# prac01s - THE UNIVERSITY OF SYDNEY PURE MATHEMATICS Linear...

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Unformatted text preview: THE UNIVERSITY OF SYDNEY PURE MATHEMATICS Linear Mathematics 2012 Practice session 1 — Solutions 1. Find all numbers a , b and c such that a ) a parenleftBig 1 4 7 parenrightBig + b parenleftBig 2 5 8 parenrightBig + c parenleftBig 3 6 9 parenrightBig = parenleftBig 1- 2 1 parenrightBig ; b ) a parenleftBig 1 4 7 parenrightBig + b parenleftBig 2 5 8 parenrightBig + c parenleftBig 3 6 3 parenrightBig = parenleftBig 1- 2 1 parenrightBig . Solution a ) This matrix equation is the same as the following system of simultaneous equations: a + 2 b + 3 c = 1 4 a + 5 b + 6 c =- 2 7 a + 8 b + 9 c = 1 . We use Gaussian elimination to reduce the augmented matrix to echelon form: parenleftBigg 1 2 3 1 4 5 6- 2 7 8 9 1 parenrightBigg R 2 := R 2- 4 R 1-------→ parenleftBigg 1 2 3 1- 3- 6- 6 7 8 9 1 parenrightBigg R 3 := R 3- 7 R 1-------→ parenleftBigg 1 2 3 1- 3- 6- 6- 6- 12- 6 parenrightBigg R 2 :=- 1 3 R 2------→ parenleftBigg 1 2 3 1 1 2 2- 6- 12- 6 parenrightBigg R 3 :=- 1 6 R 3------→ parenleftBigg 1 2 3 1 1 2 2 1 2 1 parenrightBigg R 3 := R 3- R 2-------→ parenleftBigg 1 2 3 1 1 2 2- 1 parenrightBigg The last line reads a + 0 b + 0 c =- 1 , so the original equation has no solution. b ) This matrix equation is the same as the following system of simultaneous equations: a + 2 b + 3 c = 1 4 a + 5 b + 6 c =- 2 7 a + 8 b + 3 c = 1 . We use Gaussian elimination to reduce the augmented matrix to echelon form: parenleftBigg 1 2 3 1 4 5 6- 2 7 8 3 1 parenrightBigg R 2 := R 2- 4 R 1-------→ parenleftBigg 1 2 3 1- 3- 6- 6 7 8 3 1 parenrightBigg R 3 := R 3- 7 R 1-------→ parenleftBigg 1 2 3 1- 3- 6- 6- 6- 18- 6 parenrightBigg R 2 :=- 1 3 R 2------→ parenleftBigg 1 2 3 1 1 2 2- 6- 18- 6 parenrightBigg R 3 :=- 1 6 R 3------→ parenleftBigg 1 2 3 1 1 2 2 1 3 1 parenrightBigg R 3 := R 3- R 2-------→ parenleftBigg 1 2 3 1 1 2 2 1- 1 parenrightBigg By back-substitution, the solution is a =- 4 , b = 4 and c =- 1 . 2. Each of the following matrices is the reduced row echelon form of an augmented matrix belong- ing to a system of linear equations in the variables x i , ( i = 1 , 2 , . . . ) . (Both systems represented here have infinitely many solutions – why?) For each augmented matrix below (i) determine the number of parameters needed to solve the system and (ii) express the solution of the system in parametric form. Math 2061: Practice session 1 — Solutions A.M. 5/1/2012 Linear Mathematics Practice session 1 — Solutions Page 2 a ) parenleftBigg 1 3 4 1- 2- 1 parenrightBigg b ) parenleftBigg 1 2- 1 3 1- 1 2 4 1 2- 1 parenrightBigg Solution The linear systems represented by each of these augmented matrices are consistent, and have infinitely many solutions because, in each case, there is at least one column (corre- sponding to a free variable), which is missing a leading 1.sponding to a free variable), which is missing a leading 1....
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## This note was uploaded on 02/06/2012 for the course MATH 2061 taught by Professor Notsure during the Three '09 term at University of Sydney.

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prac01s - THE UNIVERSITY OF SYDNEY PURE MATHEMATICS Linear...

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