I have this matrix:

1 2 -2 -1 0

2 3 -5 1 -7 0

0 1 1 -1 1 0

-1 1 5 -1 0 0

I solve for row echelon form:

1 2 -2 -1 0 0

0 1 1 -1 0 0

0 0 0 1 -3 0

0 0 0 0 0 0

I'm supposed to find the general solution of this as well a set of vectors that span the solution space.

Therefore,

x5= t1

x4= -3x5

x4= -3(t1)

x3= (t2)

x2 + x3 - x4=0

x2= 3(t1) - (t2)

x1+ 2x2 - 2x3 - x4 = 0

x1= 2(t2) - 3(t1) - 2(3(t1)-(t2))

x1= 4(t2) - 9(t1)

So I get:

x1= 4(t2) - 9(t1)

x2= 3(t1) - (t2)

x3= (t2)

x4= -3(t1)

x5= (t1)

(x1, x2, x3, x4, x5)= t1(-9, 3, 0, -3, 1) + t2 (4, -1, 1, 0, 0)

And the general solution would be:

x1 =[-9] [4]

x2 = [3] [-1]

x3 =t1[0]+t2[1]

x4 = [-3] [0]

x5 = [1] [0]

Sorry about the messy writing, I'm not sure how to write matrices on a computer. I think that is correct, can anyone verify?