M 340L Final Exam Solutions - Version 059 FINAL...

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Version 059 – FINAL – gilbert – (56540) 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points Find a basis for the Null space of the matrix A = 3 6 3 18 1 2 4 15 2 4 5 21 . 1. 2 1 0 0 , 3 0 3 1 2. 2 1 0 0 3. 3 0 3 1 4. 2 1 0 0 , 3 0 3 1 5. 2 1 0 0 , 3 0 3 1 correct 6. 3 0 3 1 Explanation: We first row reduce [ A 0 ]: rref([ A 0 ]) = 1 2 0 3 0 0 0 1 3 0 0 0 0 0 0 to identify the free variables for x in the ho- mogeneous equation A x = 0 . Thus x 1 and x 3 are basic variables, while x 2 and x 4 are free variables. So set x 2 = s and x 4 = t . Then x 1 = 2 s + 3 t, x 3 = 3 t , and Nul( A ) = Span 2 1 0 0 , 3 0 3 1 . Consequently, B = 2 1 0 0 , 3 0 3 1 is a basis for Nul( A ). 002 10.0points Find the distance from y to the plane in R 3 spanned by u 1 and u 2 when y = 5 9 5 , u 1 = 3 5 1 , u 2 = 3 2 1 . 1. dist = 6 2. dist = 4 3. dist = 2 5 4. dist = 2 10 correct 5. dist = 6 6. dist = 8 Explanation: The distance from a point y in R 3 to the subspace W = Span { u 1 , u 2 } is the distance bardbl y proj W y bardbl from y to the closest point, proj W y , in W . Now u 1 , u 2 are orthogonal because u 1 · u 2 = [ 3 5 1] 3 2 1 = 0 ,
Version 059 – FINAL – gilbert – (56540) 2 so proj W y = parenleftbigg y · u 1 bardbl u 1 bardbl 2 parenrightbigg u 1 + parenleftbigg y · u 2 bardbl u 2 bardbl 2 parenrightbigg u 2 = 35 35 u 1 28 14 u 2 = u 1 2 u 2 = 3 9 1 . Thus y proj W y = 5 9 5 3 9 1 = 2 0 6 . Consequently, the distance from y to W is bardbl y proj W y bardbl = 40 = 2 10 . 003 10.0points Find the Least Squares Regression line y = mx + b that best fits the data points ( 2 , 4) , ( 1 , 2) , (0 , 3) , (1 , 1) . 1. y = 8 5 x + 13 10 2. y = 13 10 x 8 5 3. y = 13 10 x + 8 5 4. y = 8 5 x 13 10 correct 5. y = 8 5 x + 13 10 6. y = 13 10 x 8 5 Explanation: The design matrix and list of observed val- ues for the data ( 2 , 4) , ( 1 , 2) , (0 , 3) , (1 , 1) . are given by A = 1 2 1 1 1 0 1 1 , b = 4 2 3 1 .

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