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MATH 2164 - FINAL EXAM - 12/18/
NAME:
ID:
GRADE OPTIONAL PART (CIRCLE ONE): YES NO
EXAM INSTRUCTIONS
- ONLY PENCILS AND ERASERS ALLOWED. NO CALCULATORS! If you need scratch paper, raise your hand, and I will provide it.
- Put your name and 800-ID in the appropriate blank above.
- Circle exactly one of “Yes” or “No” above (see the exam strategy section below). If you try any funny stuff, like circling both or neither, I will grade the optional part.
- Write NEATLY and LEGIBLY.
- On the True/False questions, please clearly mark T or F in the left margin (or write out “true” or “false”). If your answer is not clearly marked, you will receive no credit. Remember, you get extra credit if you provide a reason (if true) or a counterexample (or correct the statement) if false. Finally, GUESS if you do not know. NEVER leave a T/F question unanswered!
- For free response questions, you MUST SHOW YOUR WORK (unless an answer is obvious or conceptual, in which case you should supply a sentence or two explaining why).
- Unless you know otherwise, you should assume matrices are square if reference is made to determi- nants, inverses, or eigenvalues/eigenvectors.
EXAM STRATEGY The exam has 225 total points (75 points on the mandatory part, 150 points on the optional part). As a reminder, if you circled “No” above, I will substitute the average of your midterm exam scores in its place. If you circled “Yes” above, no such substitution will take place. Your final exam score will replace the lower of your two midterm scores, provided the final exam score is not the lowest of the three. You have 3 hours to finish this exam. It is roughly 1/3 new material and 2/3 old material. So if you’re writing both parts, you should spend 1 hour on the mandatory part and 2 hours on the optional part.
MANDATORY PART OF EXAM (75 points)
MANDATORY CONCEPTUAL QUESTIONS (4 points each - 20 points total).
- (T/F) If A is an n × n matrix, x is a nonzero vector, and λ is a scalar such that Ax = λx, then x is an eigenvector of A with eigenvalue λ.
- (T/F) Any basis for an n dimensional vector space must have exactly n vectors.
- (T/F) λ is an eigenvalue of a square matrix A if and only if det(A − λI) 6 = 0.
- (T/F) [Definition: An eigenvalue λ with multiplicity m is called defective if dim(Eλ) < m.] A is diagonalizable if and only if A has no defective eigenvalues.
- (T/F) If A is an m × n matrix, then rank A + nullity A = n.
MANDATORY FREE RESPONSE QUESTIONS (55 points total).
- (10 points) Let B =
(a) (3 points) Write down the change of coordinates matrix PB (no work required).
(b) (3 points) Find [x]B if x =
. (Hint: PB[x]B = x. Go to echelon form and use back substi-
tution.)
(c) (2 points) Is B linearly independent? Why or why not? (Hint: Cite your work from part (b).)
(d) (2 points) Is B a basis for R^3? Why or why not? (Hint: Cite your answer from part (c).)
- (10 points) Suppose A =
(a) (1 points) Reduce A to an echelon form (do not go to reduced echelon form). Circle the pivots.
(b) (3 points) What is rank A? (Hint: rank is the dimension of the column space – no work required).
(b) (3 points) What is nullity A? (Hint: nullity is the dimension of the null space – no work required).
(c) (3 points) What is rank A + nullity A? Does Rank + Nullity Theorem hold? Why or why not?
- (10 points) Let A =
(a) (3 points) What are the eigenvalues of A^5? (Hint: Since A is triangular, so is A^5. Think about the diagonal of A^5 , but do not calculate A^5 .)
(b) (4 points) Obviously λ = 2 is an eigenvalue of A with multiplicity 2. Determine E 2 , the eigenspace corresponding to λ = 2. (Hint: Recall E 2 = Nul (A − 2 I).)
(c) (3 points) Explain why A is not diagonalizable. (Hint: Cite what you found out in part (b).)
- (15 points) Let A =
. Diagonalize A as A = P DP −^1 by following these steps:
(a) (3 points) Determine the eigenvalues of A and their multiplicities.
(b) (3 points) Find a basis for each eigenspace.
OPTIONAL PART OF EXAM (150 points)
CONCEPTUAL QUESTIONS (4 points each - 40 points total).
- (T/F) Ax = b has a solution if and only if b lies in the span of the columns of A.
- (T/F) If the linear system Ax = b has a free variable, then the associated homogeneous equation Ax = 0 has infinitely many solutions.
- (T/F) The set {u 1 ,... , un} is linearly independent if and only if the equation x 1 u 1 + · · · + xnun = 0 has only the trivial solution.
- (T/F) A is invertible if and only if det A 6 = 0.
- (T/F) If T : V → W is linear, then for all scalars λ and vectors u, v in V the following are true: T (u + v) = T (u) + T (v) and T (λu) = λT (u).
- (T/F) If {v 1 ,... , vn} are vectors in V , then Span{v 1 ,... , vn} is a subspace of V.
- (T/F) If A is m × n, then Col A is a subspace of Rn.
- (T/F) A square matrix A is invertible if and only if every column (or row) contains a pivot position.
- (T/F) If A and B are square matrices of equal sizes, then det(AB) 6 = det(A) det(B) in general.
- (T/F) If A = LU , then to solve Ax = b you first solve U x = y and then solve Ly = b.
FREE RESPONSE QUESTIONS (110 points total).
- (8 points) Suppose each of the following is the matrix of a linear transformation T. In each case, decide if T is 1-1 (or not) and if T is onto (or not). Circle your answers. (Calculations not necessary.)
(a) (2 points) A =
. One-to-One: YES or NO. Onto: YES or NO.
(b) (2 points) A =
[
]
. One-to-One: YES or NO. Onto: YES or NO.
(c) (2 points) A =
. One-to-One: YES or NO. Onto: YES or NO.
(d) (2 points) A =
. One-to-One: YES or NO. Onto: YES or NO.
- (12 points) Let A =
(a) (4 points) Express Nul A as the span of a set of vectors.
(b) (4 points) Express Col A as the span of a set of vectors. (Hint: Reuse your work from (a).)
(c) (4 points) Express Row A as the span of a set of vectors. (Hint: Reuse your work from (a).)
- (10 points) Let v 1 =
, v 2 =
, v 3 =
h
, B = {v 1 , v 2 , v 3 }, and u =
(a) (5 points) For what values of h is B linearly independent? (One row reduction step is enough!)
(b) (5 points) Assuming h = 1, express u as a linear combination of the vectors in B (i.e. find x 1 , x 2 , x 3 such that u = x 1 v 1 + x 2 v 2 + x 3 v 3 ). (Hint: Go to echelon form and use back-substitution.)
- (10 points) Let A =
[
]
(a) (5 points) Find the LU factorization of A (Hint: One row reduction step gives you L and U ).
(b) (5 points) Use the LU factorization found in part (a) to solve Ax =
[
]
- (10 points) Let T : R^3 → R^2 be defined by T (x 1 , x 2 , x 3 ) = (2x 1 − x 3 , x 2 + x 3 ).
(a) (5 points) Find the standard matrix A of T.
(b) (5 points) Find a vector x such that T (x) = (1, 3).
- (10 points) Let A =
(a) (6 points) Calculate A−^1.
(b) (4 points) Using A−^1 you calculated in part (a), solve Ax =
. (Hint: Use a certain matrix-
vector multiplication and not row reduction).
