# tut03s - THE UNIVERSITY OF SYDNEY PURE MATHEMATICS Linear...

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Unformatted text preview: THE UNIVERSITY OF SYDNEY PURE MATHEMATICS Linear Mathematics 2012 Tutorial 3 — Solutions 1. Determine whether the following sets are linearly independent, and whether or not they span R 3 . a ) X = braceleftBigparenleftBig 1 parenrightBig , parenleftBig 1 parenrightBig , parenleftBig 1 parenrightBigbracerightBig b ) Y = braceleftBigparenleftBig 1 2 3 parenrightBig , parenleftBig 4 5 6 parenrightBig , parenleftBig 1 2- 3 parenrightBigbracerightBig Solution a ) This is the standard basis of R 3 . It is linearly independent and it spans R 3 . b ) Let A = parenleftBig 1 4 1 2 5 2 3 6- 3 parenrightBig . This matrix reduces to parenleftBig 1 4 1 0 1 0 0 0 1 parenrightBig . Hence the only solution to the equation a parenleftBig 1 2 3 parenrightBig + b parenleftBig 4 5 6 parenrightBig + c parenleftBig 1 2- 3 parenrightBig = is a = b = c = 0 . That is, the vectors in Y are linearly independent. Similarly, there is a solution to parenleftBig x y z parenrightBig = a parenleftBig 1 2 3 parenrightBig + b parenleftBig 4 5 6 parenrightBig + c parenleftBig 1 2- 3 parenrightBig for every parenleftBig x y z parenrightBig ∈ R 3 , so Y spans R 3 . Note that this means that X is a basis of R 3 . 2. Find the column space of A = parenleftBig 1 4 1 1 2 5 1 0 3 6 1- 1 parenrightBig . Do the columns of A form a basis for R 3 ? Solution Use Gaussian elimination to check whether or not an arbitrary vector parenleftBig x y z parenrightBig belongs to Col( A ) : parenleftBigg 1 4 1 1 x 2 5 1 y 3 6 1 − 1 z parenrightBigg Row reduce −−−−−→ parenleftBigg 1 4 1 1 x 1 1 3 2 3 1 3 (2 x − y ) x − 2 y + z parenrightBigg ....
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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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tut03s - THE UNIVERSITY OF SYDNEY PURE MATHEMATICS Linear...

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