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Application of Linearity
1. Bob knows the encryption matrix
The letters that Alice is sending will be sent in groups of three, then multiplied by her encryption matrix which we will call [A]. We will call her letter message x, and the encrypted output y. The dimension calculation is:
$ (3\times 3)*(3\times 1)=(3\times 1)\, $
$ [A]* \begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix} = \begin{bmatrix} y_1\\ y_2\\ y_3 \end{bmatrix} $
Now for bob to decrypt the message, he must multiply the inverse of [A], which is also a 3X3 matrix, to the left side of both sides of the equation. This will isolate the x values, which are the original message. Bob can find the inverse of the matrix using the determinants, some Gaussian methods, but we will assume he has a calculator or matlab to do it for him.
$ [A]^{-1}*[A]* \begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix} =[A]^{-1}* \begin{bmatrix} y_1\\ y_2\\ y_3 \end{bmatrix} = \begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix} $
He knows all the y values, A, and the inverse of A, so he can find the x's, and thus the decoded message.
2. Eve knows inputs and outputs, but not the matrix
Eve technically could solve the function without knowing the matrix or its inverse using the linearity principle, but it is much easier to solve by finding the matrix, solving for it, and using its inverse to decode the method.
3. Eve wants to decode another grouping of 3
The matrix $ [A] $ can be solved in the following manner:
$ [A]*[X]=[Y] \, $
or
$ [A]* \begin{bmatrix} 1 & 0 & 1\\ 0 & 1 & 0\\ 4 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 3 \end{bmatrix} \, $
solves for the values of the output, and we know every X and every Y, so lets invert X and multiply it to the right side of both sides:
$ [A]*[X]*[X]^{-1}=[A]=[Y]*[X]^{-1}\, $
Doing this final calculation yields:
$ [A]= \begin{bmatrix} 2 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 3 \end{bmatrix} * \begin{bmatrix} 1 & 0 & 1\\ 0 & 1 & 0\\ 4 & 0 & 1 \end{bmatrix}^{-1} = \begin{bmatrix} -\frac{2}{3} & 0 & \frac{2}{3} \\ 0 & 1 & 0\\ 4 & 0 & -1 \end{bmatrix} \, $
So again, any $ [X]=[A]^{-1}*[Y] $ will give you the original message, so doing this goes as follows:
$ [X]= \begin{bmatrix} -\frac{2}{3} & 0 & \frac{2}{3} \\ 0 & 1 & 0\\ 4 & 0 & -1 \end{bmatrix}^{-1} * \begin{bmatrix} 2\\ 23\\ 3 \end{bmatrix} =\begin{bmatrix} 2\\ 23\\ 5 \end{bmatrix} \, $
These yield the letters B,W,E.
