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Unformatted text preview: Solutions to Final Examination Math 20 Introduction to Multivariable Calculus and Linear Algebra May 19, 2006 Summary Data Problem 1 2 3 4 5 6 7 Max Possible 9 9 10 10 10 10 10 Max Achieved 9 9 10 10 10 10 10 Mean 8.89 8.75 9.03 9.44 9.69 9.81 9.22 Median 9.0 9.0 9.0 10.0 10.0 10.0 10.0 Mode 9 9 9 10 10 10 10 % full credit 90% 93% 40% 80% 90% 87% 70% % no credit 0% 0% 0% 0% 0% 0% 0% Standard Deviation 0.3928 1.0897 0.9570 1.8325 0.9667 0.4606 1.8725 Correlation with Total 0.0773 0.7614 0.6707 0.4646 0.6925 0.5538 0.7940 Problem 8 9 10 11 12 Total Percent Max Possible 10 10 12 10 10 120 100.00% Max Achieved 10 10 12 10 7 117 97.50% Mean 8.94 8.89 8.83 5.14 3.83 100.47 83.73% Median 9.0 10.0 10.0 5.0 4.0 102.5 85.42% Mode 10 10 12 5 6 107 89.17% % full credit 37% 63% 23% 13% 0% 0% 0% % no credit 0% 0% 3% 7% 13% 0% 0% Standard Deviation 0.9985 1.7760 3.7896 2.6890 2.0480 11.7933 9.83% Correlation with Total 0.3349 0.7677 0.7227 0.5918 0.4886 1.0000 1.0000 1 Math 20 Solutions to Final Examination May 19, 2006 1 1. (9 Points) Let A = 1 2 1 3 B = 1 2 3 1 2 Find (i) BA Solution. The answer is BA =  1 3 1 13 3 4 . N (ii) AB T Solution. The answer is AB T = 2 8 3 3 7 7 . N (iii) A 2 Solution. The answer is A 2 = 1 8 4 7 . N –1– 2 Math 20 Solutions to Final Examination May 19, 2006 2 2. (9 Points) Let v = 1 1 w = 1 1 Find (a) v 3 w Solution. The answer is  3 2 1 . N (b) v · w Solution. The answer is 0 · 1 + 1 · 1 + 1 · = 1. N (c) k w k . Solution. The answer is p 1 2 + 1 2 + 2 = √ 2. N –2– 3 Math 20 Solutions to Final Examination May 19, 2006 3 3. (10 Points) I. (6 points) Let A = 1 1 2 4 1 1 1 1 2 2 Find the reduced row echelon form of A. Solution. We have A = 1 1 2 4 1 1 1 1 2 2  · ( 1 ) ← + 1 1 2 4 1 1 2 2  · 1 2 1 1 1 2 1 1 2 2  · ( 1 ) ← + 1 1 1 2 1 2  · ( 1 ) ← + 1 1 1 2 1  · ( 1 ) ← + N (continued) –3– 3 Math 20 Solutions to Final Examination May 19, 2006 3 II. (4 points) Suppose the system of linear equations A x = b has an augmented matrix whose reduced row echelon form is 1 2 1 4 1 1 1 2 1 2 3 Find the parametric form to the general solution. Solution. We have five variables: x 5 is free x 4 = 3 2 x 5 x 3 is free x 2 = 2 x 5 + x 3 x 1 = 4 + x 4 2 x 3 So x 1 x 2 x 3 x 4 x 5 = 4 2 x 3 + x 5 2 x 5 + x 3 x 3 3 2 x 5 x 5 = 4 2 3 + x 3  2 1 1 + x 5 1 1 2 1 ....
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 Spring '09
 MihaiBeligan
 Linear Algebra, Algebra, Multivariable Calculus, Correlation, Standard Deviation, A. Solution

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