How can Bob decrypt the message?
Bob can take the inverse of the message matrix and multiply it to the encrypted matrix. (These matrices are created from the message vector and encrypted vector.) Then the inverse of the matrix of the results can be multiplied to the (2,23,3) vector to get the corresponding letters.
Can Eve decrypt the message without finding the inverse of the secret matrix?
Eve can defiantly do it without finding the inverse, how she goes about exactly doing it I do not remeber. But I do the same in Linear algebra by not using the inverse but at this time I can not remeber what I used.
What is the decrypted message corresponding to (2,23,3)?
$ \left[\begin{array}{ccc}1 & 0 & 1 \\0 & 1 & 0 \\4 & 0 & 1 \end{array}\right]^{-1}\!\left[\begin{array}{ccc}2 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & 3 \end{array}\right] = \left[\begin{array}{ccc}-\frac{2}{3} & 0 & \frac{2}{3} \\0 & 1 & 0 \\4 & 0 & -1 \end{array}\right]\! $
$ \left[\begin{array}{ccc}-\frac{2}{3} & 0 & \frac{2}{3} \\0 & 1 & 0 \\4 & 0 & -1 \end{array}\right]^{-1}\! = \left[\begin{array}{ccc}-\frac{1}{2} & 0 & \frac{1}{3} \\0 & 1 & 0 \\2 & 0 & -\frac{1}{3} \end{array}\right]\! $
$ \left[\begin{array}{ccc}-\frac{1}{2} & 0 & \frac{1}{3} \\0 & 1 & 0 \\2 & 0 & -\frac{1}{3} \end{array}\right] \!\left[\begin{array}{ccc}2 \\ 23 \\3 \end{array}\right]\! = \left[\begin{array}{ccc}2 \\ 23 \\5 \end{array}\right]\! $
$ 2 = B , 23 = W , 5 = E $ So the o so secret message is BWE
