Precalculus, Chapter 8, 8.1, Section 8.1, Problem 73

3x+2y-z+w=0
x-y+4z+2w=25
-2x+y+2z-w=2
x+y+z+w=6
The above system of equations can be represented by the coefficient matrix A and right hand side vector b as follows:
A=[[3,2,-1,1],[1,-1,4,2],[-2,1,2,-1],[1,1,1,1]]
b=[[0],[25],[2],[6]]
The augmented matrix can be written as:
[A b]=[[3,2,-1,1,0],[1,-1,4,2,25],[-2,1,2,-1,2],[1,1,1,1,6]]
Now let's bring the above matrix in row-echelon form by performing various row operations,
Rewrite the 1st Row(R_1) as (R_1+R_3)
[[1,3,1,0,2],[1,-1,4,2,25],[-2,1,2,-1,2],[1,1,1,1,6]]
Rewrite the 2nd Row(R_2) as (R_2-R_4)
[[1,3,1,0,2],[0,-2,3,1,19],[-2,1,2,-1,2],[1,1,1,1,6]]
Rewrite the 3rd Row(R_3) as(R_3+2R_4)
[[1,3,1,0,2],[0,-2,3,1,19],[0,3,4,1,14],[1,1,1,1,6]]
Rewrite the 4th Row as (R_4-R_1)
[[1,3,1,0,2],[0,-2,3,1,19],[0,3,4,1,14],[0,-2,0,1,4]]
Rewrite the 2nd Row as (R_2+R_3)
[[1,3,1,0,2],[0,1,7,2,33],[0,3,4,1,14],[0,-2,0,1,4]]
Rewrite the 3rd Row as (R_3-3R_2)
[[1,3,1,0,2],[0,1,7,2,33],[0,0,-17,-5,-85],[0,-2,0,1,4]]
Rewrite the 4th Row as (R_4+2R_2)
[[1,3,1,0,2],[0,1,7,2,33],[0,0,-17,-5,-85],[0,0,14,5,70]]
Rewrite the 3rd Row as (R_3+R_4)
[[1,3,1,0,2],[0,1,7,2,33],[0,0,-3,0,-15],[0,0,14,5,70]]
Rewrite the 3rd Row by dividing it with -3,
[[1,3,1,0,2],[0,1,7,2,33],[0,0,1,0,5],[0,0,14,5,70]]
Rewrite the 4th Row as(R_4-14R_3)
[[1,3,1,0,2],[0,1,7,2,33],[0,0,1,0,5],[0,0,0,5,0]]
Rewrite the 4th Row by dividing it with 5,
[[1,3,1,0,2],[0,1,7,2,33],[0,0,1,0,5],[0,0,0,1,0]]
Now the above matrix is row-echelon form and we can perform back substitution on the corresponding system,
x+3y+z=2 ----- Eq:1
y+7z+2w=33 ------ Eq:2
z=5
w=0
Substitute back the value of z and w in Eq:2,
y+7(5)+2(0)=33
y+35=33
y=33-35=-2
Substitute back the value of y and z in Eq:1,
x+3(-2)+5=2
x-6+5=2
x-1=2
x=3
So the solutions are x=3,y=-2,z=5 and w=0