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Matrix question. ACA-AB=0. What's C?

  1. Mar 10, 2013 #1
    quest.JPG

    I know how to find A inverse. And I know ACA=AB.

    But how do I go through solving this? I figured it had something to do with the inverse of A. But matrix multiplication isn't the same as normal multiplication.

    Will something like

    (Inverse of A)ACA=(Inverse of A)B

    Since (Inverse of A)A=The identity Matrix

    CA=B

    Then

    CA(Inverse of A)=B(Inverse of A)

    C=B(Inverse of A)

    I'm pretty sure all of this is wrong, but even if it isn't, I'm stuck here. How can I solve this and similar questions?
     
  2. jcsd
  3. Mar 10, 2013 #2
    You did the mathematics right. Just do the matrix multiplication.

    You are unclear on a certain point. Maybe some of us could help if you explained to us the following.

    1) How did you learn to calculate the inverse of a matrix but not learn matrix multiplication?

    2) Are you asking how to do matrix multiplication?


    3) What makes you think that your mathematics is wrong?

    The logical structure of a book on matrix mathematics would be to learn matrix multiplication before multiplication before determining the inverse of a matrix. The definition of inverse requires matrix multiplication.

    That may be why no one else replied to your question. You should give a little more context to make it clear what you already know.

    A reasonable teacher would teach matrix multiplication before teaching the calculation of an inverse. Therefore, I deduce that this isn't a homework question. However, I still am not sure what you are asking.
     
  4. Mar 10, 2013 #3
    I already retook Linear Algebra two times. This is my third. I was looking through old exams and I found this question.

    How do I get rid of of B or the A inverse at the end? I know how to do matrix multiplication, but I wasn't sure if the order I was multiplying in is allowable.

    What if I multiply something from one end by a number, and from the other end by the same number?

    Like, A=B
    A*X=B*X

    Is this true? And how much does the order matter?
     
  5. Mar 10, 2013 #4

    ehild

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    A and B are specified in the problem. You determined A-1, the inverse of A. Just calculate the matrix product BA-1.

    ehild
     
  6. Mar 10, 2013 #5

    Ray Vickson

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    Yes, of course it is true. If A = B they are the same matrix! And order generally matters a lot: often UV and VU are unequal if U and V are both square matrices.
     
  7. Mar 10, 2013 #6
    Matrices form groups. Matrix multiplication is a type of group multiplication. Groups have certain general laws.

    Matrix multiplication usually does not commute. So the sequential order is critical. However, matrix multiplication does satisfy the associative law. So you can do the multiplication from the inside out.

    You derived the answer in the right order. You shouldn't multiply the matrices in any other order than what you derived.

    However, what is wrong with multiplying in that order? You have the correct order of matrices. Multiply matrices from left to right, just as in your equations.

    I am still missing your problem (obviously).
     
    Last edited: Mar 10, 2013
  8. Mar 10, 2013 #7

    rcgldr

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    Matrtix multiplication is associative but not commutative. You can add parenthesis, but you can't rearrange the order of the matrices in a matrix equation. Since you seem to be stuck, here is a hint:

    A (C A) = A B
     
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