Examination in TMA4110/TMA4115 Calculus 3, August 2013 Solution


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1 Norwegian University of Science and Technology Department of Mathematical Sciences Page of Examination in TMA40/TMA45 Calculus 3, August 03 Solution 0 0 Problem Given the matrix A a) Write the solution set of the matrix equation Ax 8 in parametric vector form. Solution. We begin by reducing the augmented matrix of the equation to its reduced echelon form We see that the last column does not contain a pivot position, so the equation is consistent. We furthermore see that the first and the second columns contain pivot position, and that the third and forth do not, so we choose x 3 and x 4 to be free variables. The reduced echelon form of the augmented matrix of the equation corresponds to the system x + x 3 x + 5x 3 + x 4 3 It follows that the complete solution of the equation is x s 0 x x x 3 3 5s t s s 5 + t 0 x 4 t where s and t are free parameters.
2 Page of b) Find an orthonormal basis for the row space Row(A). Solution. We begin by reducing A to its reduced echelon form We see that the first and second rows contain pivot position, and that the third does not. It follows that the first row v [ 0 0 ] of the reduced echelon form of A and the second row v [ 0 5 ] of the reduced echelon form of A form a basis of Row(A). We next use the GramSchmidt process in order to get an orthogonal basis of Row(A). Let u v [ 0 0 ] and u v v u u [ 0 5 ] 0 [ ] [ ] 0 0. u u 5 Then {u, u } is an orthogonal basis of Row(A). Finally, we normalize u and u in order to get an orthonormal basis of Row(A). Let x u [ ] u and x u [ ] u 0. Then {x, x } is an orthonormal basis for the row space Row(A). Problem [ ] 3 a) Diagonalize the matrix A 3 matrix D such that A P DP ). (that is, find an invertible matrix P and a diagonal Solution. The characteristic polynomial of A is det(a λi) λ 3 3 λ ( λ) 3 λ λ 8, and the eigenvalues of A are λ ± [ ] 3 3 A 4I. It follows that v 3 3 eigenvalue 4. [ ] ± 6 { 4. is an eigenvector of A corresponding to the
3 [ ] 3 3 A + I. 3 3 It follows that v eigenvalue. Thus, if we let P b) Make a change of variable, [ ] [ ] [ ] 4 0 and D, then A P DP 0. [ x x ] B [ y y Page 3 of is an eigenvector of A corresponding to the ], that transforms the quadratic form x + 6x x + x into [ ] a quadratic [ ] form with no crossproduct term (that is, find a matrix B x y such that if B, then x + 6x x + x ay + by for a suitable choice of x constants a and b). y Solution. The idea is to let B be an orthogonal matrix whose columns are eigenvectors of A. The eigenvectors v and v we found in a) are orthogonal to each other so we just have to normalize them. Since v v [ ], it follows that if we let B P, then B is an orthonormal matrix, so B T B (in fact, B T B B). Thus, if we let [ ] [ ] [ ] x y (y + y ) B, then (y y ) x y x + 6x x + x (We could also just had let ) [ ] (y + y ) + 6 (y + y ) [ ] (y y ) + (y y ) y + y y + y + 3y 3y + y y y + y 4y y. [ x x ] P [ y y ] [ ] y + y, then y y x + 6x x + x (y + y ) + 6(y + y )(y y ) + (y y ) y + y y + y + 6y 6y + y y y + y 8y 4y. Problem 3 Let y (t) t and y (t) t ln(t) be two solutions of the differential equation t y ty + y 0 on the interval (0, ).
4 Page 4 of a) Show that the set {y (t), y (t)} is a fundamental set of solutions for the above equation on the interval (0, ). Solution. Let us first check that y and y are solutions to the differential equation t y ty + y 0 (this is not really necessary, since it is stated in the problem that they are). We have that y (t) and that y (t) 0, so t y (t) ty (t) + y (t) 0 t + t 0, and y (t) ln(t) +, y (t) /t, so t y (t) ty (t) + y (t) t t ln(t) t + t ln(t) 0. Thus, y and y are solutions to the differential equation t y ty + y 0. To show that {y (t), y (t)} is a fundamental set of solutions for the equation, we compute the Wronskian of y and y. W (y, y )(t) y (t)y (t) y (t)y (t) t(ln(t) + ) t ln(t) t 0 for all t (0, ), so {y (t), y (t)} is a fundamental set of solutions for the differential equation t y ty + y 0. Alternatively, we could show directly that y and y are linearly independent on the interval (0, ), for suppose that c and c are constants such that c y (t)+c y (t) 0 for all t (0, ), then we get by evaluating at t that c 0 (since y () and y () 0), and then by evaluating at t exp() that c 0 (since y (exp()) exp() 0). This shows that y and y are linearly independent on the interval (0, ), and it follows that {y (t), y (t)} is a fundamental set of solutions for the differential equation t y ty + y 0. b) Find the general solution to the differential equation t y ty +y t, where 0 < t <. Solution. We will first find a particular solution by using variations of parameters. So we define a function y on the interval (0, ) by letting y(t) v (t)y (t) + v (t)y (t) for each t (0, ) where v and v are twice differentiable functions to be determined below. Then y (t) v (t)y (t) + v (t)y (y) + v (t)y (t) + v (t)y (t). We will assume that v (t)y (t) + v (t)y (y) 0 for all t (0, ). Then y (t) v (t)y (t) + v (t)y (t), y (t) v (t)y (t) + v (t)y (t) + v (t)y (t) + v (t)y (t), and t y (t) ty (t) + y(t) t (v (t)y (t) + v (t)y (t) + v (t)y (t) + v (t)y (t)) t(v (t)y (t) + v (t)y (t)) + v (t)y (t) + v (t)y (t) v (t)(t y (t) ty (t) + y (t)) + v (t)(t y (t) ty (t) + y (t)) + v (t)t y (t) + v (t)t y (t) v (t)t + v (t)t (ln(t) + ).
