s6q3 - A ± 4 1 5 18 ² = ± 21 26 22 18 ² Therefore A =...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
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:
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: A ± 4 1 5 18 ² = ± 21 26 22 18 ² Therefore A = ± 21 26 22 18 ² ± 4 1 5 18 ²-1 We find ± 4 1 5 18 ²-1 ± 4 1 5 18 ³ ³ ³ ³ 1 0 0 1 ² = ⇒ ± 1 0 0 1 ³ ³ ³ ³ 2 16 22 23 ² Thus A = ± 21 26 22 18 ² ± 2 16 22 23 ² = ± 5 6 5 12 ² 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 ² = ⇒ ± 1 0 12 23 0 1 24 5 ² thus A-1 = ± 12 23 24 5 ² To decipher multiply A-1 C = ± 12 23 24 5 ² ± 21 26 27 8 22 25 13 20 19 22 18 25 0 15 25 14 15 9 10 22 24 22 ² = ± 4 1 9 5 14 5 0 5 18 19 18 23 0 5 4 8 5 16 12 ² 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...
View Full Document

This note was uploaded on 12/20/2010 for the course MAT MATB24 taught by Professor X.jiang during the Spring '10 term at University of Toronto.

Page1 / 2

s6q3 - A ± 4 1 5 18 ² = ± 21 26 22 18 ² Therefore A =...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online