)\]
\[ = - 96 + 34 + 90\]
\[ = 28\]
Now,
\[x = \frac{D_1}{D} = \frac{- 12}{4} = - 3\]
\[y = \frac{D_2}{D} = \frac{- 4}{4} = - 1\]
\[z = \frac{D_3}{D} = \frac{28}{4} = 7\]
\[ \therefore x = - 3, y = - 1\text{ and }z = 7\]
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