- #1
Kavorka
- 95
- 0
Find a basis for the solution space of the linear system
x1-x2-2x3+x4 = 0
-3x1+3x2+x3-x4 = 0
2x1-2x2+x3 = 0
I created a matrix (not augmented, will be 0 on right side no matter what row operations) and brought it to reduced echelon form. x2 and x4 were free variables and I set them to the variables t and s respectively. I found that:
x1 = t - (1/5)s
x3 = (-2/5)s
All of this I'm fairly certain I did right, I just wanted to make sure I am stating the basis correctly because this is the first problem I've done where I've found a basis for a linear system. Stating all the x's in terms of s and t I get (forgive weird formatting not sure how to do matrices on here):
[x1]\\\\\\[1]\\\\\\\[-1/5]
[x2]\\=\t\[1]\+\s\[0]
[x3]\\\\\\[0]\\\\\\\[-2/5]
[x4]\\\\\\[0]\\\\\\\[1]
Forming the 2 dimensional basis:
[1] [-1]
[1] [0]
[0] [-2]
[0], [5]
x1-x2-2x3+x4 = 0
-3x1+3x2+x3-x4 = 0
2x1-2x2+x3 = 0
I created a matrix (not augmented, will be 0 on right side no matter what row operations) and brought it to reduced echelon form. x2 and x4 were free variables and I set them to the variables t and s respectively. I found that:
x1 = t - (1/5)s
x3 = (-2/5)s
All of this I'm fairly certain I did right, I just wanted to make sure I am stating the basis correctly because this is the first problem I've done where I've found a basis for a linear system. Stating all the x's in terms of s and t I get (forgive weird formatting not sure how to do matrices on here):
[x1]\\\\\\[1]\\\\\\\[-1/5]
[x2]\\=\t\[1]\+\s\[0]
[x3]\\\\\\[0]\\\\\\\[-2/5]
[x4]\\\\\\[0]\\\\\\\[1]
Forming the 2 dimensional basis:
[1] [-1]
[1] [0]
[0] [-2]
[0], [5]