Another problem on Matrix Algebra?

Row reduce
[1 -4 3 -4|0]
[2 -8 6 -8|0].

2R1 - R2 --> R2:
[1 -4 3 -4|0]
[0 0 0 0|0]; this is in reduced row echelon form.

So, x1 is the only pivot variable, and so x2, x3, x4 are free variables.

The last augmented matrix gives x1 - 4x2 + 3x3 - 4x4 = 0.

Rewrite this in terms of the free variables:
x1 = 4x2 - 3x3 + 4x4
x2 = x2
x3 = x3
x4 = x4.

Finally, write this in vector form:
..........[4].......[-3].....[4]
x = x2.[1]+ x3[0]+ x4[0]
..........[0].......[1]......[0]
..........[0].......[0]......[1].

I hope this helps!