1. Diagonalize the matrix A if possible, that is, find an invertible matrix P and a diagonal

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1 . Diagonalize the matrix A if possible, that is, find an invertible matrix P and a diagonal 3 9 matrix D such that A = P DP, for A = (a) P = 4, D =. 3 (b) P = 4, D =. (c) P = 4 8 4, D =. 3 (d) P =, D =. 3 (e) P = , D =. Solution. The answer is (a). Multiply A by the columns of P to check if they are linearly independent eigenvectors. Thus [ 4 ] T, [ 3 ] T and [ ] T are linearly independent eigenvectors with eigenvalues,,, respectively. Alternatively, we can find the eigenvalues as roots of det(a λi) =. With cofactor expansion on the second row we obtain ( λ) [( λ)(4 λ) + 4] = (λ + )(λ + 7λ + ) = (λ + ) (λ + ), giving eigenvalues,,. For each eigenvalue, wecompute a basis for its eigenspace For λ =, the matrix (A + I) = 7 with an eigenvector 3 [,, ] T. 3 9 For λ =, the matrix (A + I) = 3 with linearly independent eigenvectors [ ] T and [ 3 ] T. Choice (a) contains [ 4 ] T = [ ] T and 3 9 [ 3 ] T and [ ] T. All the other matrices P are not invertible or D is not diagonal.. Let W be the subspace spanned by u =, u = 3. Write y = 9 as the sum 4 y = w + z of a vector w in W and a vector z orthogonal to W.

2 (a) y = + 7 (b) y = +, (d) y = + 7 (e) y = 4 8 (c) y = 3 + 3, Solution. The answer is (a). We write y as the sum of the orthogonal projection w = ŷ of the vector y onto W, which is in W, and a vector z orthogonal to W computed as: ŷ = proj u y + proj u y = u y u u u + u y u u u = u + 4u = z = y ŷ = 7, which is orthogonal to W. 8 The decomposition of y is y = ŷ + z = =, 4 3. Find a QR factorization of the matrix A = 3 8. / / [ (a) Q = / / / 3/ ] /, R = 3/ (b) Q = / / 3 4 9, R = (c) Q = / / 3 /, R = (d) Q = 3 8 [3/ ], R = 3/ / (e) Q = 3, R = 3

3 Solution. The answer is (a). In order to find Q we need to find an orthonormal basis for the column space of A using the Gram-Schmidt process on the column vectors a = and a = 8. u = a = 3 u = a proj u a = 8 u a u = 8 + u u We normalize them to get an orthonormal basis: Then Q = v = u u = u 3 = v = u u = u = [ and R = Q T A = 3 3 ].. 3 = Find the least-square solution to the inconsistent system Ax = b, A =, b =. 3 (a) / 4/ (b) 3/7 /4 (c) 3 84 (d) 4 (e) 4

4 Solution. The answer is (a). We solve the consistent system A T Ax = A T b. 3 A T A = 4 3 =, A 9 T b = = 3 9 The matrix A T A is invertible with inverse (A T A) =. 9 4 The solution to the least-squares problem is unique and equal to (A T A) (A T b) = 9 = 3 / = / Alternatively, augment the matrix [ (A ] T A) with (A T b) and take it to reduced row echelon / form to give the same solution 4/ 4 3/7 3/7 3/7 / 9 3/ /4 4/ 4/. Let The eigenvalues of A are A =. (a) 3 ± i (b) ± i (c) ± i (d) ± i (e) + i, + i Solution. The characteristic polynomial of A is given by λ det(a λi) = det = ( λ)( λ) ( )() = λ λ + λ 4 λ + = λ λ +. Setting this equal to zero, we have the roots λ = ( ± 3 4) = ( ± i 4) = 3 ± i.. Which plot depicts the integral curves of the differential equation y = x y?

5 (a) - - (b) (c) - - (d) (e) - - Solution. The ODE is a st order linear ODE and can be solved using the method of integrating factors. The integrating factor is e x. The general solution is y(x) = x Ce x. Thus, integral curves asymptotically approach the line x + as x, and this 4 provides enough information to determine the plot.

