Matrix Ciphers - Introduction to Cryptography with Java Applets - page 146

She then multiplies row 3 by 25 (which is its own inverse mod 26), then subtracts 11 times
row 3 from both row 1 and row 2. This gives the desired solutions.

10031
01001
00110

She now knows that

a 3 (mod 26) c 1 (mod 26)

b 0 (mod 26) d 1 (mod 26)

s 1 (mod 26) t 0 (mod 26)

and so an enciphering matrix A is

30
11

and its corresponding shift vector B is

1
0

(You should test these values to ensure that A and B actually map the given plaintext to the
ciphertext.) Once our cryptanalyst has A , it is simple to compute the inverse A modulo 26

to obtain

90
17 1

E XAMPLE .

Earlier we enciphered the message

TH EE ND

to the ciphertext

LB PA RV.

Suppose the cryptanalyst knows we are using matrix ciphers of block size m = 2 with the
ordinary alphabet. She acquires both the plaintext message and the ciphertext message.
Now,

“TH”(=19 7) corresponds with “LB”(=11 1),
“EE”(=4 4) corresponds with “PA”(=15 0),
“ND”(=13 3) corresponds with “RV”(=17 21).

Using the same procedure described above, she solves the first system to obtain values
for a , b , and s , and the second system to get the values for c , d , and t . You should be able to
do this, and to verify that an enciphering matrix A is

517
415