Linear algebra: Solution space

[_Bd_]

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.

 

[mighty2000]

Hello,

Your approach is correct, but you have made a small error in your calculations. The dimension of the solution space is equal to the nullity, which is indeed the number of free variables (or parameters) in the system of equations. In this case, there are two free variables (t and s) so the dimension is 2, and the basis would be the two corresponding vectors: [-1, 1, 1, 0] and [2, -2, 0, 1].

Your mistake was in setting x_3 = t = x_2, which would actually give you only one free variable, and therefore a dimension of 1. Instead, you should have set x_2 = s and x_3 = t, which would give you two free variables and a dimension of 2.

I hope this helps clarify things for you. Let me know if you have any other questions.