Old Math 133 Final Exams
Please note that the course content varies from year to year. Some of the problems in these exams involve material that
we have not covered this year.
- April 2004 Exam.
- December 2004 Exam.
- December 2005 Exam.
Answers to December 2005 exam: 1(C), 2(E), 3(B), 4(D), 5(A), 6(C), 7(E), 8(D), 9(B), 10(C).
# 11 (a) n=(1,-1,1) (b) distance = 3*sqrt(3)
# 12 (a) A= [4/13 , -6/13 // -6/13, 9/13], (here the notation // means "begin a new row")
# 12 (b) v1= (3,2), v2= (-2,3), have eigenvalues -1 and 1 resp.
# 12 (c) A=[-5/13, -12/13 // -12/13, 5/13]. (here the notation // means "begin a new row")
# 13 (a) Let c1w1+c2w2+c3w3 = 0. Substitute for the w's. Get (c1+c2+c3)v1+(-c1+c3)v2+(c1-c2+c3)v3 = 0.
Therefore 0=(c1+c2+c3), 0=(-c1+c3), 0=(c1-c2+c3). Solve this system to show that its only
solution is c1=0, c2=0, c3=0.
# 13 (b) Find a nontrivial set (k1,k2,k3) of constants such that k1a1+k2a2+k3a3=0. (By substituting
for the a's and immitating part (a). One nontrivial solution is k1=3, k2=-6, k3=-2.)
# 14 (a) P=(1/sqrt(5))*[2, 1 // -1, 2], D=[-3, 0 // 0, 2]. (here the notation // means "begin a new row")
# 14 (b) m=-3, M=2
# 14 (c) m is achieved when X= (+/-)(1/sqrt(5)(2, -1).
M is achieved when X= (+/-)(1/sqrt(5)(1, 2).
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December 2007 Exam with Answers provided by Professor Labute.
There are a few typos in the solutions:
- 4(b), the final line should say k NOT equal to 0,-1.
- 9(a), the final line should say k NOT equal to 1.
- 9(b), in the second line P^(-1)AB should be P^(-1)AP.
- April 2008 Exam.
Answers to April 2008:
1) C. 2) C. 3) D. 4) B. 5) A. 6) B. 7) E. 8) C. 9) C. 10) A. 11) E. 12) E.
II.1) a) sqrt(17)/3 b) (8/9, 11/9, 7/9)
II.2) (a) [1 0 3 0 3 \\ 0 1 1 0 -5 \\ 0 0 0 1 -3 \\ 0 0 0 0 0 \\ 0 0 0 0 0 ] (\\ means new row)
(b) From (a), columns 1,2,4 of A will work.
(c) [1 0 3 0 3], [0 1 1 0 -5], [0 0 0 1 -3]
(d) [-3 -1 1 0 0]^T, [-3 5 0 3 1]^T
II.3) (a) 1/29 [21 -20 \\ -20 -21]
(b) 1/2 [ 5 -1 \\ -1 5]
II.4) (a) (x-3)(x-3)(x+1), eigenvalues 3,3,-1
(b) Show that (3I-A)X=0 has two parameters.
(c) D=diag(-1,3,3), P=[-2 0 2 \\ 1 1 0 \\ 1 0 1]
II.5) (a) 1/sqrt(5) [1 -2 \\ 2 1] (b) Max=6, Min=1
- You can find more old exams here.