Solve this
Question:

$2 x+3 y-z=0$

$x-y-2 z=0$

$3 x+y+3 z=0$

Solution:

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

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

$A X=O$

Here,

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

Now,

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

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

$=-2-27-4$

$=-33 \neq 0$

So, the given system of homogeneous equations has only trivial solution, which is given below:

$x=y=z=0$

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