Since Bob has knowledge of the secret matrix, he can find the inverse of the secret matrix. Using some basic knowledge of linear algebra, he can then multiply each (a,b,c) vector by the inverted secret matrix to yield the original matrix.
Since Eve does not know the secret matrix, she will have to use the idea of a basis.
The encrypted matrix:
$ \left( \begin{array}{ccc} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{array} \right) $
Can be written as a basis, using the rows of the matrix, so that other vectors can be expressed as a linear combination of the basis.
For example, the vector $ \left( \begin{array}{ccc} 2a & b & 3c \end{array} \right) $
can be written as the linear combination of the rows of the encrypted matrix.
Using the vector $ \left( \begin{array}{ccc} 10 & 8 & 6 \end{array} \right) $, we know that the coefficients used with the basis to yield the vector are (5,8,2). Then, we can use multiply those coefficients with the unencrypted vector to yield the unencrypted message. The encrypted message of (10,8,6) is (7,8,9).
Finally, (2,23,3) is the encrypted message. The coefficients used with the basis are (1,23,1) which yields (2,23,5).
