Solve this
Question:

$x+y-z=0$

$x-2 y+z=0$

$3 x+6 y-5 z=0$

Solution:

$x+y-z=0$       ….(1)

$x-2 y+z=0$       …..(2)

$3 x+6 y-5 z=0$   ….(3)

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

$\left[\begin{array}{ccc}1 & 1 & -1 \\ 1 & -2 & 1 \\ 3 & 6 & -5\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]$

$A X=O$

Here,

$A=\left[\begin{array}{ccc}1 & 1 & -1 \\ 1 & -2 & 1 \\ 3 & 6 & -5\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $O=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]$

Now,

$|A|=\left|\begin{array}{ccc}1 & 1 & -1 \\ 1 & -2 & 1 \\ 3 & 6 & -5\end{array}\right|$

$=1(10-6)-1(-5-3)-1(6+6)$

$=4+8-12$

$=0$

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

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

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

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

$A X=B$

Here,

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

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

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

$\operatorname{adj} A=\left[\begin{array}{cc}-2 & -1 \\ -1 & 1\end{array}\right]$

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

$\Rightarrow A^{-1}=\frac{1}{-3}\left[\begin{array}{cc}-2 & -1 \\ -1 & 1\end{array}\right]$

$X=A^{-1} B$

$\Rightarrow\left[\begin{array}{l}x \\ y\end{array}\right]=\frac{1}{-3}\left[\begin{array}{cc}-2 & -1 \\ -1 & 1\end{array}\right]\left[\begin{array}{c}k \\ -k\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x \\ y\end{array}\right]=\frac{1}{-3}\left[\begin{array}{c}-2 k+k \\ -k-k\end{array}\right]$

Thus, $x=\frac{k}{3}, y=\frac{2 k}{3}$ and $z=k$ (where $k$ is any real number) satisfy the given system of equations.