Solve this

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

$\left[\begin{array}{ccc}2 & -1 & 1 \\ 3 & 2 & -1 \\ 1 & 4 & 3\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]$

or, $A X=O$

where, $A=\left[\begin{array}{ccc}2 & -1 & 1 \\ 3 & 2 & -1 \\ 1 & 4 & 3\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $O=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]$

$|A|=\left|\begin{array}{ccc}2 & -1 & 1 \\ 3 & 2 & -1 \\ 1 & 4 & 3\end{array}\right|$

$=2(6+4)+1(9+1)+1(12-2)$

$=40$

$\therefore|A| \neq 0$

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

$x=y=z=0$