$2 x-y+2 z=0$
$5 x+3 y-z=0$
$x+5 y-5 z=0$
Here,
$2 x-y+2 z=0$ ....(1)
$5 x+3 y-z=0$ ....(2)
$x+5 y-5 z=0$ .....(3)
The given system of homogeneous equations can be written in matrix form as follows:
$\left[\begin{array}{ccc}2 & -1 & 2 \\ 5 & 3 & -1 \\ 1 & 5 & -5\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]$
$A X=O$
Here,
$A=\left[\begin{array}{ccc}2 & -1 & 2 \\ 5 & 3 & -1 \\ 1 & 5 & -5\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $O=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]$
Now,
$|A|=\left|\begin{array}{ccc}2 & -1 & 2 \\ 5 & 3 & -1 \\ 1 & 5 & -5\end{array}\right|$
$=2(-15+5)+1(-25+1)+2(25-3)$
$=-20-24+44$
$=0$
$\therefore|A| \neq 0$
So, the given system of homogeneous equations has non-trivial solution.
Substituting $z=k$ in eq. (1) and eq. (2), we get
$2 x-y=-2 k$ and $5 x+3 y=k$
$A X=B$
Here,
$\mathrm{A}=\left[\begin{array}{cc}2 & -1 \\ 5 & 3\end{array}\right], \mathrm{X}=\left[\begin{array}{c}x \\ y\end{array}\right]$ and $\mathrm{B}=\left[\begin{array}{c}-2 k \\ k\end{array}\right]$
$\Rightarrow\left[\begin{array}{cc}2 & -1 \\ 5 & 3\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{c}-2 k \\ k\end{array}\right]$
$|A|=\left|\begin{array}{cc}2 & -1 \\ 5 & 3\end{array}\right|$
$=(3 \times 2+1 \times 5)$
$=11$
So, $A^{-1}$ exists.
We have
$\operatorname{adj} A=\left[\begin{array}{cc}3 & 1 \\ -5 & 2\end{array}\right]$
$A^{-1}=\frac{1}{|A|}$ adj $A$
$\Rightarrow A^{-1}=\frac{1}{11}\left[\begin{array}{cc}3 & 1 \\ -5 & 2\end{array}\right]$
$X=A^{-1} B$
$\Rightarrow\left[\begin{array}{l}x \\ y\end{array}\right]=\frac{1}{11}\left[\begin{array}{cc}3 & 1 \\ -5 & 2\end{array}\right]\left[\begin{array}{c}-2 k \\ k\end{array}\right]$
$=\frac{1}{11}\left[\begin{array}{c}-6 k+k \\ 10 k+2 k\end{array}\right]$
Thus, $x=\frac{-5 k}{11}, y=\frac{12 k}{11}$ and $z=k$ (where $k$ is any real number) satisfy the given system of equations.
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