Math 118, Fall 2014 Final Exam

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1 Math 8, Fall 4 Final Exam True or false Please circle your choice; no explanation is necessary True There is a linear transformation T such that T e ) = e and T e ) = e Solution Since T is linear, if T e ) = e, then T e ) = T e ) = e e True If A is a 4 3 matrix and the reduced row echelon form of A has three leading ones, then the equation A x = b is consistent for all vectors b in R 4 Solution Since the rank of A is 3, the image of A is a 3-dimensional subspace of R 4 Since the image is not all of R 4, there exists vectors b in R 4 for which the equation A x = b is inconsistent 3 4 True False If an invertible matrix A is diagonalizable, then A is also diagonalizable Solution If A = SDS, where D is diagonal, then A = SDS ) = S ) D S = SD S Since D is diagonal, so is D In fact, the diagonal entries of D are just the reciprocals of the diagonal entries of D) True False If v is an eigenvector of a matrix A, with eigenvalue λ, then v is in the kernel of the matrix A λi, where I is the identity matrix Solution Since A v = λ v = λi v we have A λi) v = So v is in the kernel of the matrix A λi True False If A = SBS, then A n = SB n S for any positive integer n Solution This follows from direct calculation, since SBS ) n = SBS ) SBS ) SBS ) = SB S S ) B S S ) S S ) BS = SBIBI IBS = SB n S True False If v is in the kernel of a matrix A, then v is in the kernel of BA, for any matrix B for which BA is defined Solution If v is in the kernel of A, then A v =, and so BA) v = BA v = B =

2 7 True If v is in the image of a matrix A, then v is in the image of AB, for any matrix B for which AB is defined Solution A simple counterexample is to take a non-zero matrix A and a non-zero vector v in the image of A, and then take B the appropriately { sized) zero matrix Then AB is the zero matrix, which has image } 8 True 9 True True True If b is in the image of A, then A x = b has a unique least-squares solution Solution If b is in the image of A, then the equation A x = b is consistent, ie, it has an actual solution, not just a least squares solution In this case a least-squares solution is just an actual solution to the equation However, if the equation has more than one solution, which is certainly possible, then it has more than one least-squares solution If a square matrix is invertible, then it is diagonalizable [ ] Solution A counterexample is the horizontal shear given by A ==, which is invertible, since det A =, but not diagonalizable, since its only eigenvalue λ = has geometric multiplicity The surface described by x y z = is a hyperboloid of one sheet Solution Since the surface does not intersect the yz-plane where x = ), this is a hyperboloid of two sheets If the contours of a function fx, y) are parallel lines, then the surface described by z = fx, y) is a plane Solution A counterexample is fx, y) = e x+y The contours are lines of the form x + y = ln k, ie, all have slope, but the surface is not a plane For a plane, the contours are evenly spaced parallel lines True False Suppose fa, b) = If the x = a trace of f is concave up at a, b) and the y = b trace of f is concave down at a, b), then a, b) is a saddle Solution Any critical point that is neither a local minimum nor a local maximum is a saddle In this case, since the graph is at a local minimum in one trace, and a local maximum in the other, it cannot be either a local minimum or a local maximum, and so must be a saddle Alternately, we can use the discriminant D = f xx f yy fxy Since f xx > and f yy < at a, b), Da, b) < regardless of the value of f xy

3 3 True False The function fx, y) = x + y has an absolute minimum value subject to the constraint y = x Solution Although the constraint curve is unbounded, it is closed, and so there must be a point on the curve that is closest to the origin; at this point fx, y) = x + y takes on an absolute minimum value subject to the constraint 4 True Suppose fx, y) is a continuous function Then f must have both an absolute minimum value and an absolute maximum value on the region x + y < Solution The region x + y < is not closed since it does not contain the boundary curve x + y = So there is no guarantee that f has an absolute minimum value and an absolute maximum value on the region For example, if f is a non-constant linear function, for example fx, y) = x + 3y, then the absolute minimum and maximum values of f on the region x + y occur on the boundary We can get arbitrarily close to these values by taking points that are strictly inside the circle, but we there are no points strictly inside the circle where f takes on these values True False If fx, y) is a continuous and differentiable function, and fa, b), then the value of f increases most steeply with respect to distance) in the direction of fa, b) Solution In general, the directional derivative of f at a point a, b) in the direction of a unit vector u is given by f u a, b) = fa, b) u = fa, b) cos θ, where θ is the angle between fa, b) and u From this expression we can see that f u a, b) is greatest when θ =, ie, when u is in the same direction as fa, b) Let k be a real number, and consider the following system of equations 3x + k y = 9 x + 3y = k Find all values of k for which the system has no solutions, exactly one solution, and infinitely many solutions Show your work, and write your answers in the blanks provided Note that the answer for one or more of these could be no such value of k exists [ 3 k 9 3 k ] [ ] 3 k Solution Starting with the augmented matrix, we can flip the rows to get 3 k, 9 [ ] 3 k then subtract three times the first row from the second to get k If k 9 9 3k 9, ie, if [ k ±3, then ] the system has a unique solution If k = 3, the transformed augmented matrix becomes 3 3, and the system has infinitely many solutions If k = 3, the transformed augmented matrix becomes [ ], and the system has no solutions 3

