Math 2076-004, Fall 2024 Exam III PDF

Math 2076-004, Fall 2024 Exam III Name: Answer Key No work means no credit, unless specified otherwise. This exam contains 8 problems on 4 pages. All problems have the same weight. Use the backs of the exam pages for work, i...

Math 2076-004, Fall 2024 Exam III Name: Answer Key No work means no credit, unless specified otherwise. This exam contains 8 problems on 4 pages. All problems have the same weight. Use the backs of the exam pages for work, if necessary. 1. Let     3 2 1 2 A = 0 −1 2 ; b = −1 1 −3 4 3 (a) Compute det(A). Solution: Using the co-factor expansion along the first column, we have det(A) = 3(−1 · 4 − 2(−3)) + 1(2 · 2 − 1(−1)) = 6 + 5 = 11 (b) Consider the linear system Ax = b. Use Cramer’s Rule to find x2 (i.e., the second component of the solution vector x ). You must show the use of the Rule; solutions by other methods will lose credit. det(A2 (b)) Solution: Cramer’s Rule says that x2 = det(A) , where A2 (b) is the matrix A with the second column replaced by b. Thus,   3 2 −1 A2 (b) = 0 −1 2 , 1 3 4 det(A2 (b)) = (1st column) = 3(−4 − 6) + 1(4 + 1) = −25, and −25 25 x2 = = − 11 11 (c) Use the co-factor formula for the inverse to find the (2, 1) -rd entry of the inverse matrix A−1 , meaning the entry located in the second row and first column of A−1. Do not compute the full inverse; use the formula indicated. C Solution: We have (A−1 )2,1 = det(A)1,2 , where C1,2 is the co-factor corresponding to the (1, 2) -nd entry of A (not the (2, 1) -rd entry!) Thus, 1 2 (A−1 )2,1 = (−1)1+2 (0 · 4 − 2 · 1) = 11 11 2. (a) Compute the coordinate vectors of the polynomials p1 (t) = 1 + 2t − 3t3 , p2 (t) = t + t2 , p3 (t) = −2 + t, and p4 (t) = −t + 2t3 relative to the basis B = {1, t, t2 , t3 }.         1 0 −2 0  2 1  1 −1 Solution: [p1 ]B =   0 ,  [p2 ]B =  1 ,  [p3 ]B =   0 ,  [p4 ]B =    0 −3 0 0 2 (b) Using your answer to part (a), determine if the set {p1 , p2 , p3 , p4 } is a basis for P3. Fully support your conclusion. Solution: This set if polynomials is a basis for P3 if and only if the corresponding set of coordinate vectors is a basis in R4. One easy way to determine if 4 vectors form a basis in R4 is to compute the determinant of the corresponding matrix: 1 0 −2 0 1 −2 0  
2 1 1 −1 det [p1 ]B [p2 ]B [p3 ]B [p4 ]B = = (3rd row) = −1 2 1 −1 = −4. 0 1 0 0 −3 0 2 −3 0 0 2 Since the determinant is not zero, the set {p1 , p2 , p3 , p4 } is a basis in P3. Math 2076-004, Fall 2024 Exam III Name: Answer Key 3. Matrices A and B below are row equivalent (do not verify this).     1 4 8 −3 −7 1 4 8 0 5 −1 2 7 3 4 0 2 5 0 −1 A=  ∼ B=  −2 2 9 5 5 0 0 0 1 4 3 6 9 −5 −2 0 0 0 0 0 (a) State rank(A). Briefly explain.
Solution: From B we can see that A has 3 pivot columns, so rank(A) = dim Col(A) = 3
(b) State dim Nul(A). Briefly explain.
Solution: By the Rank Theorem, Null(A) = 5 − 3 = 2
(c) State dim Row(A). Briefly explain.
