# 3 pass protocol

suppose we have 3 matrix named A,B,C (2x2, 3x3, 10x10 or whatever). As we know A^(-1) A C B B^(-1)=C

So i take a file and convert it to numbers e.g. ASCII code (lets say 65,72,71...) and i add a number sequence.

Then i multiply A*C and the result (D) is send.

The other guy multiplies D*B and sends the result (E) back.

Then i multiply A^(-1)*E and send the result (F).

At last the other guy multiplies F*B^(-1) , gets the original file C, subtracts the number sequence and gets the original file.

My question is: how will a cryptanalyst break it?

Well, the most obvious way is that the cryptanalyst sees both $D$ and $E = D \times B$; he then can compute $D^{-1} \times E = D^{-1} \times D \times B = B$.
Then, he sees $F$; as he now knows $B$, he can compute $F \times B^{-1} = C$.
So, the strength of the system is in the 'number sequence' that maps $C$ back to the original file, that is, essentially a symmetric crypto system.