Problem #1

Bob can decrypt the message by splitting the encrypted text into a 3x3 matrix, then multiplying the encrypted by the inverse of the secret message matrix. The secret message is also a 3x3 matrix.

Problem #2

No she cannot figure out the secret message without finding the inverse of the secret matrix. As of right now, I cannot think of another way to decrypt the message without using the inverse of the matrix.

Problem #3

Step by Step equations to finding the ecrypted message:

$ \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} x \begin{bmatrix} 1 & 0 & 4 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{bmatrix} $

$ \begin{bmatrix} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{bmatrix} x \begin{bmatrix} 1 & 0 & 4 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix}^{-1} = \begin{bmatrix} \frac{-2}{3} & 0 & \frac{2}{3} \\ 0 & 1 & 0 \\ 4 & 0 & -1 \end{bmatrix} $

$ \begin{bmatrix} \frac{-2}{3} & 0 & \frac{2}{3} \\ 0 & 1 & 0 \\ 4 & 0 & -1 \end{bmatrix}^{-1} = \begin{bmatrix} \frac{1}{2} & 0 & \frac{1}{3} \\ 0 & 1 & 0 \\ 2 & 0 & \frac{1}{3} \end{bmatrix} $

Now that you have the decrypting matrix, you will be able to decrypt any secret message.

Example:

Decrypting matrix * secret message = decrypted secret message

$ \begin{bmatrix} \frac{1}{2} & 0 & \frac{1}{3} \\ 0 & 1 & 0 \\ 2 & 0 & \frac{1}{3} \end{bmatrix} x \begin{bmatrix} 2 \\ 23 \\ 3 \end{bmatrix} = \begin{bmatrix} 2 \\ 23 \\ 5 \end{bmatrix} $

The decrypted code is BWE.

Alumni Liaison

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett