Solve this

Solution:

These equations can be written as

$x+y+0 z=1$

$x+0 y+z=-6$

 

$x-y-2 z=3$

$D=\left|\begin{array}{ccc}1 & 1 & 0 \\ 1 & 0 & 1 \\ 1 & -1 & -2\end{array}\right|$

$=1(0+1)-1(-2-1)+0(-1-0)$

$=4$

$D_{1}=\left|\begin{array}{ccc}1 & 1 & 0 \\ -6 & 0 & 1 \\ 3 & -1 & -2\end{array}\right|$

$=1(0+1)-1(12-3)+0(6-0)$

$=-8$

$D_{2}=\left|\begin{array}{ccc}1 & 1 & 0 \\ 1 & -6 & 1 \\ 1 & 3 & -2\end{array}\right|$

$=1(12-3)-1(-2-1)+0(3+6)$

$=12$

$D_{3}=\left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & 0 & -6 \\ 1 & -1 & 3\end{array}\right|$

$=1(0-6)-1(3+6)+1(-1-0)$

Now,

$x=\frac{D_{1}}{D}=\frac{-8}{4}=-2$

$y=\frac{D_{2}}{D}=\frac{12}{4}=3$

 

$z=\frac{D_{3}}{D}=\frac{-16}{4}=-4$

$\therefore x=-2, y=3$ and $z=-4$