Solve system of linear equations, using matrix method.

Solution:

The given system of equations can be written in the form of AX = B, where

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

Now,

$|A|=1(1+3)+1(2+3)+1(2-1)=4+5+1=10 \neq 0$

Thus, A is non-singular. Therefore, its inverse exists.

Now, $A_{4}=4, A_{12}=-5, A_{13}=1$

$A_{21}=2, A_{22}=0, A_{23}=-2$

$A_{31}=2, A_{32}=5, A_{33}=3$

$\therefore A^{-1}=\frac{1}{|A|}($ adj $A)=\frac{1}{10}\left[\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3\end{array}\right]$

$\therefore X=A^{-1} B=\frac{1}{10}\left[\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3\end{array}\right]\left[\begin{array}{l}4 \\ 0 \\ 2\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{10}\left[\begin{array}{c}16+0+4 \\ -20+0+10 \\ 4+0+6\end{array}\right]$

$=\frac{1}{10}\left[\begin{array}{c}20 \\ -10 \\ 10\end{array}\right]$

$=\left[\begin{array}{c}2 \\ -1 \\ 1\end{array}\right]$

Hence, $x=2, y=-1$, and $z=1$.