# praticemidterm - SCORE/xx Math 470 Communications and...

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SCORE/xx: Math 470 Communications and Cryptography NAME: PRACTICE MIDTERM I (more problems and solutions to be posted soon...) Please show your work and write only in pen . Notes are forbidden. Calculators, and all other electronic devices, are forbidden. Brains are encouraged, but at most one (your own, please!) may be used per exam. 1: Please solve the following linear system (modulo 26) for x and y : 5 x + 2 y = 3 7 x + y = 7 The safest thing to do when solving any equation is to first see if there are any solutions at all. For linear systems mod an integer, division is a pain, so it’s helpful to use a method that allows you to use a smaller number of divisions. For 2 × 2 systems, this is easy via the adjoint formula for matrix inverse. In particular, we can rewrite our system in terms of matrices as follows: bracketleftbigg 5 2 7 1 bracketrightbiggbracketleftbigg x y bracketrightbigg = bracketleftbigg 3 7 bracketrightbigg and then multiply both sides by the inverse of the matrix (assuming it’s invertible): bracketleftbigg x y bracketrightbigg = bracketleftbigg 5 2 7 1 bracketrightbigg - 1 bracketleftbigg 3 7 bracketrightbigg Since det bracketleftbigg 5 2 7 1 bracketrightbigg = 9 has no non-trivial factor in common with 26, we can then compute bracketleftbigg 5 2 7 1 bracketrightbigg - 1 = 1 9 bracketleftbigg 1 2 7 5 bracketrightbigg via the usual 2 × 2 matrix inverse formula. (This formula has the advantage of needing only 1 reciprocal.) By hand (or a lucky guess, or the Extended Euclidean Algorithm) we see that ( 9)( 3) 1 mod 26. So then bracketleftbigg 5 2 7 1 bracketrightbigg - 1 = 3 bracketleftbigg 1 2 7 5 bracketrightbigg = bracketleftbigg 3 6 21 15 bracketrightbigg = bracketleftbigg 23 6 21 11 bracketrightbigg and thus bracketleftbigg x y bracketrightbigg = bracketleftbigg 23 6 21 11 bracketrightbiggbracketleftbigg 3 7 bracketrightbigg = bracketleftbigg 69 + 42 63 + 77 bracketrightbigg = bracketleftbigg 111 140 bracketrightbigg = bracketleftbigg 19 10 bracketrightbigg = bracketleftbigg 7 10 bracketrightbigg .