Solution: This is the number of pivot rows, so dim Row(A) = 3. (d) Find a basis for Col(A). Solution: Matrix B shows that the pivot columns of A are the 1st, 2nd, and 4th. Therefore, they          1 4 −3    −1 2  3 (and not the pivot columns of B !) form a basis for Col(A) :   ,   ,          −2 2 5  3 6 −5   (e) Find a basis for Row(A). Solution: The basis for Row(A) is formed by the non-zero rows of any echelon form of A, for example B. Thus, the 1st, 2nd, and 3rd rows of B (and not of A !) form a basis we need: {(1, 4, 8, 0, 5), (0, 2, 5, 0, −1), (0, 0, 0, 1, 4)} (It’s okay to write these as column vectors.) 4. (a) Assume that the null space of a matrix A is a three-dimensional subspace of R8. (i) How many columns does A have? Explain. Solution: The vectors in the null space of A are of length 8, thus A has 8 columns. (ii) What is rank(A)? Explain. Solution: By the Rank Theorem, rank(A) = 8 − 3 = 5. (b) Suppose that G is an 7 × 7 matrix and the equation Gx = b is consistent for every b in R7. Is λ = 0 an eigenvalue of G? Explain. Solution: Since the equation Gx = b is consistent for every b in R7 , the columns of G span R7 , which means that G is an invertible matrix. Therefore, its null space is trivial, and so λ = 0 cannot be an eigenvalue of G (otherwise, there would be a non-zero vector in its null space). Page 2 Math 2076-004, Fall 2024 Exam III Name: Answer Key 5. Let M = {m1 , m2 } and N = {n1 , n2 } be bases for a vector space V, and suppose m1 = 2n1 − n2 , m2 = n1 + 3n2 (a) Find PN ←M , the change-of-coordinates matrix from M to N. Solution: We have     2 1 PN ←M = [m1 ]N , [m2 ]N = −1 3 (b) Use part (a) to find [x]N for x = 3m1 + 2m2. Solution:      2 1 3 8 [x]N = PN ←M [x]M = = −1 3 2 3 (c) Find PM←N , the change-of-coordinates matrix from N to M. Solution:  
−1   1 3 −1 PM←N = PN ←M = [m1 ]N , [m2 ]N = 7 1 2 (d) Use part (c) to find find [y]M If y = 2n1 − 3n2. Solution:      " 9# 1 3 −1 2 1 9 7 [y]M = PM←N [y]N = = = 7 1 2 −3 7 −4 − 47   −6 3 6. Let B =. 2 −1 (a) Compute the characteristic polynomial of B. Solution: −6 − λ 3 det(B − λI) = = λ2 + 7λ 2 −1 − λ (b) Compute the eigenvalue(s) of B. Give the algebraic multiplicity for each. Solution: Setting λ2 + 7λ = 0, we obtain λ = 0, λ = −7 , each of algebraic multiplicity 1 (c) Compute the eigenvectors. Clearly indicate what eigenvector(s) correspond to each eigenvalue. Solution: The eigenspace corresponding to λ = 0 is the null space for the matrix 1 − 12     −6 3 B − 0λ = ∼ (reduction) ∼. Solving the system (B − 0λ)x = 0, we obtain 2 −1 0 0 1 1 x = x2 2. Thus, the eigenvector corresponding to λ = 0 is 2. 1 1 The eigenspace corresponding to λ = −7 is the null space for the matrix     1 3 1 3 B − (−7)λ = ∼ (reduction) ∼. Solving the system (B − (−7)λ)x = 0, we obtain 2 6 0 0     −3 −3 x = x2. Thus, the eigenvector corresponding to λ = −7 is. 1 1 Page 3 Math 2076-004, Fall 2024 Exam III Name: Answer Key   3 0 0 7. Let A =  1 1 0. −2 −5 1 (a) List the eigenvalues of A. Give the algebraic multiplicity for each value. (Hint: you are not asked to, and do not have to, compute the characteristic polynomial of A. The eigenvalues are easy to find.) Solution: Since A is a triangular matrix, its eigenvalues are the entries on its main diagonal: λ = 3, algebraic multiplicity 1; λ = 1, algebraic multiplicity 2. (b) Find a basis for each eigenspace. Solution: The eigenspace corresponding to λ = 3 is the null space for the matrix A − 3I = 1 0 49  4  −9      0 0 0  1 −2 0 ∼ 0 0 29 . A basis for this null space is given by a single vector: − 29 . −2 −5 −2   0 0 0 1   2 0 0 The eigenspace corresponding to λ = 1 is the null space for the matrix A − 1I =  1 0 0 ∼ −2 −5 0     1 0 0  0  0 1 0. A basis for this null space is given by a single vector: 0. 0 0 0 1   8. True or False Mark each statement as either True, if it is always true, or False, if it is not always true. There is no partial credit in this problem. (a) A 6 × 8 matrix must have nullity greater than or equal to 2. Solution: True. A 6 × 8 matrix has at most 6 pivots, so its rank is at most 6. Therefore, it’s nullity, which is (8 − rank) must be a at least 2. (b) If the characteristic polynomial of a square matrix is (λ + 2)2 (λ − 1)(λ + 3), then the matrix is of size 3 × 3. Solution: False. This polynomial is of degree 4, so the matrix is of size 4 × 4. (c) If V is a vector space of dimension 6 and S is a set of 6 vectors in V that spans V, then S is a linearly independent in V. Solution: True. By the Basis Theorem, such a set S is automatically a basis for V, thus, it’s linearly independent. (d) The set of all 3 × 3 invertible matrices with the usual operations is a vector space. Solution: False. For an invertible matrix A, neither the matrix 0A, nor the matrix A − A is invertible, meaning this set isn’t closed under addition or multiplication by a scalar. (e) If −3 is an eigenvalue of A, then A + 3I is invertible. Solution: False. The matrix A + 3I has a non-trivial null space, so is not invertible. (f) The set of all 2 × 2 matrices with the usual operations is a vector space of dimension 4.         1 0 0 1 0 0 0 0 Solution: True. A basis is given by , , ,. 0 0 0 0 1 0 0 1 Page 4