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== Part C: Application of Linearity == | == Part C: Application of Linearity == | ||
| − | 1. Bob can decrypt the message by multiplying it (in groups of 3 numbers) by the inverse of the 3-by-3 secret matrix. | + | '''1.''' Bob can decrypt the message by multiplying it (in groups of 3 numbers) by the inverse of the 3-by-3 secret matrix. |
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| − | 2. No. <math>[Secret Message]*[Secret Matrix]=[Encrypted Message]\!</math>. Thus in order to find the secret message she must determine the inverse of the 3-by-3 secret matrix. I think. | + | '''2.''' No. <math>[Secret Message]*[Secret Matrix]=[Encrypted Message]\!</math>. Thus in order to find the secret message she must determine the inverse of the 3-by-3 secret matrix. I think. |
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| − | 3. In part 2 both a decrypted message and encrypted message are given. Thus that can be used to find the secret matrix: | + | '''3.''' In part 2 both a decrypted message and encrypted message are given. Thus that can be used to find the secret matrix: |
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<math>\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]\!</math> | <math>\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]\!</math> | ||
Revision as of 13:07, 18 September 2008
Part C: Application of Linearity
1. Bob can decrypt the message by multiplying it (in groups of 3 numbers) 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. I think.
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.
