there results
B=AC =
8 15 23
1 18 5
25 15 21
1 1 0
0 1 − 1
1 −2 4
=
2 6 19
6 9 2
17 27 11
=
B F S
F I B
Q? K
Here modulo 29 arithmetic has been used. For example,
8(1) + 15(0) + 23(1) = 31 ≡2( mod 29 )
8(1) + 15(1) + 23(−2) = − 23 ≡6( mod 29 )
8(0) + 15(−1) + 23(4) = 77 ≡19( mod 29 ).
with similar results using inner products involving the second and third row vectors
of A. Upon receiving the coded message, where you know that the matrix C was
used to make up the code, then you can decipher the message by multiplying by
C−^1 (mod 29 ), since B=AC implies that A=BC −^1 .For example,
A=BC −^1 =
2 6 19
6 9 2
17 27 11
2 − 4 − 1
−1 4 1
−1 3 1
=
8 15 23
1 18 5
25 15 21
=
H O W
A R E
Y O U
.
Here are some coded messages which were coded modulo 29 using the matrix C
above.
C T C I?
!! D! C
P F E N N
Y R L
S Y?
T V M
G Q T
! T R
P X M
U N P
Y J X
W V U
W Y L
R J Q
O
S G M
N E H
