# Linear algebra: Solution space

## Homework Statement

Find a basis for, and the dimension of, th esolution space of Ax = b

A=[1 3 -2 4]
[0 1 -1 0]
[-2 -6 4 8]

## Homework Equations

Allowed to use calculator

## The Attempt at a Solution

getting the RRE form yields:
[1 0 1 0]
[0 1 -1 0]
[0 0 0 1]
which means its a rank 3 matrix (since there are no dependent vectors. . .right?)
the nullity for a mxn matrix is n-r . . .so 4 - 3 = 1. . .that would be the nullity . . .which from what I see in the book is the same as the dimension?
it says:
If A is an m x n matrix of rank r, then the dimension of the solution space of Ax = 0 is n- r. That is
n = rank(A) + nullity(A)

so then using the RREf and setting up a system of equations I get
x_1 + 0 + x_3 + 0 = 0
0 + x_2 - x_3 + 0 = 0
0 + 0 + 0 + x_4 = 0

so then x_4 = 0 | x_3 = t = x_2 and x_1 = -t
so then my basis would be [-1, 1, 1, 0]
my dimension would be: 1

. . .the back of the book says the following:
dimension = 2
basis:
{[-1, 1, 1, 0], [ 2, -2, 0, 1]}

thank you.

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