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Thread: Matrix manipulation

  1. #1
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    Matrix manipulation

    How does the fact that A isn't a square n x n matrix change how to do this question? i know that if A is an invertible n x n matrix, the solution would be $\displaystyle x = A^{-1}b $...

    Suppose A is a 3 x 4 matrix and there exists a 4 x 3 matrix C such that $\displaystyle AC = I_{3} $ (the 3x3 identity matrix). Let b be an arbitrary vector in $\displaystyle \mathbb{R}^3$. Produce a solution of $\displaystyle Ax=b$
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by Robb View Post
    How does the fact that A isn't a square n x n matrix change how to do this question? i know that if A is an invertible n x n matrix, the solution would be $\displaystyle x = A^{-1}b $...

    Suppose A is a 3 x 4 matrix and there exists a 4 x 3 matrix C such that $\displaystyle AC = I_{3} $ (the 3x3 identity matrix). Let b be an arbitrary vector in $\displaystyle \mathbb{R}^3$. Produce a solution of $\displaystyle Ax=b$
    A matrix that is non square is not invertible.
    That's the main difference.

    As for your question.. :

    You know that $\displaystyle AC=I_3$
    Hence $\displaystyle Ax=b=I_3b=ACb$

    By the associativity of matrix multiplication, this is : $\displaystyle Ax=A(Cb)$

    Hence, a (*) solution is $\displaystyle x=Cb$



    (*) The reason why I say "a" and not "the", and your question states "a", is that it's not necessarily the only one. This is only one possible solution.


    Does it look clear to you ?
    Last edited by Moo; May 17th 2009 at 06:18 AM. Reason: missed a bee
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