5 Page 5 of Thus, if v (t)y (t) + v (t)y (y) v (t)t + v (t)t ln(t) 0 and v (t)t + v (t)t (ln(t) + ) t for all t (0, ), then y is a solution to the differential equation t y ty + y t on the interval (0, ). It is easy to see that for any t (0, ), the system v (t)t + v (t)t (ln(t) + ) t v (t)t + v (t)t ln(t) 0 has a unique solution and that this solution is v (t) /t and v (t) ln(t)/t. It follows that if we let v (t) v (t)dt ln(t)/t dt (ln(t)), and v (t) v (t)dt /t dt ln(t), then y(t) y (t)v (t) + y (t)v (t) t(ln(t)) + t(ln(t)) t(ln(t)) is a solution to the differential equation t y ty + y t on the interval (0, ). It follows that the general solution to the differential equation t y ty + y t on the interval (0, ) is where c and c are constants. y(t) t(ln(t)) + c t + c t ln(t) Problem 4 0 a) The matrix A 0 has 3 eigenvalues. One eigenvalue is + i, and 0 + i i is an eigenvector corresponding to + i. Find the other two eigenvalues and a corresponding eigenvector for each of these eigenvalues.
6 Page 6 of + i Solution. Since all the entries of A are real, and i is an eigenvector of A corresponding i to + i, it follows by complex conjugation that + i is an eigenvector of A corresponding to i. The characteristic polynomial of A is λ 0 det(a λi) λ 0 ( λ)( λ)( λ) + 0 λ λ 3 4λ 5λ. It follows that 0 is an eigenvalue of A. In order to find a corresponding eigenvector, we reduce A to an echelon form We see that is an eigenvector of A corresponding to the eigenvalue 0. b) Three tanks contain salt water. Tank holds 0 litres, tank holds 5 litres, and tank 3 holds 0 litres. Salt water flows from tank to tank at a rate of 0 litres per second, and from tank to tank 3 at a rate of 0 litres per second, and from tank 3 to tank at a rate of 0 litres per second. Suppose that initially tank contains 0 grammes of salt, tank contains 6 grammes of salt, and tank 3 contains 4 grammes of salt. Suppose also that the tanks are stirred so that the salt in each tank is evenly distributed. How much salt does each of the three tanks contains after π/ seconds? Solution. Let x (t) be the amount of salt in tank at time t, let x (t) be the amount of salt in tank at time t, and let x 3 (t) be the amount of salt in tank 3 at time t. Then x (t) x (t) + x 3 (t) x (t) x (t) x (t) x 3(t) x (t) x 3 (t)
7 Page 7 of and x (0) 0, x (0) 6, x 3 (0) 4. Thus, it follows from a) that x (t) + i i x (t) c i e ( +i)t + c + i e ( i)t + c 3 x 3 (t) where c, c and c 3 are constants satisfying + i i 0 c i + c + i + c To solve the last equation, we reduce the augmented matrix of the equation to its reduced echelon form. + i i 0 4 i + i 6 i + i 6 0 i i 8 i 4 + i i i 3 + i + i 0 i i 8 i 0 i i i Thus c c and c 3 4, and x (t) + i i 8 x (t) i e ( +i)t + + i e ( i)t +4 e t cos(t)+ e t sin(t)+ 4 x 3 (t) It follows that x (π/) 8 e π, x (π/) 4+e π and that x 3 (π/) 8, so tank contains 8 e π grammes of salt after π/ seconds, tank contains 4 + e π grammes of salt after π/ seconds, and tank 3 contains 8 grammes of salt after π/ seconds. Problem 5 You do not have to give reasons for your answers for this problem. a) For each of the following 4 statements, determine whether it is true or not.. The equation x x ( + i) is linear. True.
