Introduction to Cryptography with Coding Theory, 3rd Edition

B.6 Examples for Chapter 6

Example 21

The ciphertext

$$22\text{,}09\text{,}00\text{,}12\text{,}03\text{,}01\text{,}10\text{,}03\text{,}04\text{,}08\text{,}01\text{,}17$$

was encrypted using a Hill cipher with matrix

$$\left(\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ 7& 8& 10\end{array}\right)\text{.}$$

Decrypt it.

SOLUTION

There are several ways to input a matrix. One way is the following. A $2\times 2$ matrix $\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ can be entered as matrix(2,2,[a,b,c,d]). Type evalm(M&*N) to multiply matrices $M$ and $N\text{.}$ Type evalm(v&*M) to multiply a vector $v$ on the right by a matrix $M\text{.}$

Here is the encryption matrix.

> M:=matrix(3,3,[1,2,3,4,5,6,7,8,10])

$$\left[\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ 7& 8& 10\end{array}\right]$$

We need to invert the matrix mod 26:

> invM:=map(x->x mod 26, inverse(M))

$$\left[\begin{array}{ccc}8& 16& 1\\ 8& 21& 24\\ 1& 24& 1\end{array}\right]$$

The command map(x->x mod 26, E) takes each number in an expression E and reduces it mod 26.

This is the inverse of the matrix mod 26. We can check this ...