(New page: ==Part C: Application of linearity==) |
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==Part C: Application of linearity== | ==Part C: Application of linearity== | ||
| + | == Part 1 == | ||
| + | |||
| + | Bob can decrypt the message by using the inverse of the encrypted matrix. | ||
| + | |||
| + | == Part 2 == | ||
| + | |||
| + | No,I don't think so. But I looked at some other answers on the homework page, some do think differently. I will stick to the "no". | ||
| + | |||
| + | == Part 3 == | ||
| + | |||
| + | This is how we get the message. | ||
| + | <math> | ||
| + | \begin{bmatrix} | ||
| + | a & b & c \\ | ||
| + | d & e & f \\ | ||
| + | g & h & i | ||
| + | \end{bmatrix} | ||
| + | x | ||
| + | \begin{bmatrix} | ||
| + | 1 & 0 & 4 \\ | ||
| + | 0 & 1 & 0 \\ | ||
| + | 1 & 0 & 1 | ||
| + | \end{bmatrix} | ||
| + | = \begin{bmatrix} | ||
| + | 2 & 0 & 0 \\ | ||
| + | 0 & 1 & 0 \\ | ||
| + | 0 & 0 & 3 | ||
| + | \end{bmatrix}</math> | ||
| + | |||
| + | <math> | ||
| + | \begin{bmatrix} | ||
| + | 2 & 0 & 0 \\ | ||
| + | 0 & 1 & 0 \\ | ||
| + | 0 & 0 & 3 | ||
| + | \end{bmatrix} | ||
| + | x | ||
| + | \begin{bmatrix} | ||
| + | 1 & 0 & 4 \\ | ||
| + | 0 & 1 & 0 \\ | ||
| + | 1 & 0 & 1 | ||
| + | \end{bmatrix}^{-1} | ||
| + | = \begin{bmatrix} | ||
| + | \frac{-2}{3} & 0 & \frac{2}{3} \\ | ||
| + | 0 & 1 & 0 \\ | ||
| + | 4 & 0 & -1 | ||
| + | \end{bmatrix}</math> | ||
| + | |||
| + | <math> \begin{bmatrix} | ||
| + | \frac{-2}{3} & 0 & \frac{2}{3} \\ | ||
| + | 0 & 1 & 0 \\ | ||
| + | 4 & 0 & -1 | ||
| + | \end{bmatrix}^{-1} | ||
| + | = \begin{bmatrix} | ||
| + | \frac{1}{2} & 0 & \frac{1}{3} \\ | ||
| + | 0 & 1 & 0 \\ | ||
| + | 2 & 0 & \frac{1}{3} | ||
| + | \end{bmatrix}</math> | ||
| + | |||
| + | Apply the decrypting matrix to <math>\begin{bmatrix} | ||
| + | 2 \\ | ||
| + | 23 \\ | ||
| + | 3 \\ | ||
| + | \end{bmatrix}</math> to get BWE. | ||
Latest revision as of 17:04, 19 September 2008
Part C: Application of linearity
Part 1
Bob can decrypt the message by using the inverse of the encrypted matrix.
Part 2
No,I don't think so. But I looked at some other answers on the homework page, some do think differently. I will stick to the "no".
Part 3
This is how we get the message. $ \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} x \begin{bmatrix} 1 & 0 & 4 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{bmatrix} $
$ \begin{bmatrix} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{bmatrix} x \begin{bmatrix} 1 & 0 & 4 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix}^{-1} = \begin{bmatrix} \frac{-2}{3} & 0 & \frac{2}{3} \\ 0 & 1 & 0 \\ 4 & 0 & -1 \end{bmatrix} $
$ \begin{bmatrix} \frac{-2}{3} & 0 & \frac{2}{3} \\ 0 & 1 & 0 \\ 4 & 0 & -1 \end{bmatrix}^{-1} = \begin{bmatrix} \frac{1}{2} & 0 & \frac{1}{3} \\ 0 & 1 & 0 \\ 2 & 0 & \frac{1}{3} \end{bmatrix} $
Apply the decrypting matrix to $ \begin{bmatrix} 2 \\ 23 \\ 3 \\ \end{bmatrix} $ to get BWE.
