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?
Yes, because of the fact $ \,e=mA\, $ is linear and Eve was given three linearly independent vector responses to the system and their corresponding inputs.
Proof of Linearity
Say we have two inputs $ \,m_1\, $ and $ \,m_2\, $ yielding outputs
$ \,e_1=m_1A\, $ and
$ \,e_2=m_2A\, $, respectively.
thus,
$ \,ae_1+be_2=am_1A+bm_2A $
Now, apply $ \,am_1+bm_2\, $ to the system
$ \,(am_1+bm_2)A=am_1A+bm_2A\, $
Since the two results are equal
$ \,am_1A+bm_2A=am_1A+bm_2A\, $
the system is linear.
Main Proof
Since Eve was given
$ \,m_1=(1,0,4)\, $ yields $ \,e_1=(2,0,0)\, $
$ \,m_2=(0,1,0)\, $ yields $ \,e_2=(0,1,0)\, $
$ \,m_3=(1,0,1)\, $ yields $ \,e_3=(0,0,3)\, $
where $ \,e_1, e_2, e_3\, $ are clearly linearly independent vectors, Eve can take any encrypted message and write it as a linear combination of $ \,e_1, e_2, e_3\, $
$ \,e=ae_1+be_2+ce_3\, $
Because the system is linear, we can write the input as
$ \,m=am_1+bm_2+cm_3\, $
thus, the message has been decrypted without knowing $ \,A^{-1}\, $.
What is the Decrypted Message?
TODO
