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Unformatted text preview: Solutions to problems from section 2.9 2. Find a vector x determined by the coordinate vector [ x ] B = bracketleftbigg − 1 3 bracketrightbigg and the basis B = braceleftbiggbracketleftbigg − 2 1 bracketrightbigg , bracketleftbigg 3 1 bracketrightbiggbracerightbigg . Illustrate your answer with a figure. Solution: Let b 1 = bracketleftbigg − 2 1 bracketrightbigg and b 2 = bracketleftbigg 3 1 bracketrightbigg . Then we have [ x ] B = bracketleftbigg − 1 3 bracketrightbigg ⇒ x = − b 1 + 3 b 2 = bracketleftbigg 11 2 bracketrightbigg Here’s a picture: 4. Given that x = bracketleftbigg − 7 5 bracketrightbigg is in the subspace H with basis B = { b 1 , b 2 } where b 1 = bracketleftbigg 1 − 3 bracketrightbigg and b 2 = bracketleftbigg − 3 5 bracketrightbigg , find the Bcoordinates of x . Solution: We need to write x as a linear combination of b 1 and b 2 , so let’s reduce the appropriate augmented matrix: bracketleftbigg 1 − 3 − 7 − 3 5 5 bracketrightbigg NR 2 = R 2 +3 R 1 −→ bracketleftbigg 1 − 3 − 7 − 4 − 16 bracketrightbigg NR 2 = 1 4 R 2 −→ bracketleftbigg 1 − 3 − 7 1 4 bracketrightbigg NR 1 = R 1 +3 R 2 −→ bracketleftbigg 1 0 5 0 1 4 bracketrightbigg . Thus x = 5 b 1 + 4 b 2 so that [ x ] B = bracketleftbigg 5 4 bracketrightbigg ....
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 Fall '08
 AXENOVICH
 Linear Algebra, Vector Space, basis, NulA

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