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blhhblah
Jul7-08, 03:32 AM
I'm having trouble with Ax=b matrices. I get really confused when dealing with matrices that have a different number of rows from columns. An example is this problem:
(A is a 3x2 matrix)
A =
3 1
4 2
-5 -1

(b is a 3x1 matrix)
b =
1
0
-3
I need to find the solution set for x. Aren't there only 2 variables?
Thanks.

Defennder
Jul7-08, 03:39 AM
Yes there are only two variables but 3 equations. Construct the augmented matrix A|b and perform row reduction until you get the reduced-row echelon form of the matrix. Now you can easily extract the solutions.

blhhblah
Jul7-08, 03:52 AM
is this the reduced row echelon form?

1 0 1
0 1 -2
0 0 0

blhhblah
Jul7-08, 03:57 AM
regarding another question...

how do you solve for x given A is a (3x4) matrix and b is a (4x1) matrix?

Defennder
Jul7-08, 03:57 AM
Yeah I got that as well. Now you can read the solutions for the variables directly from the matrix.

Defennder
Jul7-08, 03:59 AM
regarding another question...

how do you solve for x given A is a (3x4) matrix and b is a (4x1) matrix?It's by using the same technique as well. But note that this time you have more unknowns than equations. What does that tell you about the nature of the solutions you will have?

blhhblah
Jul7-08, 04:02 AM
i guess all the values won't be exact, but i'm not sure how to set this up

Defennder
Jul7-08, 04:05 AM
What do you mean by "exact"? Just write out the augmented matrix and again use row-reduction as before. Note that it is entirely possible that the matrix might be inconsistent, in that case there are no solutions.

blhhblah
Jul7-08, 04:21 AM
matrix A was

2 8 4 16
1 1 2 2
1 -1 2 -2

I was able to reduce it to
1 0 2 0
0 1 0 2
0 0 0 0

but i'm not sure how to match it up to b, which is

-1
9
5
17

HallsofIvy
Jul7-08, 06:45 AM
Your first matrix was equaivalent to the set of equations
2x+ 8y+ 3z= 16, x+ y+ 2z= 2, and x- y+ 2z= -2 with b= <x, y, z>.

You have reduced it to a matrix equivalent to
x+ 2z= 0 and y= 2. What possible values of x, y, and z satisfy both of those?

sennyk
Jul7-08, 01:39 PM
Whenever I have more or fewer equations than unknowns, I use the following to obtain the least squares answer. A^T A x = A^T b, where A^T is the transpose of A.