Solve this

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.