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# tut01s - THE UNIVERSITY OF SYDNEY PURE MATHEMATICS Linear...

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Unformatted text preview: THE UNIVERSITY OF SYDNEY PURE MATHEMATICS Linear Mathematics 2012 Tutorial 1 — Solutions 1. a ) Calculate the matrix product 3 0 4 1 1 2 − 1 3 5 5 − 2 4 . b ) Hence find the solution to the system of linear equations that corresponds to the augmented matrix 3 0 4 31 1 1 2 11 − 1 3 5 9 . Solution a ) 3 0 4 1 1 2 − 1 3 5 5 − 2 4 = 31 11 9 . b ) Let A = parenleftBig 3 0 4 1 1 2- 1 3 5 parenrightBig . Then we want to find x = parenleftBig x 1 x 2 x 3 parenrightBig such that A x = parenleftBig 31 11 9 parenrightBig . From part (a) x = 5 − 2 4 is a solution. The question, then, is whether or not this is the only solution. One way to see that it is is to evaluate det A . We have det A = 13 negationslash = 0 (check!). Hence A is invertible, and the system has the unique solution x = 5 − 2 4 . We could, of course, find the solution using Gaussian elimination: 3 0 4 31 1 1 2 11 − 1 3 5 9 R 1 ↔ R 2 −−−−→ 1 1 2 11 3 0 4 31 − 1 3 5 9 R 2 := R 2- 3 R 1 −−−−−−−→ 1 1 2 11 − 3 − 2 − 2 − 1 3 5 9 R 3 := R 3 + R 1 −−−−−−−→ 1 1 2 11 − 3 − 2 − 2 4 7 20 R 2 := R 2 + R 1 −−−−−−−→ 1 1 2 11 0 1 5 18 0 4 7 20 R 3 := R 3- 4 R 2 −−−−−−−→ 1 1 2 11 0 1 5 18 0 0 − 13 − 52 R 3 :=- 1 13 R 3 −−−−−−−→ 1 1 2 11 0 1 5 18 0 0 1 4 Therefore x 3 = 4 , x 2 = 18 − 5 x 3 = − 2 , x 1 = 11 − x 2 − 2 x 3 = 5 . Using back substitution the unique solution is parenleftBig 5- 2 4 parenrightBig , as before. 2. Each of the following matrices is the reduced row echelon form of an augmented matrix be- longing to a system of linear equations in the variables x i , ( i = 1 , 2 , . . . ) . (Both the systems represented here have infinitely many solutions – why?) For each augmented matrix (i) determine the number of parameters needed to solve the system and (ii) express the solution of the system in parametric form. Math 2061: Tutorial 1 — Solutions A.M. 5/1/2012 Linear Mathematics Tutorial 1 — Solutions Page 2 a ) parenleftBigg 1 4 1 − 5 − 1 parenrightBigg b ) parenleftBigg 1 2 6 1 5 1 − 1 parenrightBigg Solution a ) i ) The variable x 3 is a parameter....
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tut01s - THE UNIVERSITY OF SYDNEY PURE MATHEMATICS Linear...

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