# prac02s - 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 2 — Solutions 1. Prove that the set S = braceleftBigparenleftBig x y z parenrightBig ∈ R 3 | 4 x − y − 5 z = 0 bracerightBig is a subspace of R 3 . Solution S is clearly non-empty, since parenleftBig parenrightBig satisfies the equation 4 x − y − 5 z = 0 and is therefore in S . Now suppose that parenleftBig a 1 a 2 a 3 parenrightBig and parenleftBig b 1 b 2 b 3 parenrightBig both belong to S . That is, 4 a 1 − a 2 − 5 a 3 = 0 and 4 b 1 − b 2 − 5 b 3 = 0 . Then 4( a 1 + b 1 ) − ( a 2 + b + 2) − 5( a 3 + b 3 ) = 0 , and so parenleftBig a 1 + b 1 a 2 + b 2 a 3 + b 3 parenrightBig ∈ S , and S is closed under addition. If k ∈ R , then 4( ka 1 ) − ( ka 2 ) − 5( ka 3 ) = k (4 a 1 − a 2 − 5 a 3 ) = 0 , and so parenleftBig ka 1 ka 2 ka 3 parenrightBig ∈ S , and S is closed under scalar multiplication. 2. The subspace Span( X ) of R 3 is a plane for the two choices of X ⊂ R 3 given below. In both cases find the equation of this plane. a ) X = braceleftBigparenleftBig 1 1 − 1 parenrightBig , parenleftBig 2 3 5 parenrightBigbracerightBig . b ) X = braceleftBigparenleftBig 1 1 − 2 parenrightBig , parenleftBig 2 3 5 parenrightBigbracerightBig . Solution a ) The vector parenleftBig x y z parenrightBig belongs to Span ( X ) if and only if a parenleftBig 1 1 − 1 parenrightBig + b parenleftBig 2 3 5 parenrightBig = parenleftBig x y z parenrightBig for some a, b ∈ R . Applying Gaussian elimination to the corresponding augmented matrix we obtain parenleftBigg 1 2 x 1 3 y − 1 5 z parenrightBigg R 2 := R 2 − R 1 R 3 := R 3 + R 1 −−−−−−−→ parenleftBigg 1 2 x 1 y − x 7 x + z parenrightBigg R 3 := R 3 − 7 R 2 −−−−−−−→ parenleftBigg 1 2 x 1 y − x 8 x − 7 y + z parenrightBigg This system of equations has a solution if and only if 8 x − 7 y + z = 0 . That is, parenleftBig x y z parenrightBig ∈ Span( X ) if and only if parenleftBig x y z parenrightBig lies on the plane 8 x − 7 y + z = 0 in R 3 . So Span( X ) is equal to the plane 8 x − 7 y + z = 0 . Note that this plane goes through the origin. b ) The vector parenleftBig x y z parenrightBig ∈ Span( X ) if and only if a parenleftBig 1 1 − 2 parenrightBig + b parenleftBig 2 3 5 parenrightBig = parenleftBig x y z parenrightBig , Math 2061: Practice session 2 — Solutions A.M. 5/1/2012 Linear Mathematics Practice session 2 — Solutions Page 2 for some a, b ∈ R . Applying Gaussian elimination we get parenleftBigg 1 2 z 1 3 y − 2 5 z parenrightBigg R 2 := R 2 − R 1 R 3 := R 3 +2 R 1 −−−−−−−→ parenleftBigg 1 2 z 1 y − x 9 2 x + 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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prac02s - THE UNIVERSITY OF SYDNEY PURE MATHEMATICS Linear...

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