Application of Linearity

Part 1

All Bob needs to do is multiply the secret message by the inverse of the 3x3 secret matrix.

Part 2

Eve needs to find the inverse of the secret matrix in order to decrypt the message. She can find the inverse, but without it she can't decrypt the message.

Part 3

$ \left[ \begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 4 & 0 & 1 \end{array} \right] \times \left[ \begin{array}{ccc} a & b & c \\ d & e & f \\ g & h & i \end{array} \right] = \left[ \begin{array}{ccc} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{array} \right] $

Therefore

$ \left[ \begin{array}{ccc} a & b & c \\ d & e & f \\ g & h & i \end{array} \right] = \left[ \begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 4 & 0 & 1 \end{array} \right]^{-1} \times \left[ \begin{array}{ccc} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{array} \right] $

SECRET MATRIX = $ \left[ \begin{array}{ccc} -\frac{2}{3} & 0 & \frac{2}{3} \\ 0 & 1 & 0 \\ 4 & 0 & -1 \end{array} \right] $

We need to take the Inverse of the Secret Matrix and multiply it by the secret message.

$ \left[ \begin{array}{ccc} -\frac{2}{3} & 0 & \frac{2}{3} \\ 0 & 1 & 0 \\ 4 & 0 & -1 \end{array} \right]^{-1} \times \left[ \begin{array}{c} 2\\ 23\\ 3\end{array} \right] = $ Decrypted 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}{c} 2\\ 23\\ 3\end{array} \right] = \left[ \begin{array}{c} 2\\ 23\\ 5\end{array} \right] $
The Secret Message is BWE

Alumni Liaison

Questions/answers with a recent ECE grad

Ryne Rayburn