8 Page 8 of. The differential equation y + 6y e 4t + 3 sin(4t) is linear. True. 3. The differential equation t y (t) + 3ty (t) 3y(t) 0 is linear. True. 4. The transformation T from R to R given by T (x) x + x is linear. False. We have for example that T () 6 4 T (). b) For each of the following 4 statements, determine whether it is true or not.. The three lines x + x 3, x + x 3 and x + x 3 have exactly one point in common. 0 0 True. The matrix 0 3 can be reduced to so (0, 3, ) is 0 the only point which belong to all three lines If a, a and b, then b belongs to Span{a, a }. 0 4 False. The matrix can be reduced to 0 4 which has a pivot position in the last column, so b does not belong to Span{a, a } If v 0, v 0 and v 3 0 0, then Span{v, v, v 3 } R 4. 0 False. A set of three vectors cannot generate R The following vectors are linearly independent: 4, 6 and False c) For each of the following 4 statements, determine whether it is true or not.. If A is an m n matrix, B is an n m matrix, and AB 0, then we must either have that A 0 or that B 0. False. We have for example that [ 0 ] [ ] 0 0.
9 Page 9 of. If C and D are n nmatrices, then it must be the case that (C + D)(C D) C D. False. We have for example that ([ ] 0 + [ ]) ([ ] 0 0 and [ ] 0 [ ] 0 [ ]) 0 [ ] 0 3. If E is an invertible matrix, then (E) E. False. (E) E. 4. If F is an matrix and the equation F x must be invertible. [ ] [ 0 0 [ ] ] [ ] 0 0 [ ] 0 [ ] 0 0 [ ] has a unique solution, then F True. If the equation F x has a unique solution, then Nul(F ) {0} (because [ ] [ ] if u Nul(F ), u 0 and F x, then F (x + u) ), and it then follows from the invertible matrix theorem that F is invertible. d) Let A be an n n matrix and k a scalar. For each of the following 4 statements, determine whether it is true or not.. det(ka) k n det(a). True. ka can be produced by multiplying each of the n rows of A by k, so it follows from Theorem 3 in Section 3. that det(ka) k n det(a).. If det(a), then det(a ) 4. True. det(a ) det(a) det(a) by Theorem 6 in Section det(a T ) det(a). True. See Theorem 5 in Section If A is invertible, then det(a) det(a ) 0. False. det(a) det(a ) det(a) det(a) 0. e) Let V be a vector space different from the zero vector space, and let v, v,..., v p be vectors in V. For each of the following 4 statements, determine whether it is true or not.. If {v, v,..., v p } is linearly independent, then it must be the case that {v, v,..., v p } is a basis for V. False. It is only true if dim(v ) p.
10 Page 0 of. If Span{v, v,..., v p } V, then some subset of {v, v,..., v p } is a basis for V. True. See Theorem 5 in Section If dim(v ) p, then {v, v,..., v p } cannot be linearly independent. False. If {v, v,..., v p } is a basis for V, then {v, v,..., v p } is linearly independent. 4. If dim(v ) p, then {v, v,..., v p } must be a basis for V. False. {v, v,..., v p } is only a basis for V if it is linearly independent or Span{v, v,..., v p } V. f) Let A be an n n matrix. For each of the following 4 statements, determine whether it is true or not.. If A is invertible and is an eigenvalue of A, then must also be an eigenvalue of A. True. If Av v, then v (A A)v A v.. If v is an eigenvector of A, then v must also be an eigenvector of A. True. If Av λv, then A v λav λ v. 3. If A has fewer than n distinct eigenvalues, [ ] then A[ cannot ] be diagonalizable. 0 0 False. is the only eigenvalue of, and is a diagonale matrix and 0 0 therefore diagonalizable. 4. If every vector in R n can be written as a linear combination of eigenvectors of A, then A must be diagonalizable. True. If every vector in R n can be written as a linear combination of eigenvectors of A, then R n has a basis consisting of eigenvectors of A, and it then follows from Theorem 5 in Section 5.3 that A is diagonalizable. g) Let v and u be vectors in R n and let W be a subspace of R n. For each of the following 4 statements, determine whether it is true or not.. The distance between v and u is u v. True.. If {u, v} is linearly independent, then [ ] u and [ v ] must be orthogonal to each other. False. We have for example that and are linearly independent but not 0 orthogonal to each other. 3. If v coincides with its orthogonal projection onto W, then v must belong to W. True. The orthogonal projection of v onto W belongs to W so if it coincides with v, then v must belong to W.
11 Page of 4. No vector in R n can belong to both W and W (the orthogonal complement of W ). False. 0 belongs to both W and W. h) Let A be an n n matrix. For each of the following 4 statements, determine whether it is true or not.. If A T A and if u and v are vectors in R n which satisfy that Au 5u and Av v, then u and v must be orthogonal to each other. True. See Theorem in Section 7... If A P DP T where P T P and D is a diagonal matrix, then A must be symmetric. True. A T (P DP T ) T P DP T A because D T D. 3. If A is symmetric and all the eigenvalues of A are positive, then the quadratic form x x T Ax must be positive definite. True. See Theorem 5 in Section If A is symmetric and λ is an eigenvalue of A, then the dimension of the eigenspace of A corresponding to λ must be equal to the multiplicity of λ as a root of the characteristic polynomial of A. True. See Theorem 3 in Section 7..
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