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The problem:

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My attempt at the solution:

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I'm not sure exactly what I need to do to find A^-1

PS: This is for a Introductory Linear algebra class.

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    This shows why e$n$cryption with matrices is not a good idea. If you know a bit of the decoded message then you can decode the whole message. In the example, knowing 4 letters is enough.2011-05-31

2 Answers 2

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$A$ takes $[0,19]$ to $[-19,-19]$ and so takes $[0,1]$ to $[-1,-1]$. $A$ takes $[21,5]$ to $[37,16]$ and so takes $[21,0]=[21,5]-5[0,1]$ to $[37,16]-5[-1,-1]=[42,21]$. So, $A$ takes $[1,0]$ to $[2,1]$. This gives you all entries of $A$. Now invert.

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    @Alex, just use linearity: $[21,0]A=([21,5]−5[0,1])A=[21,5]A−5[0,1]A=[37,16]−5[−1,−1]=[42,21]$.2011-05-31
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Let $A=\pmatrix{a&b\cr c&d\cr}$. Then $\pmatrix{a&b\cr c&d\cr}\pmatrix{0&21\cr19&5\cr}=\pmatrix{-19&37\cr-19&16}$ Can you work out $A$ from that?

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    @Virtuoso, as lhf says, $A$ takes $(0,19)$ to $(-19,-19)$ and $(21,5)$ to $(37,16)$; matrix multiplication is just a way of getting all of that into a single equation. And if you haven't learned how to multiply matrices, now is the time to do it.2011-05-31