Solve this

Solution:

Given: $x-4 y-z=11$

$2 x-5 y+2 z=39$

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

$D=\left|\begin{array}{ccc}1 & -4 & -1 \\ 2 & -5 & 2 \\ -3 & 2 & 1\end{array}\right|$

$=1(-5-4)-(-4)(2+6)+(-1)(4-15)$

$=1(-9)-(-4)(8)+(-1)(-11)=34$

$D_{1}=\left|\begin{array}{ccc}11 & -4 & -1 \\ 39 & -5 & 2 \\ 1 & 2 & 1\end{array}\right|$

$=11(-5-4)-(-4)(39-2)+(-1)(78+5)$

$=11(-9)-(-4)(37)+(-1)(83)=-34$

$D_{2}=\left|\begin{array}{ccc}1 & 11 & -1 \\ 2 & 39 & 2 \\ -3 & 1 & 1\end{array}\right|$

$=1(39-2)-11(2+6)+(-1)(2+117)$

$=1(37)-11(8)+(-1)(119)=-170$

$D_{3}=\left|\begin{array}{ccc}1 & -4 & 11 \\ 2 & -5 & 39 \\ -3 & 2 & 1\end{array}\right|$

$=1(-5-78)-(-4)(2+117)+11(4-15)$

$=1(-83)-(-4)(119)+11(-11)=272$

Now,

$x=\frac{D_{1}}{D}=\frac{-34}{34}=-1$

$y=\frac{D_{2}}{D}=\frac{-170}{34}=-5$

$z=\frac{D_{3}}{D}=\frac{272}{34}=8$

$\therefore \mathrm{x}=-1, y=-5$ and $\mathrm{z}=8$