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1. How can Bob decrypt the message? | 1. How can Bob decrypt the message? | ||
Bob can use matrix algebra to find the inverse of the 3-by-3 matrix, which he can then multiply the encrypted vector by in order to obtain the 1-by-3 decrypted vector that corresponds to the numerical place of each letter of the alphabet. | Bob can use matrix algebra to find the inverse of the 3-by-3 matrix, which he can then multiply the encrypted vector by in order to obtain the 1-by-3 decrypted vector that corresponds to the numerical place of each letter of the alphabet. | ||
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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? | ||
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3. What is the decrypted message corresponding to (2,23,3)? (Write it as a text.) | 3. What is the decrypted message corresponding to (2,23,3)? (Write it as a text.) | ||
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| + | By solving the system equations given by the vector of decrypted numbers and vector of encrypted numbers one can derive the original matrix [A] where | ||
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| + | a11 = -2/3 | ||
| + | a12 = 0 | ||
| + | a13 = 2/3 | ||
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| + | a21 = 0 | ||
| + | a22 = 1 | ||
| + | a23 = 0 | ||
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| + | a31 = 4 | ||
| + | a32 = 0 | ||
| + | a33 = -1 | ||
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| + | which can be used as matrix [A] in matlab to find the inverse matrix [B] which, when multiplied with the encrypted vector [2, 23, 3] yields | ||
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| + | A = | ||
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| + | -0.6667 0 0.6667 | ||
| + | 0 1.0000 0 | ||
| + | 4.0000 0 -1.0000 | ||
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| + | B = | ||
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| + | 0.5000 0 0.3333 | ||
| + | 0 1.0000 0 | ||
| + | 2.0000 0 0.3333 | ||
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| + | Message = | ||
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| + | 2 | ||
| + | 23 | ||
| + | 5 | ||
| + | |||
| + | which corresponds to the letters of the alphabet B W and E. | ||
| + | |||
| + | Answer = BWE | ||
Latest revision as of 14:25, 19 September 2008
1. How can Bob decrypt the message?
Bob can use matrix algebra to find the inverse of the 3-by-3 matrix, which he can then multiply the encrypted vector by in order to obtain the 1-by-3 decrypted vector that corresponds to the numerical place of each letter of the alphabet.
2. Can Eve decrypt the message without finding the inverse of the secret matrix?
Yes. Eve can solve the system of equations with the vectors she has in order to decrypt the message.
3. What is the decrypted message corresponding to (2,23,3)? (Write it as a text.)
By solving the system equations given by the vector of decrypted numbers and vector of encrypted numbers one can derive the original matrix [A] where
a11 = -2/3 a12 = 0 a13 = 2/3
a21 = 0 a22 = 1 a23 = 0
a31 = 4 a32 = 0 a33 = -1
which can be used as matrix [A] in matlab to find the inverse matrix [B] which, when multiplied with the encrypted vector [2, 23, 3] yields
A =
-0.6667 0 0.6667
0 1.0000 0
4.0000 0 -1.0000
B =
0.5000 0 0.3333
0 1.0000 0
2.0000 0 0.3333
Message =
2
23
5
which corresponds to the letters of the alphabet B W and E.
Answer = BWE
