(→2. Can Eve decrypt the message without finding the inverse of the secret matrix?) |
(→3. What is the decrypted message corresponding to (2,23,3)? (Write it as a text.)) |
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==Part C: Application of linearity== | ==Part C: Application of linearity== | ||
==1. How can Bob decrypt the message? == | ==1. How can Bob decrypt the message? == | ||
| − | Bob can get the message by multiplying the Message by the Secret | + | Bob can get the message by multiplying the Message by the Secret Matrix inverted then decoding the numbers into letters. |
| + | M*SM=E where M is message SM is the Secret Matrix and E is encrypted message. | ||
| + | M=E*SM^-1 | ||
| + | |||
==2. Can Eve decrypt the message without finding the inverse of the secret matrix? == | ==2. Can Eve decrypt the message without finding the inverse of the secret matrix? == | ||
| − | + | She can find what the secret matrix is. She can right a system of equations and solve for each component of the secret message. | |
:<math>\begin{pmatrix} 1 & 0 & 4 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{pmatrix}\begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}=\begin{pmatrix} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{pmatrix}</math> | :<math>\begin{pmatrix} 1 & 0 & 4 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{pmatrix}\begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}=\begin{pmatrix} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{pmatrix}</math> | ||
:Multiply out | :Multiply out | ||
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:<math> a+g=0, b+h=0, c+i=0 \,</math> | :<math> a+g=0, b+h=0, c+i=0 \,</math> | ||
Solving These Equations yields the Secret Matrix | Solving These Equations yields the Secret Matrix | ||
| − | :<math>\begin{pmatrix} & 0 & 4 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{pmatrix}</math> | + | :<math>\begin{pmatrix} -2/3 & 0 & 4 \\ 0 & 1 & 0 \\ 2/3 & 0 & -1 \end{pmatrix}</math> |
| + | but finding the secret message does not help her she needs to find the inverse of this matrix. This can be done without inverting the SM. Do the same method above. | ||
| + | :<math>\begin{pmatrix} 1 & 0 & 4 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{pmatrix}=\begin{pmatrix} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{pmatrix}*\begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}</math> | ||
| + | :<math>2a=1,2b=0,2c=4 \,</math> | ||
| + | :<math>d=0,e=1,f=0 \, </math> | ||
| + | :<math>3g=1,3h=0,3i=1 \, </math> | ||
| + | Solving these Equations yeild SM^-1 | ||
| + | :<math>SM^{-1}=\begin{pmatrix} 1/2 & 0 & 2 \\ 0 & 1 & 0 \\ 1/3 & 0 & 1/3 \end{pmatrix}</math> | ||
| + | The answer to the question is yes and no. You need the inverse of the matrix to decrypt the matrix, but you do not need to find the inverse directly. You can solve the a system of equations like above. It turns out that this matrix is the inverse of the secret matrix. | ||
| + | |||
| + | ==3. What is the decrypted message corresponding to (2,23,3)? (Write it as a text.)== | ||
| + | <math>E*SM^{-1}=M \,</math> | ||
| + | :<math>\begin{pmatrix} 2 & 23 & 3\end{pmatrix}*\begin{pmatrix} 1/2 & 0 & 2 \\ 0 & 1 & 0 \\ 1/3 & 0 & 1/3 \end{pmatrix}=\begin{pmatrix} 2 & 23 & 5\end{pmatrix}</math> | ||
| + | "BWE" | ||
Latest revision as of 09:15, 18 September 2008
Contents
Part C: Application of linearity
1. How can Bob decrypt the message?
Bob can get the message by multiplying the Message by the Secret Matrix inverted then decoding the numbers into letters. M*SM=E where M is message SM is the Secret Matrix and E is encrypted message. M=E*SM^-1
2. Can Eve decrypt the message without finding the inverse of the secret matrix?
She can find what the secret matrix is. She can right a system of equations and solve for each component of the secret message.
- $ \begin{pmatrix} 1 & 0 & 4 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{pmatrix}\begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}=\begin{pmatrix} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{pmatrix} $
- Multiply out
- $ a+4g=2, b+4h=0, c+4i=0 \, $
- $ d=0, e=1, f=0 \, $
- $ a+g=0, b+h=0, c+i=0 \, $
Solving These Equations yields the Secret Matrix
- $ \begin{pmatrix} -2/3 & 0 & 4 \\ 0 & 1 & 0 \\ 2/3 & 0 & -1 \end{pmatrix} $
but finding the secret message does not help her she needs to find the inverse of this matrix. This can be done without inverting the SM. Do the same method above.
- $ \begin{pmatrix} 1 & 0 & 4 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{pmatrix}=\begin{pmatrix} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{pmatrix}*\begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} $
- $ 2a=1,2b=0,2c=4 \, $
- $ d=0,e=1,f=0 \, $
- $ 3g=1,3h=0,3i=1 \, $
Solving these Equations yeild SM^-1
- $ SM^{-1}=\begin{pmatrix} 1/2 & 0 & 2 \\ 0 & 1 & 0 \\ 1/3 & 0 & 1/3 \end{pmatrix} $
The answer to the question is yes and no. You need the inverse of the matrix to decrypt the matrix, but you do not need to find the inverse directly. You can solve the a system of equations like above. It turns out that this matrix is the inverse of the secret matrix.
3. What is the decrypted message corresponding to (2,23,3)? (Write it as a text.)
$ E*SM^{-1}=M \, $
- $ \begin{pmatrix} 2 & 23 & 3\end{pmatrix}*\begin{pmatrix} 1/2 & 0 & 2 \\ 0 & 1 & 0 \\ 1/3 & 0 & 1/3 \end{pmatrix}=\begin{pmatrix} 2 & 23 & 5\end{pmatrix} $
"BWE"
