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== Part 3 == | == Part 3 == | ||
| − | Using matlab | + | Using matlab, the secret matrix can be found in the following way: |
| + | |||
| + | :<br> | ||
| + | |||
| + | :<math> | ||
| + | \begin{bmatrix} | ||
| + | 2 & 0 & 0 \\ | ||
| + | 0 & 1 & 0 \\ | ||
| + | 0 & 0 & 3 | ||
| + | \end{bmatrix} | ||
| + | \div | ||
| + | \begin{bmatrix} | ||
| + | 1 & 0 & 4 \\ | ||
| + | 0 & 1 & 0 \\ | ||
| + | 1 & 0 & 1 | ||
| + | \end{bmatrix} | ||
| + | = | ||
| + | \begin{bmatrix} | ||
| + | S11 & S12 & S13 \\ | ||
| + | S21 & S22 & S23 \\ | ||
| + | S31 & S32 & S33 | ||
| + | \end{bmatrix} | ||
| + | |||
| + | </math> | ||
| + | <br> | ||
| + | |||
| + | The result is the secret matrix | ||
| + | <br> | ||
| + | :<math> | ||
| + | |||
| + | \begin{bmatrix} | ||
| + | -2/3 & 0 & 7/3 \\ | ||
| + | 0 & 1 & 0 \\ | ||
| + | 1 & 0 & -1 | ||
| + | \end{bmatrix} | ||
| + | |||
| + | </math> | ||
| + | |||
| + | <br> | ||
| + | |||
| + | Finally, applying the secret matrix to the decrypted message: | ||
| + | |||
| + | |||
| + | :<br> | ||
| + | |||
| + | :<math> | ||
| + | \begin{bmatrix} | ||
| + | 2 & 23 & 3 | ||
| + | \end{bmatrix} | ||
| + | \cdot | ||
| + | \begin{bmatrix} | ||
| + | -2/3 & 0 & 7/3 \\ | ||
| + | 0 & 1 & 0 \\ | ||
| + | 1 & 0 & -1 | ||
| + | \end{bmatrix} | ||
| + | = | ||
| + | \begin{bmatrix} | ||
| + | S11 & S12 & S13 | ||
| + | \end{bmatrix} | ||
| + | |||
| + | </math> | ||
| + | <br> | ||
Revision as of 09:56, 19 September 2008
Part 1
Since Bob does not know a specific output given a specific input, he must use the inverse of the secret matrix.
Part 2
I believe that Eve will be able to decrypt the message because I think that the system is both linear and time invariant. The system is linear because it behaves like the picture in the link below: (the system is the secret matrix, a and b are scalars that are multiplied)
Media:301_schematic_ECE301Fall2008mboutin.pdf
The system is obviously time invariant because it doesn't matter what time you evaluate the encrypted code with the secret matrix, it will always give you the same output. I'm not exactly sure how she would do it, I need some time to think about it. Does anyone have any ideas? Or is this completly wrong?
Part 3
Using matlab, the secret matrix can be found in the following way:
- $ \begin{bmatrix} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{bmatrix} \div \begin{bmatrix} 1 & 0 & 4 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix} = \begin{bmatrix} S11 & S12 & S13 \\ S21 & S22 & S23 \\ S31 & S32 & S33 \end{bmatrix} $
The result is the secret matrix
- $ \begin{bmatrix} -2/3 & 0 & 7/3 \\ 0 & 1 & 0 \\ 1 & 0 & -1 \end{bmatrix} $
Finally, applying the secret matrix to the decrypted message:
- $ \begin{bmatrix} 2 & 23 & 3 \end{bmatrix} \cdot \begin{bmatrix} -2/3 & 0 & 7/3 \\ 0 & 1 & 0 \\ 1 & 0 & -1 \end{bmatrix} = \begin{bmatrix} S11 & S12 & S13 \end{bmatrix} $