4 3 Consider the all ones matrix A defined by A = a) Find the rank of A, and a basis of the image of A Solution The image of A is spanned by the columns of A In this case, since all of the columns are identical, the image is -dimensional The rank of A is, and a basis for the image of A is just b) Find the dimension of the kernel of A, and a basis of the kernel of A Solution By the rank-nullity theorem, rank A + dimker A) = Since rank A =, it must be the case that dimker A) = To find a basis of ker A, we use Gauss-Jordan elimination to reduce A to rref A = Letting x = a, x 3 = b, x 4 = c, x = d, and x = e, then x = a b c d e, and so we have the following parameterization of ker A ker A = a + b + c + d + e : a, b, c, d, e R So a basis for the kernel of A is,,,, 4

5 c) Find the characteristic polynomial of A Hint: The vector v = is an eigenvector of A Solution Let v = Then A v = = v, so λ = is an eigenvalue of A From part b) above we see that the kernel of A has dimension, so λ = is an eigenvalue of A with algebraic multiplicity at least Since the sum of the algebraic multiplicities of the eigenvalues of A must equal, the algebraic multiplicity of λ = must be, and the algebraic multiplicity of λ = must be five Thus, p A λ) = λ) λ) = λ λ ) d) True or False? The matrix A above is the matrix of orthogonal projection onto image of A Please explain your answer carefully Solution The statement is false While ker A = im A), which is characteristic of orthogonal projections, an orthogonal projection has eigenvalues and only Since λ = is an eigenvalue of A, A cannot be an orthogonal projection 3 4 Let P be the plane through the origin in R 3 spanned by v = and v = 3 a) Find an equation for P of the form ax + by + cz = d Solution A normal vector for the plane is n = v v = for the plane is 4x + 8y 4z =, ie, x y + z = i j k 3 3 = 4 8 So the equation 4 b) Find the matrix corresponding to reflection in R 3 over the plane P You may leave your answer as a product of matrices and inverses of matrices Solution As always, the key to finding the matrix of a linear transformation is to pick a useful basis of the domain Since we are looking at reflection over a plane, good choices are vectors in the plane which are fixed by the reflection) and vectors perpendicular to the plane which are sent to their opposite) The vectors v = v 3 = 4 n = is orthogonal to the plane and v = 3 3 are in the plane, and the vector Now we have a convenient basis v, v, v 3 ) to work with In particular, thisis an eigenbasis, since A v = v, A v = v, and A v 3 = v 3 Thus A = SDS, where S = 3 and 3

6 D = In each part, find the matrix of the linear transformation R R described Please simplify your answers completely; no explanation is necessary a) The linear transformation which first reflects over the line y = x, then reflects over the y-axis Solution A good principle in most of these questions is that the matrix of a linear transformation T : R R is [ T e ) T e ) ] Let s think about what happens to e and e one step [ at a time ] e is first sent to e, then fixed, while e is first sent to e, then to e So the matrix is b) The linear transformation which first rotates clockwise by 4, then projects onto the line y = x [ ] / Solution e is first sent to the vector, which is orthogonal to the line y = x, and so / [ [ ] / is then sent to e is first sent to the vector / ], and then fixed, since is along the / / [ / line y = x So the matrix is ] / [ [ 3 c) The linear transformation which sends to and 4] ] [ ] [ 3 to ] [ ] [ ] 3 3 Solution If A is the matrix we re looking for, then we have A = So A = 4 [ ] [ ] [ ] [ ]) [ ] [ ] [ ] /7 /7 /7 /7 = 4 7 = = 4 3 4/7 3/7 /7 /7 Consider the following scenario, in which companies A and B are competing for market share Let at) = proportion of the population that prefers company A s product at time t bt) = proportion of the population that prefers company B s product at time t where t is measured in years Suppose where r is a constant between and at + ) = 7 at) + r bt) bt + ) = 3 at) + r) bt), [ ] 7 r a) Show that λ = is an eigenvalue of the matrix A =, and find a corresponding 3 r eigenvector Note that this eigenvector may involve the constant r [ ] 3 r Solution A I = is not invertible, since its determinant is, 3r, 3r, and 3 r so λ = is an eigenvalue of A, [ To ] find an eigenvector, we need to find a vector in the kernel of r A I One such vector is v = 3

