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Question
x − y + z = 3
2x + y − z = 2
− x − 2y + 2z = 1
Solution
Using the equations we get
\[D = \begin{vmatrix}1 & - 1 & 1 \\ 2 & 1 & - 1 \\ - 1 & - 2 & 2\end{vmatrix}\]
\[ \Rightarrow 1\left( 2 - 2 \right) + 1\left( 4 - 1 \right) + 1\left( - 4 + 1 \right) = 0\]
\[ D_1 = \begin{vmatrix}3 & - 1 & 1 \\ 2 & 1 & - 1 \\ 1 & - 2 & 2\end{vmatrix}\]
\[ \Rightarrow 3\left( 2 - 2 \right) + 1\left( 4 + 1 \right) + 1\left( - 4 - 1 \right) = 0\]
\[ D_2 = \left| \begin{array}1 & 3 & 1 \\ 2 & 2 & - 1 \\ - 1 & 1 & 2\end{array} \right|\]
\[ \Rightarrow 1\left( 4 + 1 \right) - 3\left( 4 - 1 \right) + 1\left( 2 + 2 \right) = 0\]
\[ D_3 = \begin{vmatrix}1 & - 1 & 3 \\ 2 & 1 & 2 \\ - 1 & - 2 & 1\end{vmatrix}\]
\[ \Rightarrow 1\left( 1 + 4 \right) + 1\left( 2 + 2 \right) + 3\left( - 4 + 1 \right) = 0\]
Here,
\[D = D_1 = D_2 = D_3 = 0\]
Thus, the system of linear equations has infinitely many solutions.
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