Finding the null space of a 3×4 matrix

[3.141592654]

Homework Statement

Determine the null space of the following matrix:

A = [1 1 -1 2
2 2 -3 1
-1 -1 0 -5]

Homework Equations

$Ax=0$ where $x = (x_{1}, x_{2}, x_{3}, x_{4})^{T}$

The Attempt at a Solution

If I put the system Ax=0 into augmented form:

1 1 -1 2 | 0
2 2 -3 1 | 0
-1 -1 0 -5 | 0

By row reduction I get the following row echelon form:

1 1 -1 2 | 0
0 0 1 3 | 0
0 0 0 0 | 0

So if

$x = (x_{1}, x_{2}, x_{3}, x_{4})^{T} = (-5t-s, s, -3t, t)^{T}$

$= t(-5, 0, -3, 1)^{T} + s(-1, 1, 0, 0)^{T}$

$= Span[ (-5, 0, -3, 1)^{T}, (-1, 1, 0, 0)^{T} ]$

My book has the answer:

$x = (-1, 1, 0, 0)^{T}, Span[ (-5, 0, -3, 1)^{T}]$

Have I gone wrong somewhere or are these answers equivalent? I can't see it if they are...

 

[Mark44]

Homework Statement

Determine the null space of the following matrix:

A = [1 1 -1 2
2 2 -3 1
-1 -1 0 -5]

Homework Equations

$Ax=0$ where $x = (x_{1}, x_{2}, x_{3}, x_{4})^{T}$

The Attempt at a Solution

If I put the system Ax=0 into augmented form:

1 1 -1 2 | 0
2 2 -3 1 | 0
-1 -1 0 -5 | 0

By row reduction I get the following row echelon form:

1 1 -1 2 | 0
0 0 1 3 | 0
0 0 0 0 | 0

So if

$x = (x_{1}, x_{2}, x_{3}, x_{4})^{T} = (-5t-s, s, -3t, t)^{T}$

$= t(-5, 0, -3, 1)^{T} + s(-1, 1, 0, 0)^{T}$

$= Span[ (-5, 0, -3, 1)^{T}, (-1, 1, 0, 0)^{T} ]$

The above looks fine.

My book has the answer:

$x = (-1, 1, 0, 0)^{T}, Span[ (-5, 0, -3, 1)^{T}]$

They have the same vectors you have, but their notation is screwed up. The nullspace here is two-dimensional, so it takes two vectors to span it.

Have I gone wrong somewhere or are these answers equivalent? I can't see it if they are...