Linear algebra exam. The attached files are past exams for reference
FINAL EXAM (120 min) MATH 26500 SPRING 2015 Student Name Student ID Instructor Section number (3-digits, as in the table below) Mark Test 01 and Section Number on your scantron ! There are 25 questions on this exam. Each question is worth 8 points. Exam Rules 1. Students may not open the exam until instructed to do so. 2. Students must obey the orders and requests by all proctors, TAs, and lecturers. 3. No student may leave in the first 20 min or in the last 10 min of the exam. 4. Books, notes, calculators, or any electronic devices are not allowed on the exam, and they should not even be in sight in the exam room. Students may not look at anybody else’s test, and may not communicate with anybody else except, if they have a question, with their TA or lecturer. 5. After time is called, the students have to put down all writing instruments and remain in their seats, while the TAs will collect the scantrons and the exams. 6. Any violation of these rules and any act of academic dishonesty may result in severe penalties. Additionally, all violators will be reported to the Office of the Dean of Students. I have read and understand the exam rules stated above: STUDENT SIGNATURE: Section Numbers: Bauman, Patricia: 071 Dadarlat, Marius: 033 Gabrielov, Andrei: 031 Ulrich, Bernd: 121 Zhilan Julie Feng: 061 Noparstak, Jakob: 173 Ma, Linquan: 021 (10:30am) and 072 (9:30am) Li, Dan 022 (02:30pm) and 052 (01:30pm) Kelleher, Daniel: 131 (04:30pm) and 141 (03:30pm) Gnang, Edinah 032 (09:30am) and 051 (10:30am) Xu, Xiang 132 (MWF 03:30pm) and 162 (MWF 04:30pm) Zhang, Xu: 171 (TR 01:30pm) and 172 (TR 12:00pm) 1 (1) If we solve the equation: [ a+ b c− a c+ 2a 3 ] = [ 3 −4 11 3 ] for a, b and c, the values of a and b are: A. a = 2, b = 1 B. a = 7, b = −4 C. a = 5, b = −2 D. a = −1, b = 4 E. There is no solution (2) Let A = 1 −1 5 −2 3 1 4 1 2 1 −3 8 . Compute the (2,3) entry of ATA. A. 7 B. 14 C. −3 D. −14 E. −24 2 (3) Which of the following sets of vectors span R3? A. 12 1 , 11 1 , 0−1 0 B. 12 3 , 32 1 C. 12 3 , 01 3 , 11 0 D. 12 3 , 11 1 , 00 0 , 10 −1 E. 11 0 , 01 1 , 10 1 , 10 −1 (4) Which of the following is a basis for the subspace of R3 spanned by S = 12 1 , 32 1 , 22 1 , 12 −1 ? A. 12 1 , 32 1 B. 12 1 , 32 1 , 22 1 C. 12 1 , 22 1 D. 12 1 , 32 1 , 12 −1 E. 12 1 , 32 1 , 22 1 , 12 −1 3 (5) If L : R2 → R3 is a linear transformation such that L ([ 1 0 ]) = 11 2 , L([ 1 1 ]) = 23 2 and L([ 1 2 ]) = ab c , then a+ b+ c is equal to: A. 8 B. 10 C. 7 D. 4 E. 5 (6) If y = ax+ b is the least square fit line for the points (−1, 3), (0, 2), (1, 4), find a+ b. A. 2 B. 7/2 C. 3 D. 5 E. 9/2 4 (7) Consider the matrix A = t 1 t t 0 1 1 t 0 0 −1 t 0 0 t 1 , where t is a real number. Then A is nonsingular if and only if A. t 6= 0 B. t 6= 0 and t 6= 1 C. t 6= 1 and t 6= −1 D. t 6= 0, t 6= 1, and t 6= −1 E. t is any real number (8) For a nonsingular 6× 6 matrix A, the determinant of the adjoint matrix adj(A) is A. det(A) B. det(A5) C. det(A4) D. det(A2) E. det(A−1) 5 (9) Let V be the set of all strictly positive numbers in R, and let ⊕ and � be defined by a⊕ b = ab, for any a,b in V (that is, a, b are strictly positive numbers), c� a = ac, for any a in