(→Decrypted Message Corresponding to (2,23,3)) |
(→Decrypted Message Corresponding to (2,23,3)) |
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This gives us the result: <math>\left[\begin{array}{ccc}-\frac{2}{3} & 0 & \frac{2}{3} \\0 & 1 & 0 \\4 & 0 & -1 \end{array}\right]\!</math> | This gives us the result: <math>\left[\begin{array}{ccc}-\frac{2}{3} & 0 & \frac{2}{3} \\0 & 1 & 0 \\4 & 0 & -1 \end{array}\right]\!</math> | ||
| − | Now, we can take the inverse of the result: <math>\left[\begin{array}{ccc}-\frac{2}{3} & 0 & \frac{2}{3} \\0 & 1 & 0 \\4 & 0 & -1 \end{array}\right]^{-1} \!</math> = <math>\left[\begin{array}{ccc}-\frac{1}{2} & 0 & \frac{1}{3} \\0 & 1 & 0 \\2 & 0 & -\frac{1}{3} \end{array}\right] | + | Now, we can take the inverse of the result: <math>\left[\begin{array}{ccc}-\frac{2}{3} & 0 & \frac{2}{3} \\0 & 1 & 0 \\4 & 0 & -1 \end{array}\right]^{-1} \!</math> = <math>\left[\begin{array}{ccc}-\frac{1}{2} & 0 & \frac{1}{3} \\0 & 1 & 0 \\2 & 0 & -\frac{1}{3} \end{array}\right] \!</math> |
Revision as of 17:35, 18 September 2008
How will Bob decrypt the message?
The first thing Bob needs to do is to break the encrypted message (vector form) into smaller vectors. Then, Bob can multiply them by the inverse of the encryption matrix. The product of these will give the final output / hidden message.
Can Eve decrypt the message without taking the inverse of the secret matrix?
No, the message cannot be decrypted without taking the inverse of the matrix. However, since she was the creator of the decryption system, she should be able to figure it out.
Decrypted Message Corresponding to (2,23,3)
First, we take the product of the two matricies as follows: $ \left[\begin{array}{ccc}2 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & 3 \end{array}\right]\times\left[\begin{array}{ccc}1 & 0 & 1 \\0 & 1 & 0 \\4 & 0 & 1 \end{array}\right]^{-1}\! $
This gives us the result: $ \left[\begin{array}{ccc}-\frac{2}{3} & 0 & \frac{2}{3} \\0 & 1 & 0 \\4 & 0 & -1 \end{array}\right]\! $
Now, we can take the inverse of the result: $ \left[\begin{array}{ccc}-\frac{2}{3} & 0 & \frac{2}{3} \\0 & 1 & 0 \\4 & 0 & -1 \end{array}\right]^{-1} \! $ = $ \left[\begin{array}{ccc}-\frac{1}{2} & 0 & \frac{1}{3} \\0 & 1 & 0 \\2 & 0 & -\frac{1}{3} \end{array}\right] \! $
