Solve this

Question:

$3 x+y-2 z=0$

$x+y+z=0$

$x-2 y+z=0$

Solution:

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

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

or, $A X=O$

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

Now,

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

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

$=9-0+6$

$=15 \neq 0$

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

$x=y=z=0$

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