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Therefore, we can get the original message back if we multiply the encrypted message by <math>\,A^{-1}\,</math>, given that the inverse of <math>\,A\,</math> exists. | Therefore, we can get the original message back if we multiply the encrypted message by <math>\,A^{-1}\,</math>, given that the inverse of <math>\,A\,</math> exists. | ||
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| + | == Can Eve Decrypt the Message Without Finding the Inverse of A? == | ||
Revision as of 20:49, 17 September 2008
Some defines:
$ \,m=\left[ \begin{array}{ccc} x & y & z \end{array} \right] \, $ is the message
$ \,A=\left[ \begin{array}{ccc} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array} \right] \, $ is the encryption matrix
$ \,e=\left[ \begin{array}{ccc} s & t & u \end{array} \right] \, $ is the encrypted message
How can Bob Decrypt the Message?
We have the equation
$ \,e=mA\, $
which is how the message is being encrypted. If we multiply both sides by the inverse of $ \,A\, $, we get
$ \,eA^{-1}=mAA^{-1}=mI=m\, $
Therefore, we can get the original message back if we multiply the encrypted message by $ \,A^{-1}\, $, given that the inverse of $ \,A\, $ exists.
