(→Part 2) |
(→Part 3) |
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=== Part 3 === | === Part 3 === | ||
| + | |||
| + | We can also solve the system quicker by finding the inverse of the "secret" matrix. | ||
| + | |||
| + | In order to find that, we multiply the encrypted vector matrix with the inverse of the message matrix. | ||
| + | |||
| + | <math> | ||
| + | \begin{bmatrix} | ||
| + | 2 & 0 & 0 \\ | ||
| + | 0 & 1 & 0 \\ | ||
| + | 0 & 0 & 3 | ||
| + | \end{bmatrix} | ||
| + | \cdot | ||
| + | \begin{bmatrix} | ||
| + | 1 & 0 & 4 \\ | ||
| + | 0 & 1 & 0 \\ | ||
| + | 1 & 0 & 1 | ||
| + | \end{bmatrix} | ||
| + | </math> <sup>-1</sup> | ||
| + | |||
| + | |||
| + | Matlab will tell us that the resultant matrix, the "secret" matrix, is: | ||
Revision as of 15:15, 16 September 2008
Application of Linearity
Part 1
All Bob has to do to decrypt his message is:
1. Take the "message" vector and divide it into columns with three rows, such that the first, second, and third elements are in the first column, the fourth, fifth, and sixth elements are in the second column, etc.
2. Multiply the inverse of the "special" matrix that she sent him with each segment of the message vector.
3. Combine all the segments in the correct order
4. Convert the numbers back to letters using Alice's system such that A=1, B=2...
That's it!
Part 2
Eve should be able to figure out the message without knowing the "secret" matrix by using linearity properties.
$ \,\ a*\mathbf{x_1} + b*\mathbf{x_2} = A*\mathbf{y_1} + B*\mathbf{y_2} $
She can take the separate segments and solve for the constants $ a, b, A, $ and $ B $.
This should tell us what the system does on a basis.
(Of course, it would be easier if she had the "secret" matrix)
Part 3
We can also solve the system quicker by finding the inverse of the "secret" matrix.
In order to find that, we multiply the encrypted vector matrix with the inverse of the message matrix.
$ \begin{bmatrix} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{bmatrix} \cdot \begin{bmatrix} 1 & 0 & 4 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix} $ -1
Matlab will tell us that the resultant matrix, the "secret" matrix, is:
