Solve this

Solution:

Given: $2 x-3 y-4 z=29$

$-2 x+5 y-z=-15$

$3 x-y+5 z=-11$

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

$=2(25-1)+3(-10+3)-4(2-15)$

$=2(24)+3(-7)-4(-13)$

$=79$

$D_{1}=\left|\begin{array}{ccc}29 & -3 & -4 \\ -15 & 5 & -1 \\ -11 & -1 & 5\end{array}\right|$

$=29(25-1)+3(-75-11)-4(15+55)$

$=29(24)+3(-86)-4(70)$

$=158$

$D_{2}=\left|\begin{array}{ccc}2 & 29 & -4 \\ -2 & -15 & -1 \\ 3 & -11 & 5\end{array}\right|$

$=2(-75-11)-29(-10+3)-4(22+45)$

$=2(-86)-29(-7)-4(67)$

$=-237$

$D_{3}=\left|\begin{array}{ccc}2 & -3 & 29 \\ -2 & 5 & -15 \\ 3 & -1 & -11\end{array}\right|$

$=2(-55-15)+3(22+45)+29(2-15)$

$=2(-70)+3(67)+29(-13)$

$=-316$

Now,

$x=\frac{D_{1}}{D}=\frac{158}{79}=2$

$y=\frac{D_{2}}{D}=\frac{-237}{79}=-3$

$z=\frac{D_{3}}{D}=\frac{-316}{79}=-4$

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