V and c in R. Which of the following statements are true? (i) For any a,b in V and any c in R, c� (a⊕ b) belongs to V . (ii) Under the given operations, the 0 element is the real number 1. (iii) There is at least one element a in V for which there is no element −a such that (−a)⊕a = 0. A. (i) only B. (ii) only C. (i) and (ii) only D. (i) and (iii) only E. All of them (10) Let S = {v1,v2,v3}, where v1 = 20 −3 , v2 = 12 −2 , v3 = 1−2 −1 . Which of the following statements are true? (i) A basis for span S is {v1,v2}. (ii) The vector u = 0−4 1 belongs to span S. (iii) S is a linearly dependent set. A. (i) only B. (iii) only C. (i) and (iii) only D. (ii) and (iii) only E. All of them 6 (11) Let W be the vector space spanned by the vectors: u1 = [ 1 0 0 1 ] , u2 = [ 0 1 1 0 ] , u3 = [ 1 1 0 1 ] , u4 = [ 1 1 1 1 ] . Apply the Gram-Schmidt process to the vectors u1,u2,u3,u4 (in this order) to find an orthonormal basis w1,w2,w3,w4 of W . What is w3? A. [ 0 1 −1 0 ] B. [ 0 0 0 0 ] C. [ 0 1/ √ 2 1/ √ 2 0 ] D. [ 0 1/ √ 2 −1/ √ 2 0 ] E. [ 1− √ 2 −1/ √ 2 − √ 2 1− √ 2 ] (12) Consider the subspace W of R 4 : W = span {[ 1 1 1 1 ] , [ 0 0 3 1 ] , [ −1 −1 2 0 ] , [ 1 1 4 2 ]} . What is the dimension of W⊥? A. 0 B. 1 C. 2 D. 3 E. 4 7 (13) Consider the homogeneous linear system a+ 2b+ 3c+ 4d+ 5e = 0 a + 2c + 3e = 0 b+ 2c+ 3d = 0 Then the dimension of the solution space is A. 0 B. 1 C. 2 D. 3 E. 4 (14) Suppose A is a 3× 5 matrix such that rank(A) = 3. Which of the following is TRUE? A. The rank of AT is 5 B. The nullity of AT is 2 C. Ax = 0 only has trivial solution D. The rows of A are linearly dependent E. The columns of A are linearly dependent 8 (15) The eigenvectors of [ 1 2 2 1 ] are [ 1 1 ] and [ −1 1 ] with eigenvalues 3 and −1 respectively. If x1(t), x2(t) satisfy x1(0) = 1, x2(0) = 3 and[ x′1(t) x′2(t) ] = [ 1 2 2 1 ] [ x1(t) x2(t) ] compute x1(1) + x2(1). A. 4e3 B. 4e3−2e−1 C. e3 − 2e−1 D. e3 − e−1 E. e3 − 4e−1 (16) The dimension of the vector space of all 4× 4 symmetric matrices with real entries is equal to: A. 6 B. 8 C. 9 D. 10 E. 12 9 (17) What is the characteristic polynomial of A = 1 2 30 1 2 0 2 1 A. (λ− 1)(λ+ 2) B. (λ− 1)(λ− 1)(λ+ 1) C. (λ− 1)(λ+ 1)(λ− 3) D. (λ− 1)(λ+ 1)(λ+ 3) E. (λ+ 1)(λ+ 1)(λ− 3) (18) Let A = PDP−1 where P = ( 2 1 1 1 ) , D = ( −1 0 0 3 ) and P−1 = ( 1 −1 −1 2 ) . Then A3 equals A. ( −29 56 −28 55 ) B. ( −29 56 −28 57 ) C. ( −29 55 −28 54 ) D. ( −24 56 −25 57 ) E. ( −24 56 −25 55 ) 10 (19) Find the eigenvalues and associated eigenvectors of the following matrix: A = [ 2 2 5 −1 ] A. λ1 = −3, λ2 = 4, x1 = [ 2 −5 ] , x2 = [ 1 −1 ] B λ1 = −3, λ2 = 4, x1 = [ 2 −5 ] , x2 = [ 1 1 ] C. λ1 = −3, λ2 = −4, x1 = [ 2 −5 ] , x2 = [ 1 −1 ] D. λ1 = 3, λ2 = 4, x1 = [ 2 −5 ] , x2 = [ 1 1 ] E. λ1 = 3, λ2 = −4, x1 = [ 2 −5 ] , x2 = [ 1 −1 ] (20) Which of the following matrices is not diagonalizable? A. [ 0 0 0 0 ] B. [ 1 0 0 0 ] C. [ 1 1 0 0 ] D. [ 1 1 0 1 ] E. [ 1 1 1 1 ] 11 (21) For which values of a does Gaussian elimination applied to A = a 2 3a a 4 a a a fail to give three pivots (leading 1’s)? A. 1, 2, 3 B. 0, 1, 3 C. 0, 2, 3 D. 0, 3, 4 E. 0, 2, 4 (22) If A = 2 −1 0−1 2 −1 0 −1 2 has inverse A−1 = a b ca1 b1 c1 a2 b2 c2 , then b+ c must be equal to: A. 3/4 B. 5/4 C. 1/4 D. −3/4 E. 7/4