Vtyjkyuk

We assume that the message $$!IWGVIEX!ZRADRYD$$ has been encrypted
using Hill Cipher and with the correspondence $A=0,...,Z=25, _=26, ? =27,!=28$. We know that the last five letters of plaintext are sender's signature $MARIA$.

We want to find the deciphering matrix and read the message.

My attempt: At first we suppose that $m=2, P=C=BbbZ_29$ and $K=mathrmGL_2(BbbZ_29)$. By the hypothesis, we know that $MARIA leftrightarrow ADRYD iff (12,0,17,8,0)leftrightarrow (0,3,17,24,17)$.

This implies that, if we assume that the key matrix is $Ain mathrmGL_2(BbbZ_29)$, the diciphering function is
$$ E_A: (BbbZ_29)^2 longrightarrow (BbbZ_29)^2,\
(12,0)mapsto (12,0)cdot A =(0,3), \
(17,8) mapsto (17,8)cdot A=(17,24) $$
so,
$
beginpmatrix
12 & 0 \
17 & 8
endpmatrix cdot A = beginpmatrix
0 & 3 \
17 & 24
endpmatrix $.
We define $B:=beginpmatrix
12 & 0 \
17 & 8
endpmatrix$. Then, $det B=9 in U_29 iff B in mathrmGL(BbbZ_29)$. Also, $mathrmadj B= beginpmatrix
8 & 0 \
12 & 12
endpmatrix $ and $9^-1=13 in U_29$. So,
$$B^-1=13 beginpmatrix
8 & 0 \
12 & 12
endpmatrix = beginpmatrix
9 & 0 \
11 & 11
endpmatrix.$$ And from this,
$$A = beginpmatrix
9 & 0 \
11 & 11
endpmatrix cdot beginpmatrix
0 & 3 \
17 & 24
endpmatrix = beginpmatrix
0 & 27 \
13 & 7
endpmatrix in mathrmGL(BbbZ_29).
$$

But if we use this matrix as a key and encipher the word $MARIA$, we will not take the word $ADRYD$ that we expect, so something is wrong.
Is this method correct? Do I miss something?

Thank you.

You have to split the ciphertext in pairs, and your ciphertext corresponds to the following pairs of elements of $mathbbZ_29$:

(28, 8), (22, 6), (21, 8), (4, 23), (28, 25), (17, 0), (3, 17), (24, 3)

And the last 4 letters of your assumed plain text are "ARIA" (two pairs) that equal (0, 17), (8, 0) We cannot use the M, as then we'd have to guess the letter before M too, as encryption is done by pairs. But 4 unknowns with 4 equations should suffice. So if we denote by $E$ the $2 times 2$ encryption matrix we have that these 2 pairs must become the final two pairs of the cipher text, or written out as a matrix equation $EP = C$ where $P$ is a matrix of plain texts, and $C$ the corresponding ciphertexts:

$$E cdot beginbmatrix 0 & 8 \ 17 & 0 endbmatrix =
beginbmatrix 3 & 24\ 17 & 3 endbmatrix$$

As the $P$ matrix is invertible, we compute $E$ as $CP^-1$ (using mod $29$ arithmetic and the easy formula for inverses of $2 times 2$ matrices).
Then compute $D$ as $E^-1$ and you'll find a nice plaintext indeed ending in ARIA; I checked and it works out.