Part 1

Bob can multiply the inverse of the secret matrix by the encrypted matrix.

Part 2

No but she has what is necessary to find the inverse matrix.

Part 3

This process yields the decrypting matrix.

$ \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} x \begin{bmatrix} 1 & 0 & 4 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{bmatrix} $

$ \begin{bmatrix} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{bmatrix} x \begin{bmatrix} 1 & 0 & 4 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix}^{-1} = \begin{bmatrix} \frac{-2}{3} & 0 & \frac{2}{3} \\ 0 & 1 & 0 \\ 4 & 0 & -1 \end{bmatrix} $

$ \begin{bmatrix} \frac{-2}{3} & 0 & \frac{2}{3} \\ 0 & 1 & 0 \\ 4 & 0 & -1 \end{bmatrix}^{-1} = \begin{bmatrix} \frac{1}{2} & 0 & \frac{1}{3} \\ 0 & 1 & 0 \\ 2 & 0 & \frac{1}{3} \end{bmatrix} $

Now apply the decrypting matrix to $ \begin{bmatrix} 2 \\ 23 \\ 3 \\ \end{bmatrix} $ to get BWE.

Alumni Liaison

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang