Solving for a Matrix

**demode** · March 20th 2011, 12:30 AM

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?

**tonio** · March 20th 2011, 02:32 AM

Originally Posted by demode

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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