Difference between revisions of "HW3.C Caleb Tan ECE301Fall2008mboutin" - Rhea

Latest revision as of 09:14, 18 September 2008

Info from question

$ \left[ \begin{array}{ccc} 1 & 0 & 4 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{array} \right] \times \left[ \begin{array}{ccc} X \end{array} \right] = \left[ \begin{array}{ccc} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{array} \right] $

1. How can Bob decrypt the message?

Since Bob already knows the matrix [X] and the encrypted matrix, all he has to do to find the original matrix is to invert [X] and multiply with the encrypted matrix.

2. Can Eve decrypt the message without finding the inverse of the secret matrix?

The easiest way would be for her to find the inverse of the secret matrix. I think she can still find it somehow using linearity but I really can't remember how to do it.

3. What is the decrypted message corresponding to (2,23,3)?

First we find the matrix [X]:

$ [X] = \left[ \begin{array}{ccc} 1 & 0 & 4 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{array} \right]^{-1} \times \left[ \begin{array}{ccc} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{array} \right] $

Using matlab,

$ [X] =\left[ \begin{array}{ccc} \frac{-2}{3} & 0 & 4 \\ 0 & 1 & 0 \\ \frac{2}{3} & 0 & -1 \end{array} \right] $

Finally, the decrypted message can be found by multiplying the encrypted message with the inverted matrix [X]

$ [X]^{-1} =\left[ \begin{array}{ccc} \frac{1}{2} & 0 & 2\\ 0 & 1 & 0 \\ \frac{1}{3} & 0 & \frac{1}{3} \end{array} \right] $

$ [A] = [X]^{-1} \times \left[ \begin{array}{ccc} 2\\ 23\\ 3\end{array} \right] = \left[ \begin{array}{ccc} 7\\ 23\\ \frac{5}{3}\end{array} \right] $

I used MATLAB. Not sure where it went wrong and why my answer is different from others. Anybody who knows care to help me?