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exam_2_20F_solutions

exam_2_20F_solutions - Name Katie Student ID...

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Name Student ID TA/Section (circle): Katie 12pm-C01 1pm-C02 2pm-C03 Mary 3pm-C04 8pm-C05 9pm-C06 Math 20F, Winter 2010, Midterm Exam 2 Solutions Show all of your work and justify your answers to receive full credit. Write your answers and work clearly and legibly; no credit will be given for illegible solutions. Go back and check your answers if you finish early. # Points Score 1 6 2 3 3 3 4 2 5 2 6 2 7 2 Σ 20
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1. (6 points) Let A = 1 1 2 3 0 2 1 4 11 3 1 3 2 8 4 1 1 2 14 4 (a) Find a basis for Col A . (b) Find a basis for Row A . (c) Find a basis for Nul A . Answer: We need to row reduce to echelon form to determine a basis for Col A and Row A , and we need reduced echelon form to determine a basis for Nul A . 1 1 2 3 0 2 1 4 11 3 1 3 2 8 4 1 1 2 14 4 1 1 2 3 0 0 1 0 5 3 0 2 0 11 4 0 2 0 11 4 1 0 2 8 3 0 1 0 5 3 0 0 0 1 2 0 0 0 1 2 1 0 2 8 3 0 1 0 5 3 0 0 0 1 2 0 0 0 0 0 1 0 2 0 19 0 1 0 0 13 0 0 0 1 2 0 0 0 0 0 The pivots are circled. The pivot columns are columns 1, 2 and 4. The pivot columns of matrix A form a basis for Col A , so Basis for Col A = 1 2 1 1 , 1 1 3 1 , 3 11 8 14 The nonzero rows of an echelon form of A form a basis for Row A , so Basis for Row A = { (1 , 0 , 2 , 0 , 19) , (0 , 1 , 0 , 0 , 13) , (0 , 0 , 0 , 1 , 2) } To find the null space we look at x 1 + 2 x 3 + 19 x 5 = 0 x 2 + 13 x 5 = 0 + x 4 2 x 5 = 0 where x 3 and x 5 are free variables which leads to x 1 = 2 x 3 19 x 5 x 2 = 13 x 5 x 3 = x 3 = x = 2 x 3 0 x 3 0 0 + 19 x 5 13 x 5 0 2 x 5 x 5 = x 3 2 0 1 0 0 + x 5
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