Some definitions and explainations
1.Let A be the 3x3 matrix used to encrypt 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] \, $
Let B be the 3x3 matrix for the unencrypted message.
$ \,B=\left[ \begin{array}{ccc} b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} & b_{23} \\ b_{31} & b_{32} & b_{33} \end{array} \right] \, $
Note:this can also be written as $ \,B=\left[ \begin{array}{ccc} B_{1} & B_{2} & B_{3} \end{array} \right] \, $
Correspondingly,
2.
Some defines:
$ \,m=\left[ \begin{array}{ccc} x & y & z \end{array} \right] \, $ is the message
$ \,e=\left[ \begin{array}{ccc} s & t & u \end{array} \right] \, $ is the encrypted message
Contents
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 $ for any $ \,a,b\in \mathbb{R}\, $
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 that for the system
$ \,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\, $
$ \,\exists a,b,c\in \mathbb{R}\, $ such that $ \,e=ae_1+be_2+ce_3\, $, for any $ \,e\in \mathbb{R}^{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?
The given encrypted message is
$ \,e=(2,23,3)\, $
This can be rewritten as a linear combination of the given system result vectors
$ \,e=ae_1+be_2+ce_3\, $
$ \,e=(2,23,3)=1\cdot (2,0,0)+23\cdot (0,1,0)+1\cdot (0,0,3)\, $
Because the system s linear, we can write the input as
$ \,m=am_1+bm_2+cm_3\, $
$ \,m=1\cdot (1,0,4)+23\cdot (0,1,0)+1\cdot (1,0,1)=(2,23,5)\, $
Therefore, the unencrypted message is "BWE".