7 b) For what values of r does company A wind up with the larger market share in the long term? Explain your answer carefully Solution The sum of the eigenvalues of A is the trace, which is 7 r Since λ = is one eigenvalue, the other is λ = 7 r Since r, we have 3 λ 7 In all cases, λ < The general solution of the system is xt) [ = c] λ t v +c λ t v Since λ =, and λ <, r in the long term the system approaches c v = c So company A winds up with the larger 3 market share precisely when r > 3 7 Consider the surface given by z = fx, y), where fx, y) = x + xy + y 3 8y a) At the point where x = and y =, at what rate with respect to distance) is the height of the surface above the xy-plane changing in the direction of the vector v = i + j? Solution fx, y) = x+y, x+3y 8, and so f, ) = 4 i 3 j To compute the directional derivative in the direction of v = i + j we first convert v into the unit vector u = i + ) j Then f u, ) = f, ) u = 4 i 3 j i + ) j = 3 = b) At the point where x = and y =, in which directions) is the height of the surface not changing at all? Give your answers) in the form of a unit vector Solution The height does not change in directions orthogonal to that of the gradient vector There are two unit vectors orthogonal to f, ), namely, 3 i + 4 j and 3 i 4 j c) Find the equation of the tangent plane to the graph of fx, y) at the point where x = and y = Solution The tangent plane to the graph of fx, y) at, ) is given by z = f, )+f x, )x ) + f y, )y ) = + 4x ) 3y ) d) Find the equation of the tangent line to the level curve of fx, y) passing through the point, ) Solution The gradient vector f, ) = 4 i 3 j is orthogonal to this tangent line, and so the equation of the line is 4x ) 3y ) =, ie, y = x ) Alternately, the slope of the tangent line must be the negative reciprocal of the slope of the line determined by f, ), which is 3 4 So the slope of the tangent line is 4 3, from which we again have y = x ) 8 Consider the function fx, y, z) = x x xy + y y + z 4 4z Find the critical points of this function, and classify each critical point as a local maximum, a local minimum, or a saddle Solution Setting f = gives x y = x + y = 4x 3 4 = The third equation gives z =, while the first two form the linear system x y = x + y = Adding twice the first equation to the second gives 3x = 3, and so x =, from which y = Sp the only critical point is,, ) To classify this point, we compute the Hessian matrix Hx, y, z) = 7

8 z At,, ), this is H,, ) = For this matrix D =, D = 3, and D 3 = 3 Since all leading principal minors are positive,,, ) is a local minimum 9 Suppose milk production at a dairy farm is modeled by M = fc, F, T ) gallons of milk per day, where C is the number of dairy cows, F is the daily feed ration per cow, measured in pounds, and T is the average temperature of the milking shed, in degrees Fahrenheit The farm currently has 3 cows, each cow is given pounds of feed per day, the average temperature of the milking shed is 7 F, and at these levels the farm is producing gallons of milk per day Suppose that, according to the model, dm dc = 8, dm df dm =, and dt = a) The farmer decides to sell two cows, reducing the herd to 8, and at the same time up the daily feed ration to pounds per cow, while keeping the temperature at 7 F Use a linear approximation to estimate the resulting change in milk production Solution The linear approximation of fc, F, T ) based at 3,, 7) is LC, F, T ) = f3, )+ f C 3,, 7)C 3) + f F 3,, 7)F ) + f T 3,, 7)T 7) = + 8C 3) + F ) T 7) If C is decreased to 8 and at the same time F is increased to and T is held constant, the daily milk production will change by approximately 8 ) + ) ) = gallons, ie, it will decrease by gallons b) Suppose the farmer wants to maintain the original level of milk production, while still selling two cows and upping the daily feed ration to pounds per cow By approximately how much should he raise or lower the temperature of the milking shed? Solution We want the approximate change 8 ) + ) T 7) =, and so T = = 3 4 The temperature in the milking shed should be lowered by approximately 3 4 F a) Describe in words and/or sketch the surface given by the equation x + y = Solution This is a circular cylinder of radius, with axis the z-axis b) Describe in words and/or sketch the intersection of the surface x + y = with the plane z = x y Solution This is an ellipse c) Suppose we wish to find the absolute maximum and minimum values of fx, y, z) = x +3x+y +z, if they exist, subject to the dual constraints x + y = and z = x y Rather than using the two-constraint form of Lagrange multipliers, we can instead use the equation z = x y to rewrite the function as hx, y) = fx, y, x y) = x + x + y y + Use the method of Lagrange multipliers to find the absolute maximum and minimum values, if they exist, of h subject to the constraint x + y = If these values do not exist, explain why not Solution First of all, since h is continuous and the constraint curve x + y = is closed and bounded, h must take on absolute maximum and minimum values subject to the constraint 8

9 Setting gx, y) = x + y =, and then f = λ g, we have x + = λx y = λy x + y = Notice that y cannot be zero since then the second equation would read = Assuming λ, we can divide the first equation by the second to get x+ y = x y, from which xy + y = xy x, ie, x = y Substituting this expression into the constraint, we have y) + y =, ie, y = ± So we have two points,, ) and, ) If λ =, then the first equation gives x = and the second gives y = ; this point is not on the constraint So we have only two points to check h, ) = =, while h, ) = = + So the absolute maximum value of h subject to the two constraints is +, occurring at, ), and the absolute minimum value is, occurring at, ) 9