# General Solution/ Vectors

• Oct 14th 2008, 08:06 PM
Hellreaver
General Solution/ Vectors
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?
• Oct 15th 2008, 04:06 PM
ThePerfectHacker
Quote:

Originally Posted by Hellreaver
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

Get it to row-reduced echelon form first.