This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 ....
View
Full
Document
This note was uploaded on 01/28/2010 for the course MATH 307 taught by Professor Axenovich during the Fall '08 term at Iowa State.
 Fall '08
 AXENOVICH

Click to edit the document details