System of equations: finding a plane

[Catchfire]

Homework Statement

Solve the following system of equations:

2x1 - 2x2 -3x3 = -2
3x1 -3x2 -2x3 + 5x4 = 7
x1 - x2 -2x3 -x4 = -3

Homework Equations

The Attempt at a Solution

Ok so I solved the system and got:

x1 -x2 = 5
x3 = 4
x4 = 0

so I've got a point (5,0,4,0) but the answer is r(1,1,0,0) + s(-3,0,-2,1) + (5,0,4,0)

How do I get the other vectors?

 

[tiny-tim]

Ok so I solved the system and got:

x1 -x2 = 5
x3 = 4
x4 = 0

how?

 

[HallsofIvy]

x1= 5, x2= 0 is one pair of numbers that satisfy x1- x2= 5.

However, your "solution" is wrong. For example, x1= 2, x2= 0, x3= 2, x4= 1 satisfy the initial equation but none of x1- x2= 5, x3= 4, or x4= 0 are satisfied.

 

[Catchfire]

2x1 - 2x2 -3x3 = -2
3x1 -3x2 -2x3 + 5x4 = 7
x1 - x2 -2x3 -x4 = -3

x1 - x2 -2x3 -x4 = -3
4x3 + 8x4 = 16
x3 + 2x4= 4

x1 - x2 -2x3 -x4 = -3
x3 + 2x4 = 4

x1 - x2 +3x4 = 5
x3 + 2x4 = 4

x1 = x2 - 3x4 + 5
x3 = -2x4 + 4

Looks like I missed adding the bolded term. This is correct now I hope.

 

[Catchfire]

How do I find r(1,1,0,0) and s(-3,0,-2,1)?

 

[tiny-tim]

2x1 - 2x2 -3x3 = -2
3x1 -3x2 -2x3 + 5x4 = 7
x1 - x2 -2x3 -x4 = -3

x1 - x2 -2x3 -x4 = -3
4x3 + 8x4 = 16
x3 + 2x4= 4

not following that

 

[HallsofIvy]

Yes, that is correct. And now you are saying that any (x1, x2, x3, x4) satisfying those four equations can be written as (x1, x2, x3, x4)= (x2- 3x4+ 5, x2, -2x4+ 4, x4)= (x2, x2, 0, 0)+ (-3x4, 0, -2x4, x4)+ (5, 0, 4, 0)= x2(1, 1, 0, 0)+ x4(-3, 0, -2, 1)+ (5, 0, 4, 0).

 

[Catchfire]

AHHH thank you so much, I was pulling my hair out over here. I knew it had to be something pretty straight forward.