(New page: ==Application of Linearity== ===Part 1=== All Bob needs to do is multiply the secret message by the inverse of the 3x3 secret matrix. ===Part 2===) |
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===Part 2=== | ===Part 2=== | ||
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| + | Eve needs to find the inverse of the secret matrix in order to decrypt the message. She can find the inverse, but without it she can't decrypt the message. | ||
| + | |||
| + | ===Part 3=== | ||
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| + | |||
| + | <math> \left[ \begin{array}{ccc} | ||
| + | 1 & 0 & 1 \\ | ||
| + | 0 & 1 & 0 \\ | ||
| + | 4 & 0 & 1 \end{array} \right] \times | ||
| + | \left[ \begin{array}{ccc} | ||
| + | a & b & c \\ | ||
| + | d & e & f \\ | ||
| + | g & h & i \end{array} \right] = | ||
| + | \left[ \begin{array}{ccc} | ||
| + | 2 & 0 & 0 \\ | ||
| + | 0 & 1 & 0 \\ | ||
| + | 0 & 0 & 3 \end{array} \right] | ||
| + | </math> | ||
| + | :Therefore | ||
| + | <math>\left[ \begin{array}{ccc} | ||
| + | a & b & c \\ | ||
| + | d & e & f \\ | ||
| + | g & h & i \end{array} \right] = | ||
| + | \left[ \begin{array}{ccc} | ||
| + | 1 & 0 & 1 \\ | ||
| + | 0 & 1 & 0 \\ | ||
| + | 4 & 0 & 1 \end{array} \right]^{-1} \times | ||
| + | \left[ \begin{array}{ccc} | ||
| + | 2 & 0 & 0 \\ | ||
| + | 0 & 1 & 0 \\ | ||
| + | 0 & 0 & 3 \end{array} \right] | ||
| + | </math> | ||
| + | :SECRET MATRIX = <math>\left[ \begin{array}{ccc} | ||
| + | -\frac{2}{3} & 0 & \frac{2}{3} \\ | ||
| + | 0 & 1 & 0 \\ | ||
| + | 4 & 0 & -1 \end{array} \right] | ||
| + | </math> | ||
| + | We need to take the Inverse of the Secret Matrix and multiply it by the secret message. | ||
| + | |||
| + | <math>\left[ \begin{array}{ccc} | ||
| + | -\frac{2}{3} & 0 & \frac{2}{3} \\ | ||
| + | 0 & 1 & 0 \\ | ||
| + | 4 & 0 & -1 \end{array} \right]^{-1} \times | ||
| + | \left[ \begin{array}{c} | ||
| + | 2\\ | ||
| + | 23\\ | ||
| + | 3\end{array} \right] = </math> Decrypted Message | ||
| + | :<math>\left[ \begin{array}{ccc} | ||
| + | \frac{1}{2} & 0 & \frac{1}{3} \\ | ||
| + | 0 & 1 & 0 \\ | ||
| + | 2 & 0 & \frac{1}{3} \end{array} \right] \times | ||
| + | \left[ \begin{array}{c} | ||
| + | 2\\ | ||
| + | 23\\ | ||
| + | 3\end{array} \right] = | ||
| + | \left[ \begin{array}{c} | ||
| + | 2\\ | ||
| + | 23\\ | ||
| + | 5\end{array} \right] | ||
| + | </math> | ||
| + | :The Secret Message is BWE | ||
Latest revision as of 07:47, 18 September 2008
Application of Linearity
Part 1
All Bob needs to do is multiply the secret message by the inverse of the 3x3 secret matrix.
Part 2
Eve needs to find the inverse of the secret matrix in order to decrypt the message. She can find the inverse, but without it she can't decrypt the message.
Part 3
$ \left[ \begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 4 & 0 & 1 \end{array} \right] \times \left[ \begin{array}{ccc} a & b & c \\ d & e & f \\ g & h & i \end{array} \right] = \left[ \begin{array}{ccc} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{array} \right] $
- Therefore
$ \left[ \begin{array}{ccc} a & b & c \\ d & e & f \\ g & h & i \end{array} \right] = \left[ \begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 4 & 0 & 1 \end{array} \right]^{-1} \times \left[ \begin{array}{ccc} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{array} \right] $
- SECRET MATRIX = $ \left[ \begin{array}{ccc} -\frac{2}{3} & 0 & \frac{2}{3} \\ 0 & 1 & 0 \\ 4 & 0 & -1 \end{array} \right] $
We need to take the Inverse of the Secret Matrix and multiply it by the secret message.
$ \left[ \begin{array}{ccc} -\frac{2}{3} & 0 & \frac{2}{3} \\ 0 & 1 & 0 \\ 4 & 0 & -1 \end{array} \right]^{-1} \times \left[ \begin{array}{c} 2\\ 23\\ 3\end{array} \right] = $ Decrypted Message
- $ \left[ \begin{array}{ccc} \frac{1}{2} & 0 & \frac{1}{3} \\ 0 & 1 & 0 \\ 2 & 0 & \frac{1}{3} \end{array} \right] \times \left[ \begin{array}{c} 2\\ 23\\ 3\end{array} \right] = \left[ \begin{array}{c} 2\\ 23\\ 5\end{array} \right] $
- The Secret Message is BWE
