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Applying the method from part 2, we know that: | Applying the method from part 2, we know that: | ||
<math>[2,23,3]=a[2,0,0] + b[0,1,0] + c[0,0,3]</math> | <math>[2,23,3]=a[2,0,0] + b[0,1,0] + c[0,0,3]</math> | ||
| − | so | + | so |
| + | |||
<math>2=2a</math>, <math>23=1b</math>, and <math>3=3c</math>. | <math>2=2a</math>, <math>23=1b</math>, and <math>3=3c</math>. | ||
therefore, a=1, b=23, and c=1. | therefore, a=1, b=23, and c=1. | ||
Revision as of 17:51, 18 September 2008
Part 1
Bob can decrypt the message by using the inverse of the secret matrix. Multiplying the encrypted vector three at a time by the inverse matrix will yield the original vector.
Part 2
Eve can decrypt the message without finding the secret matrix because the system in linear. The three vectors of the encrypted vector form a basis for all vectors because they are linearly independent. In other words, [encrypted vector]$ =a[2,0,0] + b[0,1,0] + c[0,0,3] $ where a,b,c are real numbers.
Part 3
Applying the method from part 2, we know that: $ [2,23,3]=a[2,0,0] + b[0,1,0] + c[0,0,3] $ so
$ 2=2a $, $ 23=1b $, and $ 3=3c $. therefore, a=1, b=23, and c=1.
Now apply the same coefficients to the known message that yielded the encrypted basis, [1,0,4,0,1,0,1,0,1].
$ 1[1,0,4] + 23[0,1,0] + 1[1,0,1]=[1+1,23,4+1]=[2,23,5] $
[2,23,5] corresponds to BWE.
