Linear Algebra: Decoding cryptograms

Question

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.

Answer 1

$A$ 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.

Answer 2

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?