s6q3 - A 4 1 5 18 = 21 26 22 18 Therefore A = 21 26 22 18 4...

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University of Toronto at Scarborough MAT B24S Fall 2007 Solution 6: Question 3 1) To break a code given the message: UVZR.YH VOYYMNTOSIJXVV We get the numerical equivalents and form vectors with 2 coordinates each. UV ZR .Y H V O Y Y MN TO SI 21 , 22 26 , 18 27 , 25 8 , 0 22 , 15 25 , 25 13 , 14 20 , 15 19 , 9 JX V V 10 , 24 22 , 22 We place the above vectors in R 2 as columns of a matrix in the order in which they appear. Thus we form the matrix: C = ± 21 26 27 8 22 25 13 20 19 22 18 25 0 15 25 14 15 9 10 22 24 22 ² If the first word is DEAR then the numerical equivalent to it is DE AR 4 , 5 1 , 18 The enciphering matrix, A , can be thought of as the matrix representation of a linear transformation T relative to the standard basis. Thus T ( ± 4 5 ² ) = ± 21 22 ² and T ( ± 1 18 ² ) = ± 26 18 ² Easier Method: If T is a linear transformation with standard matrix representation A , then we have: A ± 4 5 ² = ± 21 22 ² and A ± 1 18 ² = ± 26 18 ² . Equivalently:
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Unformatted text preview: A 4 1 5 18 = 21 26 22 18 Therefore A = 21 26 22 18 4 1 5 18 -1 We nd 4 1 5 18 -1 4 1 5 18 1 0 0 1 = 1 0 0 1 2 16 22 23 Thus A = 21 26 22 18 2 16 22 23 = 5 6 5 12 This is the enciphering matrix. To decipher we need its inverse which is found by row reducing 5 6 1 0 5 12 0 1 = 1 0 12 23 0 1 24 5 thus A-1 = 12 23 24 5 To decipher multiply A-1 C = 12 23 24 5 21 26 27 8 22 25 13 20 19 22 18 25 0 15 25 14 15 9 10 22 24 22 = 4 1 9 5 14 5 0 5 18 19 18 23 0 5 4 8 5 16 12 Thus, we get the numbers and the text: 4 , 5 1 , 18 0 , 19 9 , 18 0 , 23 5 , 0 14 , 5 5 , 4 0 , 8 DE AR S IR W E NE ED H 5 , 12 160 EL P The message is: DEAR SIR WE NEED HELP...
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s6q3 - A 4 1 5 18 = 21 26 22 18 Therefore A = 21 26 22 18 4...

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