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Question
x + y = 1
x + z = − 6
x − y − 2z = 3
Solution
These equations can be written as
x+ y + 0z = 1
x + 0y + z = − 6
x − y − 2z = 3
\[D = \begin{vmatrix}1 & 1 & 0 \\ 1 & 0 & 1 \\ 1 & - 1 & - 2\end{vmatrix}\]
\[ = 1(0 + 1) - 1( - 2 - 1) + 0( - 1 - 0)\]
\[ = 4\]
\[ D_1 = \begin{vmatrix}1 & 1 & 0 \\ - 6 & 0 & 1 \\ 3 & - 1 & - 2\end{vmatrix}\]
\[ = 1(0 + 1) - 1(12 - 3) + 0(6 - 0)\]
\[ = - 8\]
\[ D_2 = \begin{vmatrix}1 & 1 & 0 \\ 1 & - 6 & 1 \\ 1 & 3 & - 2\end{vmatrix}\]
\[ = 1(12 - 3) - 1( - 2 - 1) + 0(3 + 6)\]
\[ = 12\]
\[ D_3 = \begin{vmatrix}1 & 1 & 1 \\ 1 & 0 & - 6 \\ 1 & - 1 & 3\end{vmatrix}\]
\[ = 1(0 - 6) - 1(3 + 6) + 1( - 1 - 0)\]
\[ = - 16\]
\[ \text{ Now } , \]
\[x = \frac{D_1}{D} = \frac{- 8}{4} = - 2\]
\[y = \frac{D_2}{D} = \frac{12}{4} = 3\]
\[z = \frac{D_3}{D} = \frac{- 16}{4} = - 4\]
\[ \therefore x = - 2, y = 3 \text{ and } z = - 4\]
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