Examiners: D. Burbulla, W. Cluett, S. Cohen, S. Liu, M. Pugh, S. Uppal, R. Zhu. Aids permitted: Casio FX-991 or Sharp EL-520 calculator.

Transcription

1 University of Toronto Faculty of Applied Science and Engineering Solutions to Final Examination, December 7 Duration: and / hrs First Year - CHE, CIV, CPE, ELE, ENG, IND, LME, MEC, MMS MATHF - Linear Algebra Examiners: D. Burbulla, W. Cluett, S. Cohen, S. Liu, M. Pugh, S. Uppal, R. Zhu Exam Type: A. Aids permitted: Casio FX-99 or Sharp EL-5 calculator. General Comments: The only questions with failing averages were Questions 4 and 7. In Question 4, part (c), the second choice was done much better than the first choice. In Question 7, the common mistakes were (i) thinking the given basis was an orthogonal basis of S, and (ii) finding an orthogonal basis of S and then claiming it was a basis for S. But dim(s ) is only. Questions, and 5 were very well done. In Question 6 many students failed to find orthonormal eigenvectors and hence, did not get an orthogonal matrix P. In Question, many students did not seem aware of the Normal Equations, and tried to find the least squares line from first principles, which is the hard way! Breakdown of Results: registered students wrote this test. The marks ranged from.75% to %, and the average was 7.6%. There were ten perfect papers. Some statistics on grade distribution are in the table on the left, and a histogram of the marks (by decade) is on the right. Grade % Decade % 9-% 4.9% A 35.9% -9%.% B 3.9% 7-79% 3.9% C 9.% 6-69% 9.% D.6% 5-59%.6% F.6% 4-49% 6.% 3-39% 3.% -9%.% -9%.% -9%.%

2 . avg: 9.3/ Let x 3 4 ; let u, u, u 3, u 4. Show that { u, u, u 3, u 4 } is an orthogonal set. Show your work! Then write x as a linear combination of u, u, u 3 and u 4. Solution: show that each pair of vectors u i, u j, i j is orthogonal. There are six dot products you must calculate:. u u +. u u u u u u u u u 3 u Now let x c u + c u + c 3 u 3 + c 4 u 4. Using the results from Chapter 7 we can solve for c i directly: Thus c i x u i u i. c ; c ; c ; c 4 4. OR you can solve a system of linear equations for c i by reducing an augmented matrix: ; and as before, c 5, c, c 3, c 4. Either way x 5 u u + u 3 u 4.

3 . avg: 9.3/.(a) 6 marks Find all values of a for which the matrix A 3 a a a is not invertible. Solution: use the fact that a matrix A is not invertible if and only if det(a). We have 3 det a a a + a + 3a a a a + (a )(a + 5). a Then det(a) (a )(a + 5) a or a 5, and so A is not invertible if a or a 5/..(b) 4 marks Let A ; v. Show that v is an eigenvector of A. What is the corresponding eigenvalue? Solution: use the defining property of an eigenvector: A v v. Thus v is an eigenvector of the matrix A with corresponding eigenvalue λ. 3

4 3. avg: 6.4/ Let x and y be two non-zero vectors in R n. (a) 5 marks Prove that if x + y x + y then x and y are orthogonal. Illustrate this geometrically. Solution: x + y x + y x + y y ( x + y) ( x + y) x x + y y x x x + x y + y y x x + y y x y, x + y x + y so x and y are orthogonal. The Pythagorean Theorem (b) 5 marks Prove that if x and y are orthogonal, then x+ y x y. Illustrate this geometrically. Solution: if x y, then x + y x x + x y + y y x + y and x y x x x y + y y x + y Thus x + y x y ; taking square roots, we get x + y x y. y x + y x y x The diagonals of a rectangle have the same length. 4

