(New page: == Part C: Application of linearity == 1. How can Bob decrypt the message? In order to decrypt the message, Bob must call upon his linear algebra skillz. He has been given the 3-by-3 spe...) |
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---- | ---- | ||
| + | 2. Can Eve decrypt the message without finding the inverse of the secret matrix? | ||
| + | |||
For Eve to crack the code, all she needs to do is take the inverse of the input matrix and multiply it by the output matrix. This will give her the (Until Recently) Secret Matrix with which she can start her career at NSA. | For Eve to crack the code, all she needs to do is take the inverse of the input matrix and multiply it by the output matrix. This will give her the (Until Recently) Secret Matrix with which she can start her career at NSA. | ||
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\begin{bmatrix} | \begin{bmatrix} | ||
| − | + | Coded Matrix | |
\end{bmatrix} | \end{bmatrix} | ||
</math> | </math> | ||
| Line 47: | Line 49: | ||
\end{bmatrix} | \end{bmatrix} | ||
</math> | </math> | ||
| + | |||
| + | |||
| + | ---- | ||
| + | 3. What is the decrypted message corresponding to (2,23,3)? | ||
| + | |||
| + | Following Sneaky Eve-y's example, we first find the secret matrix. | ||
| + | |||
| + | <math> | ||
| + | \begin{bmatrix} | ||
| + | 1 & 0 & 4 \\ | ||
| + | 0 & 1 & 0 \\ | ||
| + | 1 & 0 & 1 | ||
| + | \end{bmatrix} | ||
| + | </math> <sup>-1</sup> | ||
| + | <math> \cdot | ||
| + | |||
| + | \begin{bmatrix} | ||
| + | 2 & 0 & 0 \\ | ||
| + | 0 & 1 & 0 \\ | ||
| + | 0 & 0 & 3 | ||
| + | \end{bmatrix} | ||
| + | </math> | ||
| + | = | ||
| + | <math> | ||
| + | \begin{bmatrix} | ||
| + | -2/3 & 0 & 4 \\ | ||
| + | 0 & 1 & 0 \\ | ||
| + | 2/3 & 0 & -1 | ||
| + | \end{bmatrix} | ||
| + | </math> | ||
| + | |||
| + | Then we multiply our crypted message matrix by the inverse of the secret matrix. | ||
| + | |||
| + | |||
| + | <math> | ||
| + | \begin{bmatrix} | ||
| + | 2 & 23 &3 | ||
| + | \end{bmatrix} | ||
| + | \cdot | ||
| + | \begin{bmatrix} | ||
| + | -2/3 & 0 & 4 \\ | ||
| + | 0 & 1 & 0 \\ | ||
| + | 2/3 & 0 & -1 | ||
| + | \end{bmatrix} | ||
| + | </math> <sup>-1</sup> | ||
| + | = | ||
| + | <math> | ||
| + | \begin{bmatrix} | ||
| + | 2 & 23 & 5 | ||
| + | \end{bmatrix} | ||
| + | </math> | ||
| + | |||
| + | Which in letters is | ||
| + | |||
| + | B , W , E | ||
| + | |||
| + | which is probably code for something... | ||
Latest revision as of 15:48, 19 September 2008
Part C: Application of linearity
1. How can Bob decrypt the message?
In order to decrypt the message, Bob must call upon his linear algebra skillz. He has been given the 3-by-3 special decoder matrix that he will be able to multiply the inverse of to the coded message matrix he received via email. The result will be the decoded message.
Visually,
$ \begin{bmatrix} Coded Matrix \end{bmatrix} \cdot \begin{bmatrix} Secret Matrix \end{bmatrix} $ -1 = $ \begin{bmatrix} Uncoded Matrix \end{bmatrix} $
2. Can Eve decrypt the message without finding the inverse of the secret matrix?
For Eve to crack the code, all she needs to do is take the inverse of the input matrix and multiply it by the output matrix. This will give her the (Until Recently) Secret Matrix with which she can start her career at NSA.
Again, visually,
$ \begin{bmatrix} Uncoded Matrix \end{bmatrix} $ -1
$ \cdot \begin{bmatrix} Coded Matrix \end{bmatrix} $
=
$ \begin{bmatrix} Secret Matrix \end{bmatrix} $
3. What is the decrypted message corresponding to (2,23,3)?
Following Sneaky Eve-y's example, we first find the secret matrix.
$ \begin{bmatrix} 1 & 0 & 4 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix} $ -1 $ \cdot \begin{bmatrix} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{bmatrix} $ = $ \begin{bmatrix} -2/3 & 0 & 4 \\ 0 & 1 & 0 \\ 2/3 & 0 & -1 \end{bmatrix} $
Then we multiply our crypted message matrix by the inverse of the secret matrix.
$ \begin{bmatrix} 2 & 23 &3 \end{bmatrix} \cdot \begin{bmatrix} -2/3 & 0 & 4 \\ 0 & 1 & 0 \\ 2/3 & 0 & -1 \end{bmatrix} $ -1
=
$ \begin{bmatrix} 2 & 23 & 5 \end{bmatrix} $
Which in letters is
B , W , E
which is probably code for something...
