University of Toronto Solutions to MAT88HF TERM TEST of Tuesday, October 3, Duration: minutes Only aids permitted: Casio 6, Sharp 5, or Texas Instrument 3 calculator. Instructions: Make sure this test contains 6 sheets with questions on both sides. Do not tear any pages from this test. Present your solutions to all questions in the space provided. The value for each question is indicated in parantheses beside the question number. Total Marks: 6 General Comments about the Test:. Since more than a third of the class got A, and almost two-thirds received at least B, the results on this test are generally quite good. But if your test mark is much less than 3 times your quiz mark, your performance is slipping, and you may be in trouble, since things always seem much harder in Chapter 4.. It was quite amazing how many students could not solve Question correctly. 3. Questions,, 5, 6, 7, 8 and 9 can all be double-checked; these questions should all have been aced. 4. det(a + B) = det(a) + det(b) is false, but this was used by some students in some of their calculations, for example in Questions 5 and (c). 5. There are still students putting = between reduced matrices, or using implication ( ) incorrectly. This is just throwing away marks. Breakdown of Results: 987 students wrote this test. The marks ranged from 6.7% to %, and the average was 7.4%. Some statistics on grade distributions are in the table on the left, and a histogram of the marks (by decade) is on the right. Grade % Decade % 9-%.7% A 38.4% 8-89% 6.7% B 4.6% 7-79% 4.6% C 9.% 6-69% 9.% D 9.8% 5-59% 9.8% F 8.% 4-49% 5.% 3-39%.% -9%.7% -9%.% -9%.%
. [4 marks Find the inverse of the matrix A = Solution : use the adjoint formula. Since det(a) = ( ) 3 =, The 9 cofactors of A are: [ C = det A = adj(a) = adj(a). det(a) [ C = det [ C 3 = det = = = [ [ [ C = det C = det C 3 = det = = = [ [ [ C 3 = det C 3 = det C 33 = det = = = So A = adj(a) = [Cij T = Solution : use the Gaussian algorithm. (A I) = = (I A )
. [6 marks Solve the following system of linear equations x + x + x 3 + x 4 x 5 = 5 x x + x 3 x 4 + x 5 = x + 4x + 5x 3 + 4x 4 7x 5 = 3 by first finding the reduced row-echelon form of its augmented matrix. Solution: reduce the augmented matrix: 5 4 5 4 7 3 5 5/3 8/3 Let x 3 = s, x 4 = t, x 5 = u be parameters. Then x 7/3 s + u/3 7/3 x x 3 x 4 = 8/3 s t + 5u/3 s t = 8/3 + s x 5 u 5 3 3 3 5 8 3 3 3 5 8 /3 7/3 5/3 8/3 + t + u /3 5/3 } {{ } Not required; just neater.
3. [5 marks Let Calculate det(a). A = 3 3 3 Solution: use properties of determinants to simplify your work. 3 4 6 det 3 3 = det 7 8 = ( )( )4 det 3 7 = det 4 6 4 5 5 [ 4 = ( ) 3 det 5 5 [ 4 = 5 det 4 6 7 8 3 7 = (5)( 3) = 5
4. [6 marks Find all the values of c for which the system of equations has no solution. x + c x + c x 3 = x x + x 3 = 4 c x + c x + x 3 = Solution: let A be the coefficient matrix. If det(a), then A is invertible and the system will have a unique solution. c c det = + 3c c = (3c + )(c ). c c If c =, then the system is x + x + x 3 = x x + x 3 = 4 x + x + x 3 = which has infinitely many solutions, since the first and third equations are identical. If c = /3, then the system is equivalent to the system 3x x x 3 = 6 x x + x 3 = 4 x + x 3x 3 = 6 which is inconsistent, since the second plus the third equation gives x x 3 =, while the first plus the third equation gives 4x 4x 3 =. Answer: c = /3
5. [5 marks Find the eigenvalues of the matrix 3 4 A = 5 5 Solution: need the characteristic polynomial of A. Accept either det(λi A) or det(a λi). Working with the former: λ 3 4 λ λ det λ + = det λ + 5 λ 5 3 λ λ 3 = (λ )(λ 3) det = (λ )(λ 3) det = (λ )(λ 3)(λ + ) = (λ )(λ 3)(λ ); so the eigenvalues of A are λ =, λ =, λ 3 = 3. λ + λ + Alternate Approach: simply expanding det(λi A) gives det(λi A) = λ 3 6λ + λ 6 which has to be factored! If the expansion is incorrect then the factoring my be very difficult!
