| Line 8: | Line 8: | ||
<br>No, Eve would have to find the inverse of the matrix in order to decode the messages. | <br>No, Eve would have to find the inverse of the matrix in order to decode the messages. | ||
<br> However, She does have enough information to determine the encryption vector. | <br> However, She does have enough information to determine the encryption vector. | ||
| + | <math> | ||
\begin{bmatrix} | \begin{bmatrix} | ||
2 & 0 & 0 \\ | 2 & 0 & 0 \\ | ||
| Line 31: | Line 32: | ||
</math> | </math> | ||
| − | + | ||
| − | + | Now, we take the inverse of the "secret" matrix and multiply it by the (column) vector in order to find the secret code:<br> | |
| + | |||
| + | The inverse of the "secret" matrix:<br> | ||
| + | <math> | ||
| + | \begin{bmatrix} | ||
| + | \frac{1}{2} & 0 & \frac{1}{3} \\ | ||
| + | 0 & 1 & 0 \\ | ||
| + | 2 & 0 & \frac{1}{3} \\ | ||
| + | \end{bmatrix} | ||
| + | </math> | ||
| + | <br> <br> <br> <br> | ||
| + | <math> | ||
| + | \begin{bmatrix} | ||
| + | \frac{1}{2} & 0 & \frac{1}{3} \\ | ||
| + | 0 & 1 & 0 \\ | ||
| + | 2 & 0 & \frac{1}{3} \\ | ||
| + | \end{bmatrix} | ||
| + | * | ||
| + | \begin{bmatrix} | ||
| + | 2\\ | ||
| + | 23\\ | ||
| + | 3 \\ | ||
| + | \end{bmatrix} | ||
| + | = | ||
| + | \begin{bmatrix} | ||
| + | 2\\ | ||
| + | 23\\ | ||
| + | 5\\ | ||
| + | \end{bmatrix} | ||
| + | |||
| + | </math> | ||
| + | |||
| + | <br> <br> | ||
| + | |||
| + | The encrypted message is: '''(B, W, E)'''... which is actually not really a message, just a jumble of letters... | ||
Revision as of 08:43, 18 September 2008
Homework3 Part C
How can Bob decode the message?
Bob can decode the message by breaking the message into segments of three,
and then multiplying each 3 bit segment by the inverse of the encryption vector.
Can Eve decrypt it without finding the inverse of the encryption matrix?
No, Eve would have to find the inverse of the matrix in order to decode the messages.
However, She does have enough information to determine the encryption vector.
$ \begin{bmatrix} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{bmatrix} \cdot \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 4 & 0 & 1 \end{bmatrix} $ -1
Matlab will tell us that the resultant matrix, the "secret" matrix, is:
$ \begin{bmatrix} -\frac{2}{3} & 0 & \frac{2}{3} \\ 0 & 1 & 0 \\ 4 & 0 & -1 \\ \end{bmatrix} $
Now, we take the inverse of the "secret" matrix and multiply it by the (column) vector in order to find the secret code:
The inverse of the "secret" matrix:
$ \begin{bmatrix} \frac{1}{2} & 0 & \frac{1}{3} \\ 0 & 1 & 0 \\ 2 & 0 & \frac{1}{3} \\ \end{bmatrix} $
$ \begin{bmatrix} \frac{1}{2} & 0 & \frac{1}{3} \\ 0 & 1 & 0 \\ 2 & 0 & \frac{1}{3} \\ \end{bmatrix} * \begin{bmatrix} 2\\ 23\\ 3 \\ \end{bmatrix} = \begin{bmatrix} 2\\ 23\\ 5\\ \end{bmatrix} $
The encrypted message is: (B, W, E)... which is actually not really a message, just a jumble of letters...
