x

_{1}- 2x

_{3}- 3x

_{4}= 0

x

_{2}- 3x

_{3}+ x

_{4}= 0

2x

_{1}- x

_{2}- x

_{3}- 7x

_{4}= 0

I got a solution of x = x

_{3}*(2, 3, 1, 0). I could show closure, but how do you find a basis with just the one vector?

I need to find the set of solutions to a system of linear equations, show that they are a vector space of R^4, and find a basis for them. The equations are:

x_{1} - 2x_{3} - 3x_{4} = 0

x_{2} - 3x_{3} + x_{4} = 0

2x_{1} - x_{2} - x_{3} - 7x_{4} = 0

I got a solution of x = x_{3}*(2, 3, 1, 0). I could show closure, but how do you find a basis with just the one vector?

x

x

2x

I got a solution of x = x

- Martingale
**Posts:**333**Joined:**Mon Mar 30, 2009 1:30 pm**Location:**USA-
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you should have two vectors that form your basisI need to find the set of solutions to a system of linear equations, show that they are a vector space of R^4, and find a basis for them. The equations are:

x_{1}- 2x_{3}- 3x_{4}= 0

x_{2}- 3x_{3}+ x_{4}= 0

2x_{1}- x_{2}- x_{3}- 7x_{4}= 0

I got a solution of x = x_{3}*(2, 3, 1, 0). I could show closure, but how do you find a basis with just the one vector?

(2,3,1,0) is correct

and the other is (3,-1,0,1)