Difference between revisions of "HW3.C Phil Cannon ECE301Fall2008mboutin" - Rhea

Latest revision as of 15:01, 18 September 2008

Part C: Application of Linearity

1. Bob can decrypt the message by multiplying it (in groups of 3) by the inverse of the 3-by-3 secret matrix.

2. No. $ [Secret Message]*[Secret Matrix]=[Encrypted Message]\! $. Thus in order to find the secret message she must determine the inverse of the 3-by-3 secret matrix. However, there probably is some roundabout way she could decrypt it without technically finding the inverse.

3. In part 2 both a decrypted message and encrypted message are given. Thus that can be used to find the secret matrix:
$ \left[ \begin{array}{ccc}2 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & 3 \end{array} \right] \times \left[ \begin{array}{ccc}1 & 0 & 1 \\0 & 1 & 0 \\4 & 0 & 1 \end{array} \right]^{-1} = \left[ \begin{array}{ccc}-\frac{2}{3} & 0 & \frac{2}{3} \\0 & 1 & 0 \\4 & 0 & -1 \end{array} \right]\! $

Then we can find the inverse: $ \left[ \begin{array}{ccc}-\frac{2}{3} & 0 & \frac{2}{3} \\0 & 1 & 0 \\4 & 0 & -1 \end{array} \right]^{-1} = \left[ \begin{array}{ccc}-\frac{1}{2} & 0 & \frac{1}{3} \\0 & 1 & 0 \\2 & 0 & \frac{1}{3} \end{array} \right] $

Then just multiply by the given matrix: $ \left[ \begin{array}{ccc}-\frac{1}{2} & 0 & \frac{1}{3} \\0 & 1 & 0 \\2 & 0 & \frac{1}{3} \end{array} \right] \left[ \begin{array}{ccc} 2 \\23\\3 \end{array} \right]=\left[ \begin{array}{ccc} 2 \\23\\5 \end{array} \right] $

Which, written as text, is BWE.