Describe every solution to Ax=0

[Cade]

Homework Statement

Describe every solution to Ax=0 where A is:
1 2 2 4 6
1 2 3 6 9
0 0 1 2 3

Homework Equations

I'm not sure.

The Attempt at a Solution

I found the echelon form of A to be:
1 2 2 4 6
0 0 1 2 3
0 0 0 0 0

Pivot variables: x1, x3
Free variables: x2, x4, x5

Finding special solutions:
x1 + 2x2 + 2x3 + 4x4 + 6x5 = 0
x3 + 2x4 + 3x5 = 0

x2 = 1, x4 = 0, x5 = 0 -> x3 = 0, x1 = -2 (-2, 1, 0, 0, 0)
x2 = 0, x4 = 1, x5 = 0 -> x3 = -2, x1 = 0 (0, 0,-2, 1, 0)
x2 = 0, x4 = 0, x5 = 1 -> x3 = -3, x1 = 0 (0, 0,-3, 0, 1)

If this is the nullspace, how do I describe every solution? I thought of writing as a linear combination of those three, but that's apparently the wrong answer, my instructor wants a single vector like (*, x2, *, x4, x5). where * is a multiple of one of the pivot variables.

Edit: Looking at the special solutions, I think (-2x2, x2, -2x4 -3x5, x4, x5) is the right answer, but I'm not sure.

I also have a second problem: With pivot variables x1, x3 and free variables x2, x4 and x5 with the solution with x3=1 (1,-1,1), is the general solution (x3,-x3,x3)? My textbook says it should be (2x3,-x3,x3).

 

[Mark44]

Homework Statement

Describe every solution to Ax=0 where A is:
1 2 2 4 6
1 2 3 6 9
0 0 1 2 3

Homework Equations

I'm not sure.

The Attempt at a Solution

I found the echelon form of A to be:
1 2 2 4 6
0 0 1 2 3
0 0 0 0 0

I would take it one more step to reduced row-echelon form.

That gets you
1 2 0 0 0
0 0 1 2 3
0 0 0 0 0

Pivot variables: x1, x3
Free variables: x2, x4, x5

Finding special solutions:
x1 + 2x2 + 2x3 + 4x4 + 6x5 = 0
x3 + 2x4 + 3x5 = 0

From the reduced row echelon form,
x1 + 2x2 = 0
x3 + 2x4 + 3x5 = 0

You can rewrite this system as
x1 = -2x2
x2 = x2
x3 = ... -2x4 - 3x5
x4 = ...x4
x5 = ... x5
Note that the free variables are written as equal to themselves, which is obviously true.

Any vector in the nullspace can be written as
<x1, x2, x3, x4, x5> = x2 *<-2, 1, 0, 0, 0> + x4 * <0, 0, -2, 1, 0> + x5 * <0, 0, -3, 0, 1>

(All vectors here are column vectors.)

x2 = 1, x4 = 0, x5 = 0 -> x3 = 0, x1 = -2 (-2, 1, 0, 0, 0)
x2 = 0, x4 = 1, x5 = 0 -> x3 = -2, x1 = 0 (0, 0,-2, 1, 0)
x2 = 0, x4 = 0, x5 = 1 -> x3 = -3, x1 = 0 (0, 0,-3, 0, 1)

If this is the nullspace, how do I describe every solution? I thought of writing as a linear combination of those three, but that's apparently the wrong answer, my instructor wants a single vector like (*, x2, *, x4, x5). where * is a multiple of one of the pivot variables.

The nullspace is the set of all linear combinations of these three vectors.

Edit: Looking at the special solutions, I think (-2x2, x2, -2x4 -3x5, x4, x5) is the right answer, but I'm not sure.

Maybe, but this seems to me to be an unusual way to do it. Since the nullspace for this problem is three-dimensional, a more natural way to present a basis for this subspace (of R⁵) is with three vectors that span it.

I also have a second problem: With pivot variables x1, x3 and free variables x2, x4 and x5 with the solution with x3=1 (1,-1,1), is the general solution (x3,-x3,x3)? My textbook says it should be (2x3,-x3,x3).

This doesn't make sense to me. Any solution should have five coordinates.

 

[Cade]

Thanks, I wanted to write the general solution as a linear combination of the solutions, but my instructor wouldn't accept it and we haven't learned well, so I was confused over whether it was right or wrong.

I do not know why the second question only has 3 variables in its solutions when it has 5 variables. Hopefully, its a misprint.