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Unformatted text preview: Name: ID#: Solutions to Midterm III Math 20 Introduction to Multivariable Calculus and Linear Algebra May 5, 2006 Rules: This is a onehour exam. Calculators are not allowed. Unless otherwise stated, show all of your work. Full credit may not be given for an answer alone. You may use the backs of the pages or the extra pages for scratch work. Do not unstaple or remove pages as they can be lost in the grading process. Please do not put your name on any page besides the first page. If you like, you may put your ID number on the top of each page you write on. Hints: Read the entire exam to scan for obvious typos or questions you might have. Budget your time so that you dont run out. Problems may stretch across several pages. Relax and do well! Students who, for whatever reason, submit work not their own will ordinarily be re quired to withdraw from the College. Handbook for Students Summary Data Problem1 Problem2 Problem3 Problem4 Problem5 Problem6 Total Percent MaximumPossible 13 10 8 9 10 10 60 100.00% MaximumAchieved 13 10 8 9 10 10 59 98.33% Mean 10.81 9.78 3.58 8.42 7.67 3.69 43.94 73.24% Median 12 10 4 9 8 3 44.5 74.17% Mode 13 10 4 9 7 2 43 71.67% %fullcredit 23% 80% 10% 77% 10% 0% 0% 0.00% %nocredit 0% 0% 13% 0% 0% 3% 0% 0.00% StandardDeviation 2.6228 0.4779 2.4195 1.2775 2.1985 2.5584 6.6789 11.13% CorrelationwithTotal 0.6813 0.0571 0.5332 0.3120 0.7063 0.6346 1.0000 100.00% 1 Math 20 Solutions to Midterm III May 5, 2006 1 1. (13 Points) (i) (3 points) Let A = 10 4 24 10 Find the eigenvalues of A. Solution. The characteristic polynomial of A is ( 10 )( 10 )+ 24 ( 4 ) = 2 4 , so the eigenvalues are 2. N (ii) (3 points) Let B = 2 2 5 4 3 6 5 Is 1 an eigenvalue of B? Solution. 1 is an eigenvalue of B if B I is not invertible. This would be indicated by a row of zeros or a nonpivot column in the reduced row echelon form of B I . B I = 1 2 5 3 3 6 6 1 3 1 1 So 1 is an eigenvalue of B . Alternatively, one may find the characteristic polynomial of B , which is 3 + 2 + 4  4. Plugging in 1 to this polynomial gives zero, so 1 is an eigenvalue of B . N Finding the characteristic polynomial is the harder of the two ways to do this problem. Many mis takes were made along this route because of arithmetic errors in the determinant. Also, once you have the characteristic polynomial, testing that 1 is a root by plugging it in is much easier than factoring the polynomial and seeing if (  1 ) is a factor. (iii) (7 points) Let C = 1 1 1 3 2 4 3 The eigenvalues of C are 1 and 1 . Is C diagonalizable?...
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This note was uploaded on 11/10/2009 for the course MATH 1850 taught by Professor Mihaibeligan during the Spring '09 term at UOIT.
 Spring '09
 MihaiBeligan
 Linear Algebra, Algebra, Multivariable Calculus

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