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Q\begin{array}{l} x_{1}+x_{2}-x_{3}+2 x_{4}+x_{5}=0 \\x_{1}+2 x_{2}-x_{3}+x_{4}+x_{5}=0 \\2 x_{1}+3 x_{2}-x_{3}+2 x_{4}+x_{5}=0 \\4 x_{1}+5 x_{2}-2 x_{3}+5 x_{4}+2 x_{5}=0\end{array}Sel : \rightarrow We can write this system as A x=b\left[\begin{array}{ccccc}1 & 1 & -1 & 2 & 1 \\1 & 2 & -1 & 1 & 1 \\2 & 3 & -1 & 2 & 1 \\4 & 5 & -2 & 5 & 2\end{array}\right]\left[\begin{array}{l}x_{1} \\x_{2} \\x_{3} \\x_{4} \\x_{5}\end{array}\right]=\left[\begin{array}{l}0 \\0 \\0 \\0\end{array}\right]The Augmented Matrix\begin{array}{l}{[A \mid b]=\left[\begin{array}{ccccc|c}1 & 1 & -1 & 2 & 1 & 0 \\1 & 2 & -1 & 1 & 1 & 0 \\2 & 3 & -1 & 2 & 1 & 0 \\4 & 5 & -2 & 5 & 2 & 0\end{array}\right]} \\R_{2} \rightarrow R_{2}-R_{1}, \quad R_{3} \rightarrow R_{3}-2 R_{1}\end{array}R_{2} \rightarrow R_{2}-R_{1}, \quad R_{3} \rightarrow R_{3}-2 R_{1}, R_{4} \rightarrow R_{4}-4 R_{1}[A \mid b]=\left[\begin{array}{ccccc|c}1 & 1 & -1 & 2 & 1 & 0 \\0 & 1 & 0 & -1 & 0 & 0 \\0 & 1 & 1 & -2 & -1 & 0 \\0 & 1 & 2 & -3 & -2 & 0\end{array}\right]R_{1} \rightarrow R_{1}-R_{2}, R_{3} \rightarrow R_{3}-R_{2}, R_{4} \rightarrow R_{4}-R_{2}[A \mid b]=\left[\begin{array}{ccccc|c}1 & 0 & -1 & 3 & 1 & 0 \\0 & 1 & 0 & -1 & 0 & 0 \\0 & 0 & 1 & -1 & -1 & 0 \\0 & 0 & 2 & -2 & -2 & 0\end{array}\right]\begin{array}{c}R_{4} \rightarrow R_{4}-2 R_{3} \\{[A \mid b]=\left[\begin{array}{ccccc|c}1 & 0 & -1 & 3 & 1 & 0 \\0 & 1 & 0 & -1 & 0 & 0 \\0 & 0 & 1 & -1 & -1 & 0 \\0 & 0 & 0 & 0 & 0 & 0\end{array}\right]} \\R_{1} \rightarrow R_{1}+R_{3} \\{[A \mid b]=\left[\begin{array}{ccccc|c}1 & 0 & 0 & 2 & 0 & 0 \\0 & 1 & 0 & -1 & 0 & 0 \\0 & 0 & 1 & -1 & -1 & 0 \\0 & 0 & 0 & 0 & 0 & 0\end{array}\right]}\end{array}So Now Corresponding system of equation is\begin{array}{l}x_{1}+2 x_{4}=0 \Rightarrow x_{1}=-2 x_{4} \\x_{2}-x_{4}=0 \Rightarrow x_{2}=x_{4} \\x_{3}-x_{4}-x_{5}=0 \\\quad x_{3}=x_{4}-x_{5} \\\text { let } x_{5}=t, \quad x_{4}=5 \\\Rightarrow \quad x_{3}=5-t \\x_{2}=5 \\x_{1}=-25\end{array}hence generel salution syx=\left[\begin{array}{l}x_{1} \\x_{2} \\x_{3} \\x_{4} \\x_{5}\end{array}\right]=\left[\begin{array}{c}-2 s \\s \\s-t \\s \\t\end{array}\right]=s\left[\begin{array}{c}-2 \\1 \\1 \\1 \\0\end{array}\right]+t\left[\begin{array}{c}0 \\0 \\-1 \\0 \\1\end{array}\right] \text { Ay } ...