Decrypting the Message w/ Knowledge of the Secret Matrix
Bob can decrypt the message simply by finding the inverse of the secret matrix and multiplying each set of 3 values in the encrypted vector by the inverse matrix to get back the original message vector.
Decrypting the Message w/o Knowledge of the Secret Matrix
Even without knowledge of the secret matrix, it is possible to decrypt the message by applying the principles of linearity. Eve can write any encrypted message that she receives as a linear combination of the one encrypted message of which she has knowledge. In other words the encrypted message (2,0,0,0,1,0,0,0,3) that Eve has intercepted can be broken down into the 3 vectors <2,0,0>, <0,1,0>, and <0,0,3>. These vectors form a basis, which allows other vectors to be written in terms of this basis. For example, the vector <4,5,9> can be written in terms of the basis as follows:
<4,5,9> = 2<2,0,0> + 5<0,1,0> + 3<0,0,3>
Since the system is linear, these same coefficients can be used to determine the unencrypted equivalent of <4,5,9> by applying them to the equivalent basis formed from the original message (1,0,4,0,1,0,1,0,1) as follows:
2<1,0,4> + 5<0,1,0> + 3<1,0,1> = <5,5,11>
In conclusion, the unencrypted message <5,5,11> yields the encrypted message <4,5,9>.
Decryption of (2,23,3)
Using the method outlined above, the vector <2,23,3> can be written as the following linear combination:
<2,23,3> = 1<2,0,0> + 23<0,1,0> + 1<0,0,3>
Applying these same coefficients to the unencrypted basis yields:
1<1,0,4> + 23<0,1,0> + 1<1,0,1> = <2,23,5>
Therefore, the original unencrypted message is (2,23,5), which corresponds to bwe.