6. [7 marks Given that the eigenvalues of A = 4 6 9 9 6 4 are λ = 6 and λ =, find an invertible matrix P and a diagonal matrix D such that D = P AP. Solution: find the eigenvectors. ( 6I A O) = 6 4 6 9 6 9 6 4 6 3 3 3 3 3 3 3 3 so there are two (basic) eigenvectors corresponding to λ = 6. For example, any two of, 3, 3, 3, or their multiples, will do. 4 6 (I A O) = 9 9 6 4 6 3 3 6 3 3 4 3 3 6 6 6 3 so an eigenvector corresponding to λ = is 3/ or Then (for example) take P = 3 3 3, D = 6 6 3 ; 6 3 6 9 6 9 }{{} Columns of P must correspond to correct diagonal entries of D. ;
7. [6 marks Show that the triangle P QR with vertices P (,, ), Q(,, ), R(4,, ) is a right angle triangle. Solution : P Q = = ; P R = 4 = 3 ; QR = 4 = Observe that P Q QR = + =, so the sides P Q and QR are orthogonal. That is, the interior angle at Q is π/. Solution : Observe that P Q + QR = + + 4 + + 4 = = P R, which means the lengths of the three sides in P QR satisfy the Pythagorean Theorem, whence the triangle is a right triangle.
8. [5 marks If v = and d = 3, express v in the form v = v + v where v is parallel to d and v is orthogonal to d. Solution: use the projection formula. v = proj d v = v d d d d = 3 + 9 + + = 4 d = 4 4 Then v = v v = 4 4 = 8 5
[ [ 9. [6 marks Given that X = and X = are (basic) eigenvectors of the matrix [ A =, 4 3 find all the solutions to the following system of differential equations: where f and f are functions of x. f (x) = f (x) + f (x) f (x) = 4f (x) + 3f (x) Solution: you can find the eigenvalues by using the definition of eigenvector: [ [ AX = = = X λ = ; AX = [ 5 [ = 5 = 5X λ = 5. Or you can find them directly: [ λ det = λ 4λ 5 = (λ 5)(λ + ). 4 λ 3 Either way, the general solution is F = c X e λx + c X e λx. Thus [ [ [ f (x) = c f (x) e x + c e 5x ; that is, and f (x) = c e x + c e 5x f (x) = c e x + c e 5x.
. [6 marks Indicate if the following statements are True or False, and give a brief explanation why. (a) [ marks If A is a square matrix such that adj(a) is the zero matrix, then det(a) =. True False Solution: True. A adj(a) = det(a)i O = det(a)i det(a) = Alternate Solution: Every cofactor of A is zero, so the cofactor expansion along any row or column of A must be zero; that is det(a) =. (b) [ marks If A is a 3 3 matrix such that A T = A, then det(a) =. True False Solution: True. A T = A det(a T ) = det( A) det(a) = ( ) 3 det(a) det(a) = det(a) det(a) = (c) [ marks If A is a 3 3 matrix such that A 3 + 4A = 7I, then det(a) =. True False Solution: False. A 3 + 4A = 7I A(A + 4I) = 7I det(a(a + 4I)) = det(7i) det(a) det(a + 4I) = 7 3 det(a)
. [4 marks Suppose A is an invertible matrix. Indicate if the following statements are True or False, and give a brief explanation why. (a) [ marks A and A have the same eigenvalues. True False Solution: False. Consider A = [. It has eigenvalue λ =, repeated. But its inverse is [ / A =, / which has eigenvalue µ = /, repeated, and µ λ. (b) [ marks A and A have the same eigenvectors. True False Solution: True. Recall that A is invertible if and only if λ = is not an eigenvalue of A. Then AX = λx X = A (λx) X = λa X λ X = A X. Thus X is an eigenvector of A if and only if X is an eigenvector of A.