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Math Help - Solving for a Matrix

  1. #1
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    Solving for a Matrix

    This is from a cryptography related problem. I'm trying to find a 3x3 matrix in \mathbb{Z}_{26}, K, such that:

    K \begin{bmatrix} 0\\17\\10 \end{bmatrix} = \begin{bmatrix} 6\\2\\1 \end{bmatrix}

    K \begin{bmatrix} 15\\8\\15 \end{bmatrix} = \begin{bmatrix} 0\\15\\12 \end{bmatrix}

    K \begin{bmatrix} 19\\14\\12 \end{bmatrix} = \begin{bmatrix} 1\\22\\25 \end{bmatrix}

    So, could anyone please show me how to solve for K? I'm a bit confused, the only thing I'm given is what we get if we multiply K by 3 different vectors. Is there any simple systematic way to solve this?
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  2. #2
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    Quote Originally Posted by demode View Post
    This is from a cryptography related problem. I'm trying to find a 3x3 matrix in \mathbb{Z}_{26}, K, such that:

    K \begin{bmatrix} 0\\17\\10 \end{bmatrix} = \begin{bmatrix} 6\\2\\1 \end{bmatrix}

    K \begin{bmatrix} 15\\8\\15 \end{bmatrix} = \begin{bmatrix} 0\\15\\12 \end{bmatrix}

    K \begin{bmatrix} 19\\14\\12 \end{bmatrix} = \begin{bmatrix} 1\\22\\25 \end{bmatrix}

    So, could anyone please show me how to solve for K? I'm a bit confused, the only thing I'm given is what we get if we multiply K by 3 different vectors. Is there any simple systematic way to solve this?


    What you're given is equivalent with

    \displaystye{K\begin{pmatrix}0&15&19\\17&8&14\\10&  15&12\end{pmatrix}=\begin{pmatrix}6&0&1\\12&15&22\  \1&12&25\end{pmatrix}}.

    Now check that the LHS matrix is invertible (its determinant is a unit in \mathbb{Z}_{26}) , and multiply by

    its inverse on the right in both sides...and voila! What you get on the RHS is K.

    Tonio
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