Solve this

Solution:

Given: $3 x+y+z=2$

$2 x-4 y+3 z=-1$

$4 x+y-3 z=-11$

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

$=3(12-3)-2(-3-1)+4(3+4)$

$=27+8+28$

$=63$

$D_{1}=\left|\begin{array}{ccc}2 & 1 & 1 \\ -1 & -4 & 3 \\ -11 & 1 & -3\end{array}\right|$

$=2(12-3)+1(-3-1)-11(3+4)$

$=18-4-77$

$=-63$

$D_{2}=\left|\begin{array}{ccc}3 & 2 & 1 \\ 2 & -1 & 3 \\ 4 & -11 & -3\end{array}\right|$

$=3(3+33)-2(-6+11)+4(6+1)$

$=108-10+28$

$=126$

$D_{3}=\left|\begin{array}{ccc}3 & 1 & 2 \\ 2 & -4 & -1 \\ 4 & 1 & -11\end{array}\right|$

$=3(44+1)-2(-11-2)+4(-1+8)$

$=135+26+28$

$=189$

Now,

$x=\frac{D_{1}}{D}=\frac{-63}{63}=-1$

$y=\frac{D_{2}}{D}=\frac{126}{63}=2$

$z=\frac{D_{3}}{D}=\frac{189}{63}=3$

$\therefore x=-1, y=2$ and $\mathrm{z}=3$