How can Bob Decrypt the Message?
Let A be the 3x3 secret matrix 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 original 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] \, $
Correspondingly, let C be the crypted message
From the poblem: $ \,C = B * A\, $
So $ \,C*A^{-1} = B * A * A^{-1} = B\, $, i.e. $ \,B = C*A^{-1} $
Thus Bob can decrypt the message by finding the inverse of the secret matrix.
Can Eve decrypt the message without finding the inverse of the secret matrix?
if we write B as $ \,B=\left[ \begin{array}{ccc} B_{1} & B_{2} & B_{3} \end{array} \right] \, $ (write in conbimation of column vector)
Similarly, we write C as $ \,C=\left[ \begin{array}{ccc} C_{1} & C_{2} & C_{3} \end{array} \right] \, $
Thus,based on the multiplication of Matrix,we can have: $ \,C_1 = B*A\, $, same for $ \,C_2\, $ and $ \,C_3\, $
which means
$ \,C_1 = B*A_1 = B_1*A_1+B_2*A_2+B_3*A_3\, $
$ \,C_2 = B*A = B_1*A_1+B_2*A_2+B_3*A_3\, $
$ \,C_3 = B*A = B_1*A_1+B_2*A_2+B_3*A_3\, $
if we multiply $ B_1 $ by n,$ B_2 $ by p,$ B_3 $ by q, we have:
$ \,C_1new = B*A_1 = n*B_1*a_11+p*B_2*a_21+q*B_3*a_31\, $
$ \,C_2new = B*A_2 = n*B_1*a_21+p*B_2*a_22+q*B_3*a_23\, $
$ \,C_3new = B*A_3 = n*B_1*a_31+p*B_2*a_32+q*B_3*a_33\, $
and the reverse is also true
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".
