Precalculus, Chapter 8, 8.1, Section 8.1, Problem 69

x+2y-3z=-28
4y+2z=0
-x+y-z=-5
The equations in the matrix form can be written as,
[[1,2,-3,-28],[0,4,2,0],[-1,1,-1,-5]]
Add Row 1 and Row 3
[[1,2,-3,-28],[0,4,2,0],[0,3,-4,-33]]
Multiply Row 2 by 2 and Add it to Row 3
[[1,2,-3,-28],[0,4,2,0],[0,11,0,-33]]
Now the equations can be written as,
x+2y-3z=-28 ----- equation 1
4y+2z=0 ------ equation 2
11y=-33 ----- equation 3
From equation 3,
y=-33/11=-3
Substitute back y in equation 2,
4(-3)+2z=0
-12+2z=0
2z=12
z=12/2=6
substitute back y and z in equation 1,
x+2(-3)-3(6)=-28
x-6-18=-28
x=-28+18+6
x=-4
So the solutions are x=-4, y=-3, z=6