Solution to Final, MATH 54, Linear Algebra and Differential Equations, Fall 2014

Transcription

1 Solution to Final, MATH 54, Linear Algebra and Differential Equations, Fall 24 Name (Last, First): Student ID: Circle your section: 2 Shin 8am 7 Evans 22 Lim pm 35 Etcheverry 22 Cho 8am 75 Evans 23 Tanzer 2pm 35 Evans 23 Shin 9am 5 Latimer 24 Moody 2pm 8 Evans 24 Cho 9am 254 Sutardja Dai 25 Tanzer 3pm 26 Wheeler 25 Zhou am 254 Sutardja Dai 26 Moody 3pm 6 Evans 26 Theerakarn am 79 Stanley 27 Lim 8am 3 Hearst 27 Theerakarn am 79 Stanley 28 Moody 5pm 7 Evans 28 Zhou am 254 Sutardja Dai 29 Lee 5pm 3 Etcheverry 29 Wong 2pm 3 Evans 22 Williams 2pm 289 Cory 2 Tabrizian 2pm 9 Evans 22 Williams 3pm 4 Barrows 2 Wong pm 254 Sutardja Dai 222 Williams 2pm 22 Wheeler If none of the above, please explain: This is a closed book exam, no notes allowed. It consists of 8 problems, each worth points. We will grade all 8 problems, and count your top 6 scores. Problem Maximum Score Your Score Total Possible 6

2 Solution to Final, MATH 54, Linear Algebra and Differential Equations, Fall 24 Problem ) True or False. Decide if each of the following statements is TRUE or FALSE. You do not need to justify your answers. Write the full word TRUE or FALSE in the answer box of the chart. (Each correct answer receives 2 points, incorrect answers or blank answers receive points.) Statement Answer T T T T F ) For any inner product on R 2, if vectors u, v satisfy u =, v = and u v = 2, then u is orthogonal to v. u v 2 = u v, u v = u, u 2 + v, v 2 2 u, v so 2 = + 2 u, v so u, v =. 2) In the vector space of continuous functions on the interval [, ] with inner product f(t), g(t) = the functions cos(t) and sin(t) are orthogonal. f(t)g(t) dt cos(t) is even and sin(t) is odd so cos(t) sin(t) is odd so its integral over [, ] is the negative of its integral over [, ]. 3) If A is symmetric and U is orthogonal, then UAU is symmetric. A is symmetric so A T = A and U is orthogonal so U = U T, (U ) T = (U T ) T = U hence (UAU ) T = (U ) T A T U T = UAU. 4) If a 2 2 matrix A has eigenvalues λ, λ 2, then its characteristic polynomial is equal to χ A (t) = t 2 (λ + λ 2 )t + λ λ 2 If A has an eigenvalue λ, then it is similar to an upper triangular matrix [ ] λ t u χ A (t) = det = t λ 2 t 2 (λ + λ 2 )t + λ λ 2. [ ] λ u λ 2 so 5) Let V be the vector space of differentiable functions on the real line. The linear transformation T : V V given by T (y) = y e t y + 2y is injective. We can always find a nonzero y so that T (y) = (in fact T has a 2-dimensional null space) so T is not injective. 2

3 Solution to Final, MATH 54, Linear Algebra and Differential Equations, Fall 24 Problem 2) Multiple Choice. There is a single correct answer to each of the following questions. Determine what it is and write the letter in the answer box of the chart. You do not need to justify your answers. (Each correct answer receives 2 points, incorrect answers or blank answers receive points.) Question Answer B B C B A ) Which of the following matrices is similar to [ ] 4 6? 3 5 A) [ ] 4 5 B) [ ] 2 6 C) [ ] 5 4 D) [ ] 6 2 E) none of the preceding. The matrix s eigenvalues are 2, so it is similar to any upper triangular matrix with those diagonal entries. [ ] [ ] 2) For some basis B of the vector space R 2, the vectors u =, v = have coordinates [ ] [ ] [ ] 2 2 [u] B =, [v] B =. What is the vector w with coordinates [w] B =? A) [ ] /2 B) [ ] /2 /2 [ ] 2 2 P B = so P B = 4 C) [ ] /2 D) [ ] /2 /2 [ ] 2 and w = P 2 B [w] B. E) not determined by the data. 3

4 Solution to Final, MATH 54, Linear Algebra and Differential Equations, Fall ) For which pair of real numbers (a, b) is the matrix a rank one? 3 6 b A) (, 3) B) (2, ) C) (2, 3) D) ( 2, 3) E) none of the preceding. Need each row to be a scale of the first. 4) What is the sum of the dimensions of the null space and column space of the matrix A = ? A) 4 B) 5 C) 6 D) 7 E) 8 Dimension of domain = rank + dimension of null space. 5) For which triples of real numbers (a, b, c) does the linear system b a x b 2 = b c x 2 b 3 2 x 3 b have a solution for any b 2? b 3 A) (,, 2) B) (2,, ) C) (2, 2, ) D) (,, 2) E) none of the preceding. The matrix s determinant is b(2 a) so is nonzero and hence invertible when a 2, b. 4

