Exam in TMA4110 Calculus 3, June 2013 Solution

Transcription

1 Norwegian University of Science and Technology Department of Mathematical Sciences Page of 8 Exam in TMA4 Calculus 3, June 3 Solution Problem Let T : R 3 R 3 be a linear transformation such that T = 4, T =, T = 6 3. a) Find the standard matrix A for the linear transformation T. Solution. The standard matrix is A = T T T. The first two columns of A are given. To find T, we use that T is linear, T = T T T = =. So A = 4. b) Find a basis for the null space, Nul(A), of A, and a basis for the column space, Col(A), of A.

2 Page of 8 / Solution. By Gauss elimination, we see that A is row-equivalent to B =. Therefore, the equation Ax = has solution x = t, and is a basis for Nul(A). The pivots of B are in the first and third column, so the first and third column of A, that is 4,, is a basis for Col(A). Problem a) Find the solution of the differential equation y y = which satisfies y() = and y () =. Solution. The characteristic polynomial of the differential equation is λ λ, with roots λ =, λ =, so the general solution of the differential equation is y(t) = c e t + c e t = c + c e t. Enforcing the conditions y() = and y () = gives the linear equations c + c = c =, Which are solved by c =, c =. The solution is therefore y(t) = e t. b) Find the general solution to the differential equation y y = e t sin t. Solution. We must find find a particular solution. There are several ways to proceed here. Undetermined coefficients, variation of parameters, or even setting x = y and solving the first order equation x x = e t sin t by using an integrating factor, all lead to a solution. Here we consider the complex differential equation z z = e (+i)t. () (Notice that Im ( e (+i)t) = e t sin t.) If z p is a solution of (), then y p = Im(z p ) is a solution of the original equation y y = e t sin t. We use the method of undetermined coefficients and look for a solution z p = ce (+i)t of (). Now, z p = ( + i)ce (+i)t, z p = ( + i) ce (+i)t = ice (+i)t, and inserting into () gives ice (+i)t ( + i)ce (+i)t = e (+i)t, ( + i)ce (+i)t = e (+i)t.

3 Page 3 of 8 For this equality to hold for all t, we need to have c = +i z p = ( + i)e(+i)t solves (), and = ( + i). Therefore y p = Im(z p ) = Im ( ( + i)e (+i)t) = ( Re( + i) Im(e (+i)t ) + Im( + i) Re(e (+i)t ) ) = ( e t sin t + e t cos t ) is a partial solution to y y = e t sin t. = et (sin t + cos t), We know from a) that y h (t) = c +c e t is the general solution of the homogenous equation y y =, so it follows that y(t) = y h (t) + y p (t) = c + c e t + et (sin t + cos t) is the general solution of y y = e t sin t. Problem 3 Let u =, u =, and v =. 4 a) Find an orthogonal basis for the plane in R 3 spanned by u and u. Solution. We use the Gram Schmidt procedure w = u, w = u v w w = u w w w. Inserting u and u gives w = =. {w, w } is an orthogonal basis for the plane spanned by u and u. b) Find the distance from v to the plane in R 3 spanned by u and u. Solution. We use the orthogonal basis from a) to calculate the orthogonal projection of v onto Span{u, u } = Span{w, w }. ˆv = v w w + v w w = w w w w w w = w + w = The distance from v to Span{u, u } is given by v ˆv = = 6.

4 Page 4 of 8 Problem 4 of motion A particle moving in a plane under the influence of a force has the equation x (t) = [ ] 5 x(t), where x(t) [ ] denotes the position of the particle at the time [ t. Find x(t) assuming ] that x() =. The answer should be given in the form x(t) = e at c cos(bt) + c sin(bt) where c 3 cos(bt) + c 4 sin(bt) a, b, c, c, c 3 and c 4 are real numbers. [ ] 5 Solution. Let A =. The characteristic polynomial of A is λ + λ + 5, with complex [ ] roots λ = + i and λ i 5 = i. From the matrix A λi =, we can i [ + i see that the eigenvector corresponding to λ is v = [ i λ is therefore simply v = terms of the eigenvectors as x(t) = c e λt v + c e λt v, ]. The eigenvector corresponding to ]. The general complex solution of x (t) = Ax(t) is given in where c[ and ] c are complex numbers. To obtain the solution satisfying the initial condition x() =, we get two equations for c, c. x() = c v + c v [ ] [ ] c + c = + (c c )i. c + c Subtracting the second equation from the first, we see that c = c, and then the second equation gives that c = c =. The solution of the initial value problem is therefore x(t) = eλt v + e λt v. To get this expression on the form required, we could expand this expression using e a+bi = e a (cos b + i sin b). A quicker way is to recognise the expression on the right hand side above as

5 Page 5 of 8 (w(t) + w(t)) = Re(w(t)) with w(t) = eλt v. So x(t) = Re ( e λt v ) ( [ ]) + i = Re e ( +i)t ([ ]) e = e t Re it ( + i) e it ([ ]) (cos(t) + i sin(t))( + i) = e t Re cos(t) + i sin(t) ([ ]) cos(t) sin(t) + i( cos(t) + sin(t)) = e t Re cos(t) + i sin(t) = e t [ cos(t) sin(t) cos(t) ] Problem 5 You are given that a b c det p q r =. Use this information to compute the determinant of the matrix p q r. 5p a 5q b 5r c Give reasons for your answer. Solution. By the properties of determinants and elementary row operations 5p a 5q b 5r c det p q r = det p q r, 5p a 5q b 5r c a b c = det p q r, a b c = ( ) det p q r, = = 4.

