Part C: Application of linearity
Part 1
Bob can decrypt the message by using the inverse of the encrypted matrix.
Part 2
No,I don't think so. But I looked at some other answers on the homework page, some do think differently. I will stick to the "no".
Part 3
This is how we get the 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} $
Apply the decrypting matrix to $ \begin{bmatrix} 2 \\ 23 \\ 3 \\ \end{bmatrix} $ to get BWE.
