Solve this

Here,

$x+y-6 z=0$  ....(1)

$x-y+2 z=0$   .....(2)

$-3 x+y+2 z=0$    ...(3)

The given system of homogeneous equations can be written in matrix form as follows:

$\left[\begin{array}{ccc}1 & 1 & -6 \\ 1 & -1 & 2 \\ -3 & 1 & 2\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]$

$A X=O$

Here,

$A=\left[\begin{array}{ccc}1 & 1 & -6 \\ 1 & -1 & 2 \\ -3 & 1 & 2\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $O=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]$

Now,

$|A|=\left|\begin{array}{ccc}1 & 1 & -6 \\ 1 & -1 & 2 \\ -3 & 1 & 2\end{array}\right|$

$=1(-2-2)-1(2+6)-6(1-3)$

$=-4-8+12$

$=0$

$\therefore|A|=0$

So, the given system of homogeneous equations has non-trivial solution.

Substituting $z=k$ in eq. (1) and eq. (2), we get

$x+y=6 k$ and $x-y=-2 k$

$A X=B$

Here,

$A=\left[\begin{array}{cc}1 & 1 \\ 1 & -1\end{array}\right], X=\left[\begin{array}{c}x \\ y\end{array}\right]$ and $B=\left[\begin{array}{c}6 k \\ -2 k\end{array}\right]$

$\Rightarrow\left[\begin{array}{cc}1 & 1 \\ 1 & -1\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{c}6 k \\ -2 k\end{array}\right]$

Now,

$|A|=\left|\begin{array}{cc}1 & 1 \\ 1 & -1\end{array}\right|$

$=(1 \times-1-1 \times 1)$

$=-2$

So, A $^{-1}$ exists.

We have

$\operatorname{adj} A=\left[\begin{array}{cc}-1 & -1 \\ -1 & 1\end{array}\right]$

$\mathrm{A}^{-1}=\frac{1}{|A|} \operatorname{adj} A$

$\Rightarrow A^{-1}=\frac{1}{-2}\left[\begin{array}{cc}-1 & -1 \\ -1 & 1\end{array}\right]$

$X=A^{-1} B$

$\Rightarrow\left[\begin{array}{l}x \\ y\end{array}\right]=\frac{1}{-2}\left[\begin{array}{cc}-1 & -1 \\ -1 & 1\end{array}\right]\left[\begin{array}{c}6 k \\ -2 k\end{array}\right]$

$X=A^{-1} B$

$\Rightarrow\left[\begin{array}{l}x \\ y\end{array}\right]=\frac{1}{-2}\left[\begin{array}{cc}-1 & -1 \\ -1 & 1\end{array}\right]\left[\begin{array}{c}6 k \\ -2 k\end{array}\right]$

$=\frac{1}{-2}\left[\begin{array}{l}-6 k+2 k \\ -6 k-2 k\end{array}\right]$

Thus, $x=2 k, y=4 k$ and $z=k$ (where $k$ is any real number) satisfy the given system of equations.