6 Page 6 of 8 Problem 6 You do not have to give reasons for your answers for this problem. a) For each of the following 4 complex numbers, determine whether it lies in the first quadrant of the complex plane (i.e., both its real part and its imaginary part are nonnegative) or not.. 3 i. Not in first quadrant.. +i +3i. Not in first quadrant. The expression is equal to i. 3. e +7πi. Not in first quadrant. e +7πi = e e 7πi = e e πi = e. 4. z, where z = and Arg(z) = π 3. Not in first quadrant. arg(z ) = Arg(z) = π 3 > π b) Let A be an n n matrix, B an m n matrix, and C an n m matrix, where n m. For each of the following 4 expressions, determine whether it is well defined or not.. AB T. Defined.. BB T. Defined. 3. CB + A. Defined. 4. B A. Not defined. B is not defined. c) Let A and D be n n matrices and let b be a nonzero vector in R n. For each of the following 4 statements, determine whether it is true or not.. If the system Ax = b has more than one solution, then the system Ax = also has more than one solution. True. If x and x both solves Ax = b, then A(x x ) = Ax Ax = b b =.. If A T is non-invertible, then A is non-invertible. True. A T non-invertible rank(a T ) < n rank(a) < n A non-invertible. 3. If AD = I, then DA = I. True. A and D are square, so if AD = I then D = A. 4. If A has orthonormal columns, then A is invertible. True. If A has orthonormal columns, then A T A = I, so A T = A.

7 Page 7 of 8 d) For each of the following 4 statements, determine whether it is true or not. 3. The two vectors 5 and are orthogonal. Not true.. If x =, y = and z =, then z belongs to the orthogonal complement of Span{x, y}. 3 True. z is orthogonal to both x and y, and thus to all of Span{x, y}. 3. An m n matrix B has orthonormal columns if and only if BB T = I. Not true. (B has orthonormal columns if and only if B T B = I.) BB T = I is equivalent to A having orthonormal rows, but nonsquare matrices may have orthonormal rows but not orthonormal columns and vice versa. 4. If x is orthogonal to y and z, then x is orthogonal to y z. True. x T (y z) = x T y x T z = e) Let A be an n n matrix. For each of the following 4 statements, determine whether it is true or not.. If A is orthogonally diagonalizable, then A is symmetric. True. A = P DP T A T = (P T ) T D T P T = P DP T = A.. If A is an orthogonal matrix, then A is symmetric. Not true. Counterexample: A = [ ] is orthogonal, but not symmetric. 3. If x T Ax > for every x, then the quadratic form x T Ax is positive definite. True. By definition. 4. Every quadratic form can by a change of variable be transformed into a quadratic form with no cross-product term. True. Every quadratic form can be written x T Ax with A symmetric. When A is symmetric, it is also orthogonally diagonalizable, A = P DP T, so the change of variables defined by x = P y gives x T Ax = y T Dy. f) Let u and v be nonzero vectors in R n, and let r be a scalar. For each of the following 4 statements, determine whether it is true or not.. rv = r v, unless r =. Not true. Does not hold for r <..

8 Page 8 of 8. If u and v are orthogonal, then {u, v} is linearly independent. True. If au + bv =, then u T (au + bv) = au T u + bu T v = a u =, so a =. Similarly, v T (au + bv) = gives b =. 3. If u + v = u + v, then u and v are orthogonal. True. In general, u + v = (u + v) T (u + v) = u + u T v + v holds. 4. If u v = u + v, then u and v are orthogonal. True. This is the same as 3 where v is substituted for v. g) For each of the following 4 statements, determine whether it is true or not.. If A is a matrix, then rank(a) = dim(nul(a)). Not true. By the definition of rank, rank(a) = dim(row(a)) = dim(col(a)), dim(nul(a)) is usually different.. A 5 matrix can have a -dimensional null space. Not true. A 5 matrix has maximal rank 5. The rank theorem tells us that the rank of the matrix plus the dimension of the null space for a 5 matrix is equal to. Therefore the dimension of the null space is at least Row operations on a matrix can change its null space. Not true. This is a basic property of row operations. 4. If the matrices A and B have the same reduced echelon form, then Row(A) = Row(B). True. The row space of a matrix is spanned by the nonzero rows of its reduced echelon form. h) For each of the following 4 statements, determine whether it is true or not.. The polynomials p (t) = + t and p (t) = t are linearly independent. True. p (t)/p (t) is defined for all t, but not constant.. If A is a 3 4 matrix, then the mapping x Ax is a linear transformation from R 3 to R 4. Not true. The mapping is a linear transformation from R 4 to R If a linear transformation T : R 4 R 4 is onto R 4 (or is surjective), then T cannot be one-to-one (injective). Not true. A linear transformation T : R 4 R 4 which is surjective is in fact invertible, and therefore also injective. 4. A linear transformation S : R 5 R 4 cannot be one-to-one (injective). True. The null space of the transformation has to be at least one-dimensional, so S(x) = has infinitely many solutions.