Revision as of 20:50, 17 September 2008 by Jkubasci (Talk)

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.

Can Eve Decrypt the Message Without Finding the Inverse of A?

What is the Decrypted Message?

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