5 Solution to Final, MATH 54, Linear Algebra and Differential Equations, Fall 24 Problem 3) ) (5 points) Find the orthogonal projection of the vector b to the subspace of R 4 spanned by u, v where b =, u =, v = 2 Use Gram-Schmidt to find orthogonal basis consisting of u = w = v Then the orthogonal projection is v, u u, u u = b, u b, w u + u, u w, w w = = + 4 = 3/4 /4 /4 /4 and 5

6 Solution to Final, MATH 54, Linear Algebra and Differential Equations, Fall 24 2) (5 points) Find a least-squares approximate solution to the equation Ax = b where A = 2, b = Using the first part of the problem, we must solve 3/4 [ ] /4 /4 = x 2 y /4 The first two equations then require x = 3/4, y = /4. 6

7 Solution to Final, MATH 54, Linear Algebra and Differential Equations, Fall 24 Problem 4) ) (5 points) Find the general solution of the second order ODE y 2y 3y = Auxiliary equation is r 2 2r 3 = (r 3)(r + ) = with roots 3, so general solution is of the form y = c e 3t + c 2 e t 7

8 Solution to Final, MATH 54, Linear Algebra and Differential Equations, Fall 24 2) (5 points) Find the general solution of the second order ODE y 2y 3y = cos(t) Try y = a cos(t) + b sin(t). Then y 2y 3y = ( a cos(t) b sin(t)) 2( a sin(t)+b cos(t)) 3(a cos(t)+b sin(t)) = ( 4a 2b) cos(t) + (2a 4b) sin(t). So we must solve the linear system [ ] [ ] [ ] 4 2 a = 2 4 b The second equation implies a = 2b and then the first equation implies a = 2, b =. Using the first part of the problem, the general solution is thus y = 2 cos(t) sin(t) + c e 3t + c 2 e t 8

9 Solution to Final, MATH 54, Linear Algebra and Differential Equations, Fall 24 Problem 5) ) (5 points) Find a basis of solutions of the equation [ ] 2 y (t) = y(t) 2 Characteristic polynomial is χ A (t) = t 2 2t 3 so eigenvalues are, 3. Corresponding eigenvectors are [ ] [ ], Thus basis of solutions is ] ] e t [, e 3t [ 9

10 Solution to Final, MATH 54, Linear Algebra and Differential Equations, Fall 24 2) (5 points) Write down a 3 3 matrix A such that the equation y (t) = Ay(t) has a basis of solutions e t y (t) =, y 2 (t) = e, y 3 (t) = e 2t A = Eigenvalues are, 2, with corresponding eigenvectors,,

11 Solution to Final, MATH 54, Linear Algebra and Differential Equations, Fall 24 Problem 6) ( points) Use separation of variables to find a solution u = u(x, t) of the equation 2 u t = 2 u 2 x + u 2 satisfying u(x, ) = e x and u(x, t) as t for all x. Set u(x, t) = X(x)T (t). Then 2 u t 2 = X(x)T (t) Thus X(x)T (t) = X (x)t (t) + X(x)T (t) so T (t) T (t) = X (x) X(x) + 2 u x 2 = X (x)t (t) The left hand side is independent of x and the right hand side is independent of t so they both must be equal to some constant Thus we have the ODEs with bases of solutions T (t) T (t) = c = X (x) X(x) + T (t) ct (t) = X (x) (c )X(x) = e ( c)t, e ( c)t e ( c )x, e ( c )x The initial condition requires we take c = 2 and the solution e x. The limit condition then requires we take a solution ae ( 2)t. Thus altogether we find ae x e ( 2)t and the initial condition requires we take a = to conclude y = e x e ( 2)t.

12 Solution to Final, MATH 54, Linear Algebra and Differential Equations, Fall 24 Problem 7) Consider the Fourier series a 2 + (a n cos(nx) + b n sin(nx)) n= for the function x on the interval [ π, π]. ) (5 points) Calculate the coefficients a n, for all n. If n =, we find a n = π π π x cos(nx) dx = 2 π a = π π x cos(nx) dx If n >, using d (x sin(nx)) = sin(nx) + nx cos(nx), we integrate by parts to find dx 2 π x cos(nx) dx = 2 π nπ ([x sin(nx)] π [ cos(nx)/n] π ) = 2 n 2 π cos(nx) π If n is even and n >, we find If n is odd, we find a n = 4 n 2 π a n = 2

13 Solution to Final, MATH 54, Linear Algebra and Differential Equations, Fall 24 2) (5 points) Calculate the coefficients b n, for all n. b n =, for all n, since x is even. 3

14 Solution to Final, MATH 54, Linear Algebra and Differential Equations, Fall 24 Problem 8) The following assertions are FALSE. Provide a counterexample (4 points each) along with a clear and brief justification no longer than one sentence ( point each). ) (5 points) If A is a 2 2 symmetric matrix with positive integer entries, then any eigenvalue of A is positive or zero. [ ] 2 A = 2 χ A (t) = t 2 2t 3 so eigenvalues are, 3. 2) (5 points) Suppose T : R 2 R 3 is an injective linear transformation. For any given basis of R 3, there is a basis of R 2 such that the matrix of T takes the form [T ] = Take T to be the linear transformation [ ] x T ( ) = x y y with respect to the standard bases of R 2 and R 3. Then we can not choose an alternative basis of R 2 so that [T ] takes the desired form since is not in the image of T so can not be in the column span of [T ]. 4