6 7. Let y = φ(t) be the unique solution to the initial value problem Compute φ(). ty + y =, y() =. t (a) ln() (b) 3 (c) + ln() (d) 8 (e) Solution. The ODE is a st order linear ODE and can be solved using the method of integrating factors. The integrating factor is t. The general solution is y(t) = ln(t) + C. With t t the given initial condition, C = ln(). Therefore, φ() = C = ln(). 8. Consider the differential equation y = y(y )(y + ). Which of the following statements correctly describes the equilibrium solutions of this differential equation? (a) There are exactly three equilibrium solutions, two unstable and one stable. (b) There is exactly one equilibrium solution, which is unstable. (c) There are exactly three equilibrium solutions, two stable and one unstable. (d) There are exactly two equilibrium solutions, one stable and one unstable. (e) There is exactly one equilibrium solution, which is stable. Solution. There are three distinct roots of the degree 3 polynomial f(y) = y(y )(y + ) and hence three equilibrium solutions given by,,. f(y) is positive for y >, negative for > y >, positive for > y >, and negative for y <. Therefore, the two equilibrium solutions given by, are unstable, and the equilibrium solution given by is stable. 9. Consider a tank with a total capacity of liters which initially contains liters of water mixed together with a chemical substance at a concentration of. grams per liter. Suppose that water drains out of the tank at a rate of liters per hour, while water with a chemical concentration of. grams per liter flows into the tank at rate of liters per hour. Write the initial value problem which allows one to compute the quantity of chemicals in the water at any given time before the tank reaches full capacity. (a) (c) (e) dq = dq Q, Q() =. (b) = Q, dt +3t dt +3t Q() =. dq = + dq Q, Q() =. (d) = + Q, dt t dt 3t Q() =. dq = Q, dt +t Q() =.

7 Solution. Let Q(t) denote the amount of chemicals in the water, units given in grams. The initial condition is that at time t =, there are. = grams in the water, so Q() =. A differential equation describing the rate of change dq is given by dq = rate in rate out. dt dt The rate of chemicals flowing in is given as. =. The rate of chemicals flowing out is given as Q (we use that the total amount of water in the tank at time t is given by +3t + 3t, before the tank reaches full capacity). We conclude that the relevant initial value problem is dq dt = Q, Q() =. + 3t. The given set {x, x, x 3 } is a basis for a subspace R 3. Use the Gram-Schmidt process to produce an orthogonal basis for R 3 and then express y = the vectors in the obtained orthogonal basis. as a linear combination of x =, x =, x3 =. 3 Solution. v =, v = 3 = /3 /3 /3 v 3 = 3 3 /3 /3 = /. 3 /3 / y = If v = = /3v /v + v 3., v =, v3 = then y = /3v /v + /v 3.. Find the unique solution y = φ(t) to the initial value problem y = sin(t), y() =. y + and determine the interval in which that solution exists.

8 Solution. We use the technique of separation of variables. Integrating, we get (y + ) = cos(t) + C. Plugging in the initial condition, we get that C =. Then φ(t) = cos(t). (Note that the solution ψ(t) = + cos(t) is excluded by the initial condition, for this has ψ() =.) The interval of existence is determined by the requirement that cos(t). Since cos(±π/3) = / and cos(t) > / for π/3 < t < π/3, we get that the interval is [ π/3, π/3].. Consider the initial value problem y + t + y =, y() =. t Find the maximal interval on which this IVP is guaranteed to have a unique solution y = φ(t) (by the existence and uniqueness theorem for linear differential equations) and then find that solution. Solution. By the existence and uniqueness theorem for st order linear ODEs, the IVP has a unique solution on the open interval (, ), because the functions and are t+ t continuous on (, ) and discontinuous at t = respectively t =. Solving, we see that the integrating factor is t +, and the general solution for < t < is t + (t + ) y = dt = ln( t) + t + C t where we use that t+ = + to compute the integral. (Note that = ln( t) t t t by our constraint that t <.) Plugging in the initial condition, we see that C =. We therefore get φ(t) = ln( t) + t. t +

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