Contents
Part C: Application of linearity
1. How can Bob decrypt the message?
Bob can get the message by multiplying the Message by the Secret Matrix inverted then decoding the numbers into letters. M*SM=E where M is message SM is the Secret Matrix and E is encrypted message. M=E*SM^-1
2. Can Eve decrypt the message without finding the inverse of the secret matrix?
She can find what the secret matrix is. She can right a system of equations and solve for each component of the secret message.
- $ \begin{pmatrix} 1 & 0 & 4 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{pmatrix}\begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}=\begin{pmatrix} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{pmatrix} $
- Multiply out
- $ a+4g=2, b+4h=0, c+4i=0 \, $
- $ d=0, e=1, f=0 \, $
- $ a+g=0, b+h=0, c+i=0 \, $
Solving These Equations yields the Secret Matrix
- $ \begin{pmatrix} -2/3 & 0 & 4 \\ 0 & 1 & 0 \\ 2/3 & 0 & -1 \end{pmatrix} $
but finding the secret message does not help her she needs to find the inverse of this matrix. This can be done without inverting the SM. Do the same method above.
- $ \begin{pmatrix} 1 & 0 & 4 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{pmatrix}=\begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}*\begin{pmatrix} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{pmatrix} $
3. What is the decrypted message corresponding to (2,23,3)? (Write it as a text.)
$ E*SM^-1=M \, $
- $ \begin{pmatrix} 2 & 23 & 3\end{pmatrix}*\begin{pmatrix} -2/3 & 0 & 4 \\ 0 & 1 & 0 \\ 2/3 & 0 & -1 \end{pmatrix}^{-1}=\begin{pmatrix} 2 & 23 & 5\end{pmatrix} $
"BWE"
