$3 x+y+z=2$
$2 x-4 y+3 z=-1$
$4 x+y-3 z=-11$
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 $\mathrm{z}=3$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.