5 4. avg: 4.5/ Let A be an m k matrix; let B be a k n matrix. (a) 3 marks Show that nullity(b) nullity(a B). Recall: if X is a matrix, nullity(x) dim(null(x)). Solution: show null(b) is contained in null(a B), and the result follows by taking dimensions. x null(b) B( x) A(B( x)) A( ) x null(a B) Then dim(null(b)) dim(null(a B); or equivalently, nullity(b) nullity(a B). (b) 3 marks Use the Rank Theorem to show that rank(ab) rank(b). Solution: use the result from part (a) and the Rank Theorem: A B is an m n matrix and so nullity(b) nullity(a B) n rank(b) n rank(a B) rank(a B) rank(b). (c) 4 marks Show that if rank(a) k then A T A is invertible. (Hint: suppose A T A x.) Solution: follow the hint. A T A is k k; suppose x is in R k. A T A x x T A T A x x T (A x) T A x (A x) (A x) A x A x x, since rank(a) k nullity(a) Thus null(a T A) { }, and so A T A is invertible. OR: Show that if A is an m m invertible matrix, then x T A T A x > for all x in R m. Solution: similar to above. x T A T A x (A x) T (A x) (A x) (A x) A x. But since A is invertible, A x if and only if x. Thus for x, x T A T A x >. 5

6 5. avg: 9.3/ Solve the system of differential equations dx 6x y dt dy x + 3y dt for x and y as functions of t, given that (x, y) (5, 7) when t. 6 Solution: let A ; find the eigenvalues and eigenvectors of A. 3 λ 6 Eigenvalues: det(λi A) det λ 9λ+ (λ 4)(λ 5). So the eigenvalues λ 3 of A are λ 4 and λ 5. Eigenvectors: 4 6 { } null 4 3 null null span ; take v 5 6 { } null 5 3 null null span ; take v The general solution to the system of differential equations is x c v e 4t + c v e 5t c e 4t + c y e 5t To find c, c, let t and solve the system of linear equations 5 c 5 c + c 7 7 c c Thus the solutions to the system of differential equations is c. x e 4t + 3e 5t and y 4e 4t + 3e 5t. 6

7 6. avg: 6.63/ If A that P T AP D., find an orthogonal matrix P and a diagonal matrix D such Step : find the eigenvalues of A. λ λ 3 λ 3 λ det(λi A) det λ det λ 3 λ λ 3 λ (λ 3) det (λ 3) (λ ) λ(λ 3) Thus the eigenvalues of A are λ 3, repeated, and λ. Step : find an orthogonal basis of eigenvectors for each eigenspace. 3 null 3 null span 3, ; 3 3 null null null span 3 3. Step 3: for the columns of P, take the unit, orthogonal eigenvectors and for the diagonal entries of D take the corresponding eigenvalues: / / 6 / 3 P / 6 / 3 / / 6 / 3 and D 3 3, 7

8 } 7. avg: 4.4/ Let S span { T, T, T. (a) 3 marks Find a basis for S, the orthogonal complement of S. Solution: S null null null span Take w T as the basis for S. (b) 4 marks Find the standard matrix of the linear transformation perp S : R 4 R 4. Solution: let { e, e, e 3, e 4 } be the standard basis of R 4. Then w and perp S ( e ) proj S ( e ) e w w w perp S ( e ) proj S ( e ) e w w w w perp S ( e 3 ) proj S ( e 3 ) e 3 w w w perp S. perp S ( e 4 ) proj S ( e 4 ) e 4 w w w w Alternately, and more efficiently: x w perp S ( x) proj S ( x) w w ( x + x 4 ) x x 4, x + x 4 from which you can read off the matrix. (c) 3 marks What is the standard matrix of proj S : R 4 R 4? (You can make use of part (b).) Solution: use the fact that proj S I perp S. Thus proj S

9 . avg: 7.7/ Experimental data for the response, y, of an electronic device to an input, x in millivolts, is listed in the following table: Trial i Input x i Response y i 5 7 Find the best fitting line with equation y a + b x for the given data. Solution: let M ; y 5 7 a. The normal equations are M T M b M T y. Solve: a b a b a b a b a b 4 9. So the line that best fits the data has equation y.4 +. x. 9

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