1. Bob can decrypt the message by multiplying the encrypted message with the inverse of the secret matrix.


2. Eve can not decrypt the message without the inverse of the secret matrix. She does however have all the necessary information to find said inverse.


3. $ \left[ \begin{array}{ccc} 1 & 0 & 4 \\ 0 & 1 & 0 \\ 1 & 0 & 1\end{array}\right]\times \left[ \begin{array}{ccc} X \end{array}\right] = \left[ \begin{array}{ccc} 2 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 3\end{array}\right] $

We must find the inverse of the secret matrix to decode the message. Using the two given matrices, we find that the secret matrix is: $ \left[ \begin{array}{ccc} \frac{-2}{3} & 0 & \frac{2}{3}\\ 0 & 1 & 0\\ 4 & 0 & -1\end{array}\right] $

We must use the inverse of that to decrypt the message. The inverse is: $ \left[ \begin{array}{ccc} \frac{1}{2} & 0 & \frac{1}{3}\\ 0 & 1 & 0\\ 2 & 0 & \frac{1}{3}\end{array}\right] $

Therefore to decrypt the message:

$ \left[ \begin{array}{ccc} \frac{1}{2} & 0 & \frac{1}{3}\\ 0 & 1 & 0\\ 2 & 0 & \frac{1}{3}\end{array}\right]\times \left[ \begin{array}{ccc} 2\\ 23\\ 3\end{array}\right] = \left[ \begin{array}{ccc} 2\\ 23\\ 5\end{array}\right] $

Using the numbers as the letters of the alphabet 2,23,5 decodes to BWE.

Alumni Liaison

Meet a recent graduate heading to Sweden for a Postdoctorate.

Christine Berkesch