To find the key for a Hill cipher, we need to use linear algebra over modular arithmetic, specifically modulo 26 (since we're dealing with the English alphabet). The Hill cipher encryption works as follows:

$C = K \cdot P \mod 26$

Where:

• $C$ is the ciphertext represented as column vectors,
• $P$ is the plaintext represented as column vectors, and
• $K$ is the key matrix we are trying to find.

We are given the following information:

• The plaintext "flower" is encrypted as "BNADMH" using a Hill cipher with a 3x3 key matrix.

Step-by-Step Process:

1. Convert the plaintext and ciphertext into numbers:
   • Assign each letter of the alphabet a corresponding number (A = 0, B = 1, ..., Z = 25).

Letter	Number
F	5
L	11
O	14
W	22
E	4
R	17

For the plaintext "flower", we get the numbers: $P = [5, 11, 14, 22, 4, 17]$.

Letter	Number
B	1
N	13
A	0
D	3
M	12
H	7

For the ciphertext "BNADMH", we get the numbers: $C = [1, 13, 0, 3, 12, 7]$.

2. Group the plaintext and ciphertext into 3x1 column vectors:

   • Plaintext: $P_1 = \begin{bmatrix} 5 \\ 11 \\ 14 \end{bmatrix}, \quad P_2 = \begin{bmatrix} 22 \\ 4 \\ 17 \end{bmatrix}$
   • Ciphertext: $C_1 = \begin{bmatrix} 1 \\ 13 \\ 0 \end{bmatrix}, \quad C_2 = \begin{bmatrix} 3 \\ 12 \\ 7 \end{bmatrix}$

3. Set up the equation: Using the Hill cipher equation $C = K \cdot P \mod 26$, we can now write two matrix equations: $C_1 = K \cdot P_1 \mod 26$ $C_2 = K \cdot P_2 \mod 26$

4. Solve for the key matrix $K$: To find $K$, we need to invert the plaintext matrix $P$, and then multiply it by the ciphertext matrix $C$.

   The key matrix is given by: $K = C \cdot P^{-1} \mod 26$ Where $P$ is the matrix formed by concatenating $P_1$ and $P_2$, and $C$ is the matrix formed by concatenating $C_1$ and $C_2$.

   Let me compute these steps.

I'll first calculate the inverse of the plaintext matrix and then use it to find the key matrix.It looks like I attempted to calculate the inverse of a non-square matrix, which isn't allowed. In Hill cipher problems, the plaintext matrix must be constructed carefully, typically from 3x3 matrices since we're dealing with a 3x3 key.

Let me revisit the grouping. I'll take the first three letters from the plaintext and ciphertext to form the proper matrices, and then attempt to compute the key again.It seems that the determinant of the plaintext matrix is not invertible under modulo 26. This can happen if the determinant shares a common factor with 26 (such as 2 or 13), meaning the matrix is not suitable for use in a Hill cipher.

To resolve this, we would need to adjust the plaintext pairs or check if alternative groupings or corrections could provide an invertible matrix. Alternatively, we can apply a different method to reconstruct the key using alternative approaches.

Would you like me to try different groupings or provide more insights on how to handle non-invertible matrices?