(New page: '''1)''' Bob can decrypt the message by multiplying the encrypted message by the inverse of the 3x3 matrix that his friend gave him, if it exists. '''2)'''Yes she can! Reason being that ...) |
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Therefore, <math>a*X + b*\bar{X} \to a*Y + b*\bar{Y}</math> where <math>a, b \in \mathbb{C}</math> | Therefore, <math>a*X + b*\bar{X} \to a*Y + b*\bar{Y}</math> where <math>a, b \in \mathbb{C}</math> | ||
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| + | '''3)'''First, one has to find the matrix that was used in encoding. To find this, we take input '''<big><math>(1,0,4,0,1,0,1,0,1)</math></big>''' and turn it into matrix <math>\begin{Bmatrix}1 & 0 & 4\\ 0 & 1 & 0\\ 1 & 0 & 1\end{Bmatrix}</math>. By multiplying the previous matrix with matrix <math>\begin{Bmatrix}a \\ b\\ c\end{Bmatrix}</math> yields the output matrix <math>\begin{Bmatrix}2 &0&0\\0&1&0\\0&0&3\end{Bmatrix}</math>. | ||
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| + | <math>\begin{Bmatrix}a \\ b\\ c\end{Bmatrix} * \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}</math> which yields three equations with three unknowns. | ||
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| + | Solving for coefficients '''<big>a, b, c</big>''' | ||
Revision as of 16:22, 19 September 2008
1) Bob can decrypt the message by multiplying the encrypted message by the inverse of the 3x3 matrix that his friend gave him, if it exists.
2)Yes she can! Reason being that this is a linear system in which input $ X $ yields output $ Y $ and input $ \bar{X} $ yields output $ \bar{Y} $.
Therefore, $ a*X + b*\bar{X} \to a*Y + b*\bar{Y} $ where $ a, b \in \mathbb{C} $
3)First, one has to find the matrix that was used in encoding. To find this, we take input $ (1,0,4,0,1,0,1,0,1) $ and turn it into matrix $ \begin{Bmatrix}1 & 0 & 4\\ 0 & 1 & 0\\ 1 & 0 & 1\end{Bmatrix} $. By multiplying the previous matrix with matrix $ \begin{Bmatrix}a \\ b\\ c\end{Bmatrix} $ yields the output matrix $ \begin{Bmatrix}2 &0&0\\0&1&0\\0&0&3\end{Bmatrix} $.
$ \begin{Bmatrix}a \\ b\\ c\end{Bmatrix} * \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} $ which yields three equations with three unknowns.
Solving for coefficients a, b, c
