Precalculus, Chapter 8, 8.1, Section 8.1, Problem 74

x-4y+3z-2w=9
3x-2y+z-4w=-13
-4x+3y-2z+w=-4
-2x+y-4z+3w=-10
The above system of equations can be represented by the coefficient matrix A and right hand side matrix b as follows:
A=[[1,-4,3,-2],[3,-2,1,-4],[-4,3,-2,1],[-2,1,-4,3]]
b=[[9],[-13],[-4],[-10]]
The augmented matrix can be written as,
[[A,b]]=[[1,-4,3,-2,9],[3,-2,1,-4,-13],[-4,3,-2,1,-4],[-2,1,-4,3,-10]]
Now lets, perform the various row operations to bring the above matrix in the row-echelon form,
Rewrite the 2nd Row (R_2) as (R_2-3R_1)
[[1,-4,3,-2,9],[0,10,-8,2,-40],[-4,3,-2,1,-4],[-2,1,-4,3,-10]]
Rewrite the 3rd Row(R_3) as(R_3+4R_1)
[[1,-4,3,-2,9],[0,10,-8,2,-40],[0,-13,10,-7,32],[-2,1,-4,3,-10]]
Rewrite the 4th Row(R_4) as(R_4+2R_1)
[[1,-4,3,-2,9],[0,10,-8,2,-40],[0,-13,10,-7,32],[0,-7,2,-1,8]]
Rewrite the 2nd Row(R_2) as(2(R_2+R_3)-R_4)
[[1,-4,3,-2,9],[0,1,2,-9,-24],[0,-13,10,-7,32],[0,-7,2,-1,8]]
Rewrite the 3rd Row(R_3) as(R_3+13R_2)
[[1,-4,3,-2,9],[0,1,2,-9,-24],[0,0,36,-124,-280],[0,-7,2,-1,8]]
Rewrite the 4th Row(R_4) as (R_4+7R_2)
[[1,-4,3,-2,9],[0,1,2,-9,-24],[0,0,36,-124,-280],[0,0,16,-64,-160]]
Rewrite the 3rd Row(R_3) as (R_3-R_4)
[[1,-4,3,-2,9],[0,1,2,-9,-24],[0,0,20,-60,-120],[0,0,16,-64,-160]]
Rewrite the 3rd Row by dividing it with 20,
[[1,-4,3,-2,9],[0,1,2,-9,-24],[0,0,1,-3,-6],[0,0,16,-64,-160]]
Rewrite the 4th Row by dividing it with 16,
[[1,-4,3,-2,9],[0,1,2,-9,-24],[0,0,1,-3,-6],[0,0,1,-4,-10]]
Rewrite the 4th Row as (R_3-R_4)
[[1,-4,3,-2,9],[0,1,2,-9,-24],[0,0,1,-3,-6],[0,0,0,1,4]]
Now the matrix is in row-echelon form, and we can perform the back substitution on the corresponding system,
x-4y+3z-2w=9 ----- Eq:1
y+2z-9w=-24 ----- Eq:2
z-3w=-6 ----- Eq:3
w=4
Substitute back the value of w in Eq:3,
z-3(4)=-6
z-12=-6
z=-6+12
z=6
Substitute back the value of w and z in Eq:2,
y+2(6)-9(4)=-24
y+12-36=-24
y=-24+36-12
y=0
Substitute back the value of w,z and y in Eq:1,
x-4y+3z-2w=9
x-4(0)+3(6)-2(4)=9
x+18-8=9
x=9+8-18
x=-1
So the solutions are x=-1,y=0,z=6 and w=4