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Math Help - Left inverse of matrix

  1. #1
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    Left inverse of matrix

    Consider the matrix \mathbf{C} = \begin{pmatrix}1 & -1\\ 1&1 \\ 0&1 \end{pmatrix}

    Find a left-inverse of C.

    Attempt:

    \begin{pmatrix}x1 & y1 & z1\\ x2&y2  &z2 \end{pmatrix}\begin{pmatrix}1 & -1\\ 1&1\\ 0&1 \end{pmatrix} = \begin{pmatrix}1 &0\\ 0&1\end{pmatrix}
    \begin{pmatrix}x1 + y1 & -x1 +y1 + z1 \\ x2 + y2& -x2 + y2 + z2 \end{pmatrix}= \begin{pmatrix}1 &0\\ 0&1\end{pmatrix}

    I know that you suppose to somehow use gauss reduction, and thought about gauss reducing:

    x1 + y1 = 1
    -x1 + y1 + z1 = 0

    and

    x2 + y2 = 0
    -x2 + y2 + z2 = 1

    but didnt get any luck
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  2. #2
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    Only square matrices have inverses.

    To solve a matrix equation of the form

    \mathbf{Y} = \mathbf{AX} for \mathbf{X} where \mathbf{A} is not square, you need to premultiply both sides by \mathbf{A}^T first.


    \mathbf{Y} = \mathbf{AX}

    \mathbf{A}^T\mathbf{Y} = \mathbf{A}^T\mathbf{AX}.


    Now \mathbf{A}^T\mathbf{A} is a square matrix, so you can premultiply both sides by (\mathbf{A}^T\mathbf{A})^{-1}.

    \mathbf{A}^T\mathbf{Y} = \mathbf{A}^T\mathbf{AX}

    (\mathbf{A}^T\mathbf{A})^{-1}\mathbf{A}^T\mathbf{Y} = (\mathbf{A}^T\mathbf{A})^{-1}\mathbf{A}^T\mathbf{AX}

    (\mathbf{A}^T\mathbf{A})^{-1}\mathbf{A}^T\mathbf{Y} = \mathbf{IX}

    (\mathbf{A}^T\mathbf{A})^{-1}\mathbf{A}^T\mathbf{Y} = \mathbf{X}.
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  3. #3
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    Quote Originally Posted by Prove It View Post
    Only square matrices have inverses.
    Why is this so? In my notes it says that a matrix that is not square can have left-inverses or right-inverses, but it can't have both.
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  4. #4
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    You have 6 variables and 4 equations.
    This system has very many solutions.
    To limit number of solutions two variables may be given some values.
    Let z1=0 and z2=0
    then your matrix is

    <br />
\begin{pmatrix}1/2 & 1/2 & 0\\ -1/2 & 1/2 & 0 \end{pmatrix}<br />
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