# Linear Algebra: Decoding cryptograms

The problem:

My attempt at the solution:

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 encryption 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. – lhf May 31 '11 at 13:23

$A$ 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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"A takes [21,5] to [37,16] and so takes [21,0] to [37,16]−5[−1,−1]=[42,21]"... Please explain this, why did you -5[etc] ? – chaver May 31 '11 at 13:02
@Alex, I've just edited my answer to address that. – lhf May 31 '11 at 13:05
Sorry, I still don't understand . – chaver May 31 '11 at 13:07
@Alex, just use linearity: $[21,0]A=([21,5]−5[0,1])A=[21,5]A−5[0,1]A=[37,16]−5[−1,−1]=[42,21]$. – lhf May 31 '11 at 13:10

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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Why did you arrange it that way? This is so confusing -_- . – chaver May 31 '11 at 13:00
@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. – Gerry Myerson May 31 '11 at 13:32