Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Having difficulty with an Ax=b problem and understanding it

  1. Jul 7, 2008 #1
    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 =
    I need to find the solution set for x. Aren't there only 2 variables?
  2. jcsd
  3. Jul 7, 2008 #2


    User Avatar
    Homework Helper

    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.
  4. Jul 7, 2008 #3
    is this the reduced row echelon form?

    1 0 1
    0 1 -2
    0 0 0
  5. Jul 7, 2008 #4
    regarding another question...

    how do you solve for x given A is a (3x4) matrix and b is a (4x1) matrix?
  6. Jul 7, 2008 #5


    User Avatar
    Homework Helper

    Yeah I got that as well. Now you can read the solutions for the variables directly from the matrix.
  7. Jul 7, 2008 #6


    User Avatar
    Homework Helper

    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?
  8. Jul 7, 2008 #7
    i guess all the values won't be exact, but i'm not sure how to set this up
  9. Jul 7, 2008 #8


    User Avatar
    Homework Helper

    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.
  10. Jul 7, 2008 #9
    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

  11. Jul 7, 2008 #10


    User Avatar
    Science Advisor

    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?
  12. Jul 7, 2008 #11
    Whenever I have more or fewer equations than unknowns, I use the following to obtain the least squares answer. [itex]A^T A x = A^T b[/itex], where [itex]A^T[/itex] is the transpose of A.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook