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Question
x + y − z = 0
x − 2y + z = 0
3x + 6y − 5z = 0
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Solution
Using the equations we get
\[D = \begin{vmatrix}1 & 1 & - 1 \\ 1 & - 2 & 1 \\ 3 & 6 & - 5\end{vmatrix}\]
\[ \Rightarrow 1\left( 10 - 6 \right) - 1\left( - 5 - 3 \right) - 1\left( 6 + 6 \right) = 0\]
\[ D_1 = \begin{vmatrix}0 & 1 & - 1 \\ 0 & - 2 & 1 \\ 0 & 6 & - 5\end{vmatrix}\]
\[ \Rightarrow 0\left( 10 - 6 \right) - 1\left( 0 - 0 \right) - 1\left( 0 + 0 \right) = 0\]
\[ D_2 = \begin{vmatrix}1 & 0 & - 1 \\ 1 & 0 & 1 \\ 3 & 0 & - 5\end{vmatrix}\]
\[ \Rightarrow 1\left( 0 - 0 \right) - 0\left( - 5 - 3 \right) - 1\left( 0 - 0 \right) = 0\]
\[ D_3 = \begin{vmatrix}1 & 1 & 0 \\ 1 & - 2 & 0 \\ 3 & 6 & 0\end{vmatrix}\]
\[ \Rightarrow 1\left( 0 - 0 \right) - 1\left( 0 - 0 \right) + 0\left( 6 + 6 \right) = 0\]
\[\therefore D = D_1 = D_2\]
Hence, the system of linear equations has infinitely many solutions.
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