This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: M346 First Midterm Exam Solutions, September 18, 2009 1) In R 2 , let E = braceleftbiggparenleftbigg 1 parenrightbiggparenleftbigg 1 parenrightbiggbracerightbigg be the standard basis and let B = braceleftbiggparenleftbigg 2 3 parenrightbigg , parenleftbigg 5 7 parenrightbiggbracerightbigg be an alternate basis. a) Find P EB and P BE . P EB = ([ b 1 ] E [ bb 2 ] E ) = parenleftbigg 2 5 3 7 parenrightbigg . P BE = P 1 EB = parenleftbigg 7 5 3 2 parenrightbigg . Here we used the fact that parenleftbigg a b c d parenrightbigg 1 = 1 ad bc parenleftbigg d b c a parenrightbigg . b) If v = parenleftbigg 4 1 parenrightbigg , find [ v ] B . Since [ v ] E = parenleftbigg 4 1 parenrightbigg , [ v ] B = P BE [ v ] E = parenleftbigg 23 10 parenrightbigg . c) Solve the system of equations: 2 x 1 + 5 x 2 = 4; 3 x 1 + 7 x 2 = 1. This is the exact same problem as (b), namely writing parenleftbigg 4 1 parenrightbigg as a linear combination of parenleftbigg 2 3 parenrightbigg and parenleftbigg 5 7 parenrightbigg . The solution, as before, is x 1 = 23 , x 2 = 10 ....
View
Full Document
 Spring '08
 RAdin
 Linear Algebra, Algebra, Vector Space, basis, Arref

Click to edit the document details