Fall 2001  Math 2090  Exam 4

1. [15 points] Consider the matrix A = [ \begin{matrix}1 & 0 & 0 \\ 0 & 3 & 7 \\ 1 & 1 & −3 \end{matrix} ].

(a) Find the eigenvalues of the matrix A.

(b) Can the answer to part (a) be used to determine if the matrix A is defective? Explain.

(a) det(A − λI) = (1 − λ)(λ^{2} − 16) (e.g., expand on row 1) eigenvalues: λ_{1} = 1, λ_{2} = 4, and λ_{3} = −4

(b) Since A has distinct eigenvalues, A has three linearly independent eigenvectors, so A is non-defective.

2. [15 points] The matrix A = [\begin{matrix}1 & −3 & 1 \\ −1 & −1 & 1 \\ −1 & −3 & 3 \end{matrix} ] is diagonalizable. The characteristic polynomial of A is p(λ) = det(A − λI) = −λ^{3} + 3λ^{2} − 4 = −(λ − 2)^{2}(λ + 1).

Find a diagonal matrix D and a nonsingular matrix S so that S^{−1}AS = D.

D = [ \begin{matrix}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & −1 \end{matrix} ] S = [\begin{matrix}−3 & 1 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \end{matrix} ] (one choice of D and S)

3. [12 points] One eigenvalue of the matrix A = [\begin{matrix}0 & 10 \\ −1 & 2 \end{matrix} ] is λ_{1} = 1 + 3 i.

Find all eigenvalues and corresponding eigenvectors for this matrix.

λ_{1} = 1 + 3 i, v⃗_{1} = [ \begin{matrix}1 − 3 i \\ 1 \end{matrix} ] λ_{2} = 1 − 3 i, v⃗_{2} = [ \begin{matrix}1 + 3 i \\ 1 \end{matrix} ] (one choice of complex conjugate eigenvectors)

4. [22 points] Consider the differential equation (D + 7)(D^{2} − 4D + 5)y = e^{2x} + 3x^{2}.

(a) Find the general solution of the associated homogeneous differential equation, (D + 7)(D^{2} − 4D + 5)y = 0.

(b) Find the annihilator of the function F (x) = e^{2x} + 3x^{2}.

(c) Determine the form of a particular solution of the non-homogeneous differential equation given above.

(a) y(x) = c_{1}e^{−7x} + c_{2}e^{2x} cos(x) + c_{3}e^{2x} sin(x)  (b) (D − 2)D^{3}  (c) y_{p}(x) = Ae^{2x} + Bx^{2} + Cx + D

5. [18 points] Let A = [\begin{matrix}−2 & 3 \\ −2 & 5 \end{matrix} ], and let x⃗_{1}(t) = [ \begin{matrix}e^{4t} \\ 2e^{4t} \end{matrix} ] and x⃗_{2}(t) = [ \begin{matrix}3e^{−t} \\ e^{−t} \end{matrix} ].

(a) Verify that the function x⃗_{1}(t) is a solution of the system of differential equations x⃗′ = A x⃗.

(b) The function x⃗_{2}(t) is also a solution of the system x⃗′ = A x⃗. (Do not verify this.) Use the Wronskian to show that the set {x⃗_{1}(t), x⃗_{2}(t)} is linearly independent on the interval I = (−∞, ∞).

(c) Solve the initial value problem x⃗′ = A x⃗, x⃗(0) = [ \begin{matrix}2 \\ 9 \end{matrix} ].

(a) A x⃗_{1} = [\begin{matrix}−2 & 3 \\ −2 & 5 \end{matrix} ] [\begin{matrix}e^{4t} \\ 2e^{4t} \end{matrix}] = [\begin{matrix}4e^{4t} \\ 8e^{4t} \end{matrix}] = x′_{1} (b) W [x⃗_{1}, x⃗_{2}] = |\begin{matrix}e^{4t} & 3e^{−t} \\ 2e^{4t} & e^{−t} \end{matrix}| = −5e^{3t} ≠ 0 for all t

(c) x⃗(t) = 5x⃗_{1}(t) − x⃗_{2}(t) solves the IVP 6. [18 points] In each part below, find a fundamental set of real-valued solutions of the system

of differential equations x⃗′ = A x⃗. (You do not have to verify that it is a fundamental set.)

(a) A = [ \begin{matrix}1 & −8 \\ 2 & −7 \end{matrix} ] is defective; eigenvalues: λ_{1} = λ_{2} = −3; associated eigenvector: v⃗ = [ \begin{matrix}2 \\ 1 \end{matrix} ]

(b) A is a certain 2 × 2 matrix with real entries; One eigenvalue of A is λ_{1} = 2 + i, and a corresponding eigenvector is v⃗_{1} = [ \begin{matrix}3 + i \\ 2 \end{matrix} ].

(a) \{x⃗_{1}(t) = e^{−3t} [ \begin{matrix}2 \\ 1 \end{matrix} ] , x⃗_{2}(t) = te^{−3t} [ \begin{matrix}2 \\ 1 \end{matrix} ] + e^{−3t} [ \begin{matrix}1/2 \\ 0 \end{matrix} ]\} is one fundamental set

(b) \{x⃗_{1}(t) = e^{2t} cos(t) [ \begin{matrix}3 \\ 2 \end{matrix} ] + e^{2t} sin(t) [ \begin{matrix}1 \\ 0 \end{matrix} ] , x⃗_{2}(t) = e^{2t} sin(t) [ \begin{matrix}3 \\ 2 \end{matrix} ] + e^{2t} cos(t) [ \begin{matrix}1 \\ 0 \end{matrix} ]\} is one fundamental set

Extra Credit. Find the matrix A from problem 6, part (b).

The eigenvalues λ_{1} = 2 + i and λ_{2} = 2 − i of A are distinct, so the corresponding eigenvectors v⃗_{1} and v⃗_{2} = \overline{(v⃗_{1})} (the complex conjugate) are linearly independent. So the matrix S = [v⃗_{1}v⃗_{2}] is nonsingular, and we have AS = SD, or A = SDS^{−1}, where D is a diagonal matrix with diagonal entries λ_{1} and λ_{2}.

Thus A = SDS^{−1} = [ \begin{matrix}3 + i & 3 − i \\ 2 & 2 \end{matrix} ] [ \begin{matrix}2 + i & 0 \\ 0 & 2 − i \end{matrix} ] [\begin{matrix}2/(4 i) & (−3 + i)/(4 i) \\ −2/(4 i) & (3 + i)/(4 i) \end{matrix}] = [ \begin{matrix}5 & −5 \\ 2 & −1 \end{matrix} ]