(New page: == Problem #1 == Bob can decrypt the message by splitting the encrypted text into a 3x3 matrix, then multiplying the encrypted by the inverse of the secret message matrix. The secret mess...)
 
(Problem #3)
Line 27: Line 27:
 
0 & 1 & 0 \\
 
0 & 1 & 0 \\
 
0 & 0 & 3
 
0 & 0 & 3
 +
\end{bmatrix}</math>
 +
 +
<math>
 +
\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}</math>
 
\end{bmatrix}</math>

Revision as of 21:26, 17 September 2008

Problem #1

Bob can decrypt the message by splitting the encrypted text into a 3x3 matrix, then multiplying the encrypted by the inverse of the secret message matrix. The secret message is also a 3x3 matrix.

Problem #2

No she cannot figure out the secret message without finding the inverse of the secret matrix. As of right now, I cannot think of another way to decrypt the message without using the inverse of the matrix.

Problem #3

Step by Step equations to finding the ecrypted message:

$ \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} $

Alumni Liaison

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett