Differential Equations Quiz: Eigenvalues And Eigenvectors 2x2 Systems
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Eigenvalues And Eigenvectors 2x2 SystemsQuestion 1 of 20

A 2×22 \times 2 real matrix AA has an eigenvalue λ1=5\lambda_1 = 5 with a corresponding eigenvector v⃗1=(1−1)\vec{v}_1 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}. If the trace of AA is 22, what is the determinant of AA?

1010
22
−3-3
−15-15
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Differential Equations Quiz

Differential Equations Quiz: Eigenvalues And Eigenvectors 2x2 Systems

Practice Eigenvalues And Eigenvectors 2x2 Systems in Differential Equations with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Eigenvalues And Eigenvectors 2x2 Systems, giving you a quick way to practice the rules, question types, and explanations that matter most for Differential Equations.

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Question 1

A 2×22 \times 2 real matrix AA has an eigenvalue λ1=5\lambda_1 = 5 with a corresponding eigenvector v⃗1=(1−1)\vec{v}_1 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}. If the trace of AA is 22, what is the determinant of AA?

  1. 1010
  2. 22
  3. −3-3
  4. −15-15 (correct answer)
Explanation: The trace of a matrix is the sum of its eigenvalues. Let the eigenvalues be λ1\lambda_1 and λ2\lambda_2. We are given tr(A)=2\mathrm{tr}(A) = 2 and λ1=5\lambda_1 = 5. Therefore, λ1+λ2=5+λ2=2\lambda_1 + \lambda_2 = 5 + \lambda_2 = 2, which implies λ2=−3\lambda_2 = -3. The determinant of a matrix is the product of its eigenvalues. Thus, det⁡(A)=λ1λ2=(5)(−3)=−15\det(A) = \lambda_1 \lambda_2 = (5)(-3) = -15. The given eigenvector is extra information not needed to find the determinant.

Question 2

A student is asked to find an eigenvector for the matrix A=(2332)A = \begin{pmatrix} 2 & 3 \\ 3 & 2 \end{pmatrix} corresponding to the eigenvalue λ=−1\lambda = -1. Their work is shown below:

Step 1: Set up the equation (A−λI)v⃗=0⃗(A - \lambda I)\vec{v} = \vec{0}. ((2332)−(−1)(1001))v⃗=0⃗\left( \begin{pmatrix} 2 & 3 \\ 3 & 2 \end{pmatrix} - (-1)\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \right) \vec{v} = \vec{0}

Step 2: Simplify the matrix expression. ((2332)+(1001))v⃗=0⃗\left( \begin{pmatrix} 2 & 3 \\ 3 & 2 \end{pmatrix} + \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \right) \vec{v} = \vec{0} (3333)v⃗=0⃗\begin{pmatrix} 3 & 3 \\ 3 & 3 \end{pmatrix} \vec{v} = \vec{0}

Step 3: Write the corresponding system of linear equations for v⃗=(v1v2)\vec{v} = \begin{pmatrix} v_1 \\ v_2 \end{pmatrix}. 3v1+3v2=03v_1 + 3v_2 = 0 3v1+3v2=03v_1 + 3v_2 = 0

Step 4: From the equations in Step 3, the only possible solution is v1=0v_1=0 and v2=0v_2=0. Therefore, the only solution is the trivial solution v⃗=0⃗\vec{v} = \vec{0}.

The student's reasoning is flawed. In which step does the first conceptual error occur?

  1. Step 1
  2. Step 2
  3. Step 3
  4. Step 4 (correct answer)
Explanation: Steps 1, 2, and 3 are all arithmetically and procedurally correct. The error occurs in Step 4. The system of equations 3v1+3v2=03v_1 + 3v_2 = 0 (which is redundant) does not imply that v1v_1 and v2v_2 must both be zero. It implies that v1=−v2v_1 = -v_2. This equation defines a line of solutions in the v1v2v_1v_2-plane. Any non-zero vector satisfying this condition, such as (1−1)\begin{pmatrix} 1 \\ -1 \end{pmatrix}, is a valid eigenvector. The definition of an eigenvector requires it to be a non-zero vector.

Question 3

For the system x⃗′=Ax⃗\vec{x}' = A\vec{x} with A=(−1α−1−3)A = \begin{pmatrix} -1 & \alpha \\ -1 & -3 \end{pmatrix}, the qualitative nature of the equilibrium at the origin depends on the parameter α\alpha. What is the critical value of α\alpha at which the system's eigenvalues transition from being real and distinct to complex conjugates?

  1. α=−3\alpha = -3
  2. α=1\alpha = 1 (correct answer)
  3. α=4\alpha = 4
  4. α=0\alpha = 0
Explanation: The nature of the eigenvalues is determined by the discriminant of the characteristic polynomial. The polynomial is λ2−tr(A)λ+det⁡(A)=0\lambda^2 - \mathrm{tr}(A)\lambda + \det(A) = 0. Here, tr(A)=−1−3=−4\mathrm{tr}(A) = -1 - 3 = -4 and det⁡(A)=(−1)(−3)−(α)(−1)=3+α\det(A) = (-1)(-3) - (\alpha)(-1) = 3 + \alpha. The characteristic equation is λ2+4λ+(3+α)=0\lambda^2 + 4\lambda + (3+\alpha) = 0. The discriminant is D=b2−4ac=42−4(1)(3+α)=16−12−4α=4−4αD = b^2 - 4ac = 4^2 - 4(1)(3+\alpha) = 16 - 12 - 4\alpha = 4 - 4\alpha. The transition from real to complex eigenvalues occurs when the discriminant is zero. Setting D=0D=0 gives 4−4α=04 - 4\alpha = 0, which yields α=1\alpha=1. For α<1\alpha < 1, D>0D > 0 (real, distinct eigenvalues). For α>1\alpha > 1, D<0D < 0 (complex eigenvalues).

Question 4

A 2×22 \times 2 matrix AA has eigenvalues λ1=2\lambda_1=2 and λ2=5\lambda_2=5. If II is the 2×22 \times 2 identity matrix, what are the eigenvalues of the matrix B=A−3IB = A - 3I?

  1. 22 and 55
  2. 55 and 88
  3. 11 and −2-2
  4. −1-1 and 22 (correct answer)
Explanation: Let v⃗\vec{v} be an eigenvector of AA with corresponding eigenvalue λ\lambda. By definition, Av⃗=λv⃗A\vec{v} = \lambda\vec{v}. Now consider the action of matrix BB on v⃗\vec{v}: Bv⃗=(A−3I)v⃗=Av⃗−3Iv⃗=λv⃗−3v⃗=(λ−3)v⃗B\vec{v} = (A - 3I)\vec{v} = A\vec{v} - 3I\vec{v} = \lambda\vec{v} - 3\vec{v} = (\lambda - 3)\vec{v}. This shows that v⃗\vec{v} is also an eigenvector of BB, but with the eigenvalue λ−3\lambda - 3. Applying this to the eigenvalues of AA, the eigenvalues of BB are λ1−3=2−3=−1\lambda_1 - 3 = 2 - 3 = -1 and λ2−3=5−3=2\lambda_2 - 3 = 5 - 3 = 2.

Question 5

A 2×22 \times 2 real matrix AA has eigenvalues λ1=1\lambda_1=1 and λ2=4\lambda_2=4, with corresponding eigenvectors v⃗1=(11)\vec{v}_1=\begin{pmatrix} 1 \\ 1 \end{pmatrix} and v⃗2=(21)\vec{v}_2=\begin{pmatrix} 2 \\ 1 \end{pmatrix}. What is the top-left entry, A11A_{11}, of the matrix AA?

  1. 33
  2. −6-6
  3. 77 (correct answer)
  4. −2-2
Explanation: Let A=(abcd)A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}. The eigenvector equations are Av⃗1=λ1v⃗1A\vec{v}_1 = \lambda_1\vec{v}_1 and Av⃗2=λ2v⃗2A\vec{v}_2 = \lambda_2\vec{v}_2. For λ1=1,v⃗1=(11)\lambda_1=1, \vec{v}_1=\begin{pmatrix} 1 \\ 1 \end{pmatrix}: (abcd)(11)=1(11)\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} 1 \\ 1 \end{pmatrix} = 1 \begin{pmatrix} 1 \\ 1 \end{pmatrix}, which gives the system a+b=1a+b=1 and c+d=1c+d=1. For λ2=4,v⃗2=(21)\lambda_2=4, \vec{v}_2=\begin{pmatrix} 2 \\ 1 \end{pmatrix}: (abcd)(21)=4(21)\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} 2 \\ 1 \end{pmatrix} = 4 \begin{pmatrix} 2 \\ 1 \end{pmatrix}, which gives 2a+b=82a+b=8 and 2c+d=42c+d=4. We want to find a=A11a=A_{11}. We have a system of two equations for aa and bb: a+b=1a+b=1 and 2a+b=82a+b=8. Subtracting the first equation from the second gives (2a+b)−(a+b)=8−1(2a+b)-(a+b) = 8-1, which simplifies to a=7a=7.

Question 6

Consider the system of differential equations x⃗′=Ax⃗\vec{x}' = A\vec{x}, where A=(3k−1−1)A = \begin{pmatrix} 3 & k \\ -1 & -1 \end{pmatrix}. For what value of the parameter kk will the system have repeated eigenvalues?

  1. k=−1k = -1
  2. k=3k = 3
  3. k=4k = 4 (correct answer)
  4. k=−4k = -4
Explanation: The eigenvalues are the roots of the characteristic equation det⁡(A−λI)=0\det(A - \lambda I) = 0. For the given matrix AA, this is (3−λ)(−1−λ)−(k)(−1)=0(3-\lambda)(-1-\lambda) - (k)(-1) = 0, which simplifies to λ2−2λ−3+k=0\lambda^2 - 2\lambda - 3 + k = 0. For repeated eigenvalues, the discriminant of this quadratic equation must be zero. The discriminant is b2−4ac=(−2)2−4(1)(k−3)=4−4k+12=16−4kb^2 - 4ac = (-2)^2 - 4(1)(k-3) = 4 - 4k + 12 = 16 - 4k. Setting the discriminant to zero gives 16−4k=016 - 4k = 0, which solves to k=4k=4.

Question 7

Consider the system x′=Ax\mathbf{x}' = A\mathbf{x} where A=(5−28−3)A = \begin{pmatrix} 5 & -2 \\ 8 & -3 \end{pmatrix} . If λ1=1\lambda_1 = 1 and λ2=1\lambda_2 = 1 are the eigenvalues, which statement about the corresponding eigenvectors is correct?

  1. There is only one linearly independent eigenvector, so the matrix is defective and requires a generalized eigenvector (correct answer)
  2. There are two linearly independent eigenvectors v1=(12)\mathbf{v}_1 = \begin{pmatrix} 1 \\ 2 \end{pmatrix} and $$\mathbf{v}_2 = \begin{pmatrix} 2 \ 1 \end{pmatrix}
  3. The eigenspace has dimension 2, making the matrix diagonalizable with complete eigenspace
  4. The repeated eigenvalue guarantees two distinct eigenvectors in different eigenspaces
Explanation: For eigenvalue λ = 1, we solve (A - I)v = 0: (4−28−4)v=0 \begin{pmatrix} 4 & -2 \\ 8 & -4 \end{pmatrix}\mathbf{v} = \mathbf{0}. The second row is twice the first, so rank(A - I) = 1, making the nullity = 2 - 1 = 1. Thus there's only one linearly independent eigenvector, making the matrix defective. Choice B gives vectors that aren't eigenvectors. Choice C is wrong because the eigenspace has dimension 1. Choice D incorrectly assumes repeated eigenvalues always yield multiple independent eigenvectors.

Question 8

The matrix B=(a3−24)B = \begin{pmatrix} a & 3 \\ -2 & 4 \end{pmatrix} has eigenvalues λ1=1\lambda_1 = 1 and λ2=6\lambda_2 = 6. What is the value of aa, and what is the eigenvector corresponding to λ1=1\lambda_1 = 1?

  1. a=3a = 3; eigenvector $$\mathbf{v}_1 = \begin{pmatrix} 3 \ -2 \end{pmatrix}
  2. a=3a = 3; eigenvector $$\mathbf{v}_1 = \begin{pmatrix} 1 \ -1 \end{pmatrix}
  3. a=−1a = -1; eigenvector $$\mathbf{v}_1 = \begin{pmatrix} 3 \ 2 \end{pmatrix}
  4. a=3a = 3; eigenvector $$\mathbf{v}_1 = \begin{pmatrix} -1 \ 1 \end{pmatrix} (correct answer)
Explanation: Using trace and determinant: tr(B) = a + 4 = λ₁ + λ₂ = 1 + 6 = 7, so a = 3. For λ₁ = 1, solve (B - I)v = 0: (23−23)v=0 \begin{pmatrix} 2 & 3 \\ -2 & 3 \end{pmatrix}\mathbf{v} = \mathbf{0}. From the first row: 2v₁ + 3v₂ = 0, so v₁ = -3v₂/2. Taking v₂ = 2 gives v₁ = -3, so v1=(−32)\mathbf{v}_1 = \begin{pmatrix} -3 \\ 2 \end{pmatrix} or any scalar multiple like $$ \begin{pmatrix} -1 \ 1 \end{pmatrix}

Question 9

Given the matrix C=(04−10)C = \begin{pmatrix} 0 & 4 \\ -1 & 0 \end{pmatrix} , suppose you incorrectly compute one eigenvalue as λ=2i\lambda = 2i and find what appears to be a corresponding eigenvector $$\mathbf{v} = \begin{pmatrix} 2 \ i \end{pmatrix}

  1. The characteristic equation was set up incorrectly; the correct eigenvalues are λ=±2\lambda = \pm 2
  2. The eigenvalue is correct, but the eigenvector calculation contains an algebraic error in solving the linear system (correct answer)
  3. Both the eigenvalue and eigenvector are correct; this represents the complex solution to the system
  4. The determinant calculation was wrong; the actual eigenvalues should be λ=±4i\lambda = \pm 4i
Explanation: The characteristic polynomial is det(C - λI) = λ² + 4 = 0, giving λ = ±2i correctly. For λ = 2i, we solve (C - 2iI)v = 0: (−2i4−1−2i)v=0 \begin{pmatrix} -2i & 4 \\ -1 & -2i \end{pmatrix}\mathbf{v} = \mathbf{0}. From the first row: -2iv₁ + 4v₂ = 0, so v₁ = 2v₂/i = -2iv₂. If v₂ = 1, then v₁ = -2i, giving v=(−2i1)\mathbf{v} = \begin{pmatrix} -2i \\ 1 \end{pmatrix} , not $$ \begin{pmatrix} 2 \ i \end{pmatrix}

Question 10

Consider a 2×2 matrix DD with eigenvalues λ1=3+i\lambda_1 = 3 + i and λ2=3−i\lambda_2 = 3 - i. If one eigenvector is $$\mathbf{v}_1 = \begin{pmatrix} 1 \ 1-i \end{pmatrix}

  1. tr(D)=6\text{tr}(D) = 6, det⁡(D)=10\det(D) = 10, and $$\mathbf{v}_2 = \begin{pmatrix} 1 \ 1+i \end{pmatrix} (correct answer)
  2. tr(D)=6\text{tr}(D) = 6, det⁡(D)=8\det(D) = 8, and $$\mathbf{v}_2 = \begin{pmatrix} -1 \ 1+i \end{pmatrix}
  3. tr(D)=3\text{tr}(D) = 3, det⁡(D)=10\det(D) = 10, and $$\mathbf{v}_2 = \begin{pmatrix} 1 \ 1+i \end{pmatrix}
  4. tr(D)=6\text{tr}(D) = 6, det⁡(D)=10\det(D) = 10, and v2\mathbf{v}_2 must be computed independently
Explanation: For any 2×2 matrix, tr(D) = λ₁ + λ₂ = (3+i) + (3-i) = 6, and det(D) = λ₁λ₂ = (3+i)(3-i) = 9 + 1 = 10. Since D has real entries (complex eigenvalues come in conjugate pairs for real matrices), if λ₁ = 3+i has eigenvector v₁ = (1, 1-i)ᵀ, then λ₂ = 3-i must have eigenvector v₂ = (1, 1+i)ᵀ (the complex conjugate). Choice B has wrong determinant. Choice C has wrong trace. Choice D correctly gives trace and determinant but incorrectly suggests v₂ needs independent computation.

Question 11

Consider the family of matrices Ht=(cos⁡t−sin⁡tsin⁡tcos⁡t)H_t = \begin{pmatrix} \cos t & -\sin t \\ \sin t & \cos t \end{pmatrix} for t∈[0,2π)t \in [0, 2\pi). For which value(s) of tt does HtH_t have real eigenvalues?

  1. All values of tt give real eigenvalues since HtH_t is always symmetric
  2. t=π/2t = \pi/2 and t=3π/2t = 3\pi/2 only
  3. t=0t = 0 and t=πt = \pi only (correct answer)
  4. No values of tt give real eigenvalues except when sin⁡t=0\sin t = 0
Explanation: When you encounter eigenvalue problems involving parametric matrices, your first step should be computing the characteristic polynomial and determining when its discriminant is non-negative for real eigenvalues. For matrix HtH_t, the characteristic polynomial is det⁡(Ht−λI)=det⁡(cos⁡t−λ−sin⁡tsin⁡tcos⁡t−λ)=(cos⁡t−λ)2+sin⁡2t\det(H_t - \lambda I) = \det\begin{pmatrix} \cos t - \lambda & -\sin t \\ \sin t & \cos t - \lambda \end{pmatrix} = (\cos t - \lambda)^2 + \sin^2 t. This gives us λ2−2cos⁡t⋅λ+(cos⁡2t+sin⁡2t)=λ2−2cos⁡t⋅λ+1\lambda^2 - 2\cos t \cdot \lambda + (\cos^2 t + \sin^2 t) = \lambda^2 - 2\cos t \cdot \lambda + 1. Using the quadratic formula: λ=cos⁡t±cos⁡2t−1\lambda = \cos t \pm \sqrt{\cos^2 t - 1}. For real eigenvalues, we need cos⁡2t−1≥0\cos^2 t - 1 \geq 0, which means cos⁡2t≥1\cos^2 t \geq 1. Since −1≤cos⁡t≤1-1 \leq \cos t \leq 1, this occurs only when cos⁡2t=1\cos^2 t = 1, so cos⁡t=±1\cos t = \pm 1. This happens at t=0t = 0 (where cos⁡t=1\cos t = 1) and t=πt = \pi (where cos⁡t=−1\cos t = -1). Option A is wrong because HtH_t is not symmetric—symmetry would require −sin⁡t=sin⁡t-\sin t = \sin t, which only holds when sin⁡t=0\sin t = 0. Option B incorrectly identifies t=π/2t = \pi/2 and t=3π/2t = 3\pi/2, where cos⁡t=0\cos t = 0 and the discriminant is negative. Option D is partially correct about sin⁡t=0\sin t = 0 but misses that this condition is equivalent to cos⁡t=±1\cos t = \pm 1. Remember: For eigenvalue problems, always check the discriminant of the characteristic polynomial. Real eigenvalues require a non-negative discriminant.

Question 12

A 2×2 matrix MM has the property that both (21)\begin{pmatrix} 2 \\ 1 \end{pmatrix} and (1−1)\begin{pmatrix} 1 \\ -1 \end{pmatrix} are eigenvectors with eigenvalues 4 and -2, respectively. If x(0)=(5−1)\mathbf{x}(0) = \begin{pmatrix} 5 \\ -1 \end{pmatrix} , what is x(t)\mathbf{x}(t) for the system x′=Mx\mathbf{x}' = M\mathbf{x}?

  3. (correct answer)
Explanation: The general solution is x(t)=c1e4t(21)+c2e−2t(1−1)\mathbf{x}(t) = c_1e^{4t}\begin{pmatrix} 2 \\ 1 \end{pmatrix} + c_2e^{-2t}\begin{pmatrix} 1 \\ -1 \end{pmatrix} . Using the initial condition x(0)=(5−1)\mathbf{x}(0) = \begin{pmatrix} 5 \\ -1 \end{pmatrix} : $$c_1\begin{pmatrix} 2 \ 1 \end{pmatrix} + c_2\begin{pmatrix} 1 \ -1 \end{pmatrix} = \begin{pmatrix} 5 \ -1 \end{pmatrix}

Question 13

Two students are analyzing the matrix G=(21−14)G = \begin{pmatrix} 2 & 1 \\ -1 & 4 \end{pmatrix} . Student A finds eigenvalues λ1,2=3±i2\lambda_{1,2} = 3 \pm i\sqrt{2} and concludes the system oscillates. Student B finds the same eigenvalues but concludes the system spirals outward. Who is correct and why?

  1. Student A is correct because complex eigenvalues always indicate oscillatory motion around the origin
  2. Student A is correct for small time values, but Student B is correct for large time values
  3. Both are wrong; the eigenvalues should be real since the trace and determinant suggest real solutions
  4. Student B is correct because the positive real part (3 > 0) dominates, causing exponential growth despite oscillation (correct answer)
Explanation: When analyzing linear systems with matrix eigenvalues, the behavior depends on both the real and imaginary parts of the eigenvalues. Complex eigenvalues λ=a±bi\lambda = a \pm bi create solutions of the form eat(cos⁡(bt)+isin⁡(bt))e^{at}(\cos(bt) + i\sin(bt)), where the real part aa controls growth/decay and the imaginary part bb controls oscillation. For the eigenvalues λ1,2=3±i2\lambda_{1,2} = 3 \pm i\sqrt{2}, the real part is a=3>0a = 3 > 0 and the imaginary part is b=2b = \sqrt{2}. This means solutions have the form e3t(cos⁡(2t)+isin⁡(2t))e^{3t}(\cos(\sqrt{2}t) + i\sin(\sqrt{2}t)). The exponential term e3te^{3t} grows without bound as t→∞t \to \infty, while the trigonometric terms create oscillation. The result is a spiral that oscillates while moving outward from the origin. Choice A is wrong because complex eigenvalues don't automatically mean simple oscillation—you must check the real part's sign. Choice B incorrectly suggests the behavior changes over time; the spiral pattern is consistent for all positive times. Choice C misunderstands eigenvalue calculation—complex eigenvalues are perfectly valid when the discriminant (trace)2−4(determinant)<0(\text{trace})^2 - 4(\text{determinant}) < 0. Here, 62−4(9)=36−36=06^2 - 4(9) = 36 - 36 = 0, actually giving 36−36=036 - 36 = 0 (recalculating: tr=6\text{tr} = 6, det⁡=9\det = 9, so 36−36=036 - 36 = 0... wait, det⁡=8+1=9\det = 8 + 1 = 9, so 36−36=036 - 36 = 0). Actually, det⁡=9\det = 9, so we get complex eigenvalues as stated. Remember: for complex eigenvalues a±bia \pm bi, the sign of the real part aa determines stability—positive means unstable (outward spiral), negative means stable (inward spiral).

Question 14

For the system x′=(2−51−2)x\mathbf{x}' = \begin{pmatrix} 2 & -5 \\ 1 & -2 \end{pmatrix}\mathbf{x}, the characteristic polynomial is λ2−1=0\lambda^2 - 1 = 0. If the general solution involves terms of the form c1eλ1tv1+c2eλ2tv2c_1e^{\lambda_1 t}\mathbf{v}_1 + c_2e^{\lambda_2 t}\mathbf{v}_2, which statement about the stability and behavior is correct?

  1. The system is stable because one eigenvalue is negative, creating a stable spiral pattern
  2. The system is unstable due to one positive eigenvalue, with trajectories following hyperbolic paths (correct answer)
  3. The system exhibits oscillatory motion because the eigenvalues are purely imaginary
  4. The system is neutrally stable since the eigenvalues have zero real parts
Explanation: From λ² - 1 = 0, we get λ₁ = 1 and λ₂ = -1. Since one eigenvalue is positive (λ₁ = 1), the system is unstable - solutions grow exponentially in the direction of the corresponding eigenvector. The presence of both positive and negative real eigenvalues creates a saddle point with hyperbolic trajectories. Choice A is wrong because having one negative eigenvalue doesn't make the system stable when there's also a positive one. Choice C is incorrect because the eigenvalues are real, not imaginary. Choice D is wrong because the eigenvalues have non-zero real parts (they are ±1).

Question 15

A student claims that for any 2×2 matrix E=(pqrs)E = \begin{pmatrix} p & q \\ r & s \end{pmatrix} with eigenvalue λ\lambda, the vector $$ \begin{pmatrix} q \ \lambda - p \end{pmatrix}

  1. The claim is valid when q≠0q \neq 0 and λ\lambda is indeed an eigenvalue of EE (correct answer)
  2. The claim is valid only when EE is symmetric and has real eigenvalues
  3. The claim is never valid; eigenvectors cannot be determined by this simple formula
  4. The claim is valid for all 2×2 matrices regardless of the values of p,q,r,sp, q, r, s
Explanation: For the vector (qλ−p)\begin{pmatrix} q \\ \lambda - p \end{pmatrix} to be an eigenvector with eigenvalue λ, we need $$E\begin{pmatrix} q \ \lambda - p \end{pmatrix} = \lambda\begin{pmatrix} q \ \lambda - p \end{pmatrix}

Question 16

Let the matrix A=(4−23−3)A = \begin{pmatrix} 4 & -2 \\ 3 & -3 \end{pmatrix} have eigenvalues λ1\lambda_1 and λ2\lambda_2. Which of the following statements about the eigenvalues is necessarily true?

  1. The eigenvalues are real, with one positive and one negative. (correct answer)
  2. The eigenvalues are real and both are positive.
  3. The eigenvalues are repeated.
  4. The eigenvalues are complex conjugates with non-zero imaginary parts.
Explanation: The sum of the eigenvalues is the trace of the matrix, λ1+λ2=tr(A)=4+(−3)=1\lambda_1 + \lambda_2 = \mathrm{tr}(A) = 4 + (-3) = 1. The product of the eigenvalues is the determinant, λ1λ2=det⁡(A)=(4)(−3)−(−2)(3)=−12+6=−6\lambda_1 \lambda_2 = \det(A) = (4)(-3) - (-2)(3) = -12 + 6 = -6. Since the product is negative, the eigenvalues must have opposite signs (one positive, one negative). They must also be real, because if they were complex conjugates λ=a±bi\lambda = a \pm bi, their product would be a2+b2>0a^2+b^2 > 0. The discriminant of the characteristic polynomial λ2−(tr)λ+det⁡=0\lambda^2 - (\mathrm{tr})\lambda + \det = 0 is D=(tr)2−4(det⁡)=12−4(−6)=25>0D = (\mathrm{tr})^2 - 4(\det) = 1^2 - 4(-6) = 25 > 0, confirming the eigenvalues are real and distinct.

Question 17

For the matrix A=(2231)A = \begin{pmatrix} 2 & 2 \\ 3 & 1 \end{pmatrix}, which of the following vectors is an eigenvector?

  1. (11)\begin{pmatrix} 1 \\ 1 \end{pmatrix} (correct answer)
  2. (1−1)\begin{pmatrix} 1 \\ -1 \end{pmatrix}
  3. (32)\begin{pmatrix} 3 \\ 2 \end{pmatrix}
  4. (23)\begin{pmatrix} 2 \\ 3 \end{pmatrix}
Explanation: A vector v⃗\vec{v} is an eigenvector of AA if Av⃗=λv⃗A\vec{v} = \lambda\vec{v} for some scalar eigenvalue λ\lambda. We can test each option. For choice A: Av⃗=(2231)(11)=(2+23+1)=(44)=4(11)A\vec{v} = \begin{pmatrix} 2 & 2 \\ 3 & 1 \end{pmatrix} \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 2+2 \\ 3+1 \end{pmatrix} = \begin{pmatrix} 4 \\ 4 \end{pmatrix} = 4 \begin{pmatrix} 1 \\ 1 \end{pmatrix}. Since Av⃗=4v⃗A\vec{v} = 4\vec{v}, the vector (11)\begin{pmatrix} 1 \\ 1 \end{pmatrix} is an eigenvector with eigenvalue λ=4\lambda=4. The other options are not scalar multiples of the original vector when multiplied by AA.

Question 18

Which of the following is an eigenvector for the eigenvalue λ=2+i\lambda = 2+i of the matrix A=(11−23)A = \begin{pmatrix} 1 & 1 \\ -2 & 3 \end{pmatrix}?

  1. (11−i)\begin{pmatrix} 1 \\ 1-i \end{pmatrix}
  2. (11+i)\begin{pmatrix} 1 \\ 1+i \end{pmatrix} (correct answer)
  3. (1+i1)\begin{pmatrix} 1+i \\ 1 \end{pmatrix}
  4. (1−1−i)\begin{pmatrix} 1 \\ -1-i \end{pmatrix}
Explanation: To find an eigenvector v⃗\vec{v} for λ=2+i\lambda = 2+i, we solve (A−λI)v⃗=0⃗(A - \lambda I)\vec{v} = \vec{0}. The matrix A−λIA - \lambda I is (1−(2+i)1−23−(2+i))=(−1−i1−21−i)\begin{pmatrix} 1-(2+i) & 1 \\ -2 & 3-(2+i) \end{pmatrix} = \begin{pmatrix} -1-i & 1 \\ -2 & 1-i \end{pmatrix}. The first row gives the equation (−1−i)v1+v2=0(-1-i)v_1 + v_2 = 0, which simplifies to v2=(1+i)v1v_2 = (1+i)v_1. If we choose v1=1v_1 = 1, then v2=1+iv_2 = 1+i. So, an eigenvector is (11+i)\begin{pmatrix} 1 \\ 1+i \end{pmatrix}.

Question 19

The characteristic polynomial of the matrix A=(−182−1)A = \begin{pmatrix} -1 & 8 \\ 2 & -1 \end{pmatrix} is p(λ)=λ2+2λ−15p(\lambda) = \lambda^2 + 2\lambda - 15. Which of the following is an eigenvector of AA?

  1. (12)\begin{pmatrix} 1 \\ 2 \end{pmatrix}
  2. (2−1)\begin{pmatrix} 2 \\ -1 \end{pmatrix}
  3. (−2−1)\begin{pmatrix} -2 \\ -1 \end{pmatrix}
  4. (21)\begin{pmatrix} 2 \\ 1 \end{pmatrix} (correct answer)
Explanation: First, find the eigenvalues by solving p(λ)=0p(\lambda) = 0. λ2+2λ−15=(λ+5)(λ−3)=0\lambda^2 + 2\lambda - 15 = (\lambda+5)(\lambda-3) = 0. The eigenvalues are λ1=3\lambda_1 = 3 and λ2=−5\lambda_2 = -5. Now, find an eigenvector for either eigenvalue. Let's use λ1=3\lambda_1 = 3. We solve (A−3I)v⃗=0⃗(A-3I)\vec{v} = \vec{0}: (−1−382−1−3)(v1v2)=(−482−4)(v1v2)=(00)\begin{pmatrix} -1-3 & 8 \\ 2 & -1-3 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} -4 & 8 \\ 2 & -4 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}. Both rows give the equation −4v1+8v2=0-4v_1+8v_2=0 (or 2v1−4v2=02v_1-4v_2=0), which simplifies to v1=2v2v_1 = 2v_2. Choosing v2=1v_2=1 gives v1=2v_1=2. So, an eigenvector is (21)\begin{pmatrix} 2 \\ 1 \end{pmatrix}. This matches option D.

Question 20

Consider the system x⃗′=Ax⃗\vec{x}' = A\vec{x} where A=(1−113)A = \begin{pmatrix} 1 & -1 \\ 1 & 3 \end{pmatrix}. Which of the following describes the form of the general solution x⃗(t)\vec{x}(t)?

  1. x⃗(t)=c1v⃗1eat+c2v⃗2ebt\vec{x}(t) = c_1\vec{v}_1 e^{at} + c_2\vec{v}_2 e^{bt} for distinct real numbers a,ba, b.
  2. x⃗(t)=c1v⃗1e2t+c2v⃗2e2t\vec{x}(t) = c_1\vec{v}_1 e^{2t} + c_2\vec{v}_2 e^{2t} where v⃗1\vec{v}_1 and v⃗2\vec{v}_2 are linearly independent eigenvectors.
  3. x⃗(t)=eat(c1w⃗1cos⁡(bt)+c2w⃗2sin⁡(bt))\vec{x}(t) = e^{at}(c_1\vec{w}_1 \cos(bt) + c_2\vec{w}_2 \sin(bt)) for real aa and b≠0b \neq 0.
  4. x⃗(t)=c1v⃗1e2t+c2(tv⃗1+w⃗)e2t\vec{x}(t) = c_1\vec{v}_1 e^{2t} + c_2(t\vec{v}_1 + \vec{w})e^{2t} for some vectors v⃗1,w⃗\vec{v}_1, \vec{w}. (correct answer)
Explanation: First, we find the eigenvalues of AA. The characteristic equation is det⁡(A−λI)=(1−λ)(3−λ)−(−1)(1)=λ2−4λ+3+1=λ2−4λ+4=(λ−2)2=0\det(A-\lambda I) = (1-\lambda)(3-\lambda) - (-1)(1) = \lambda^2 - 4\lambda + 3 + 1 = \lambda^2 - 4\lambda + 4 = (\lambda-2)^2 = 0. This gives a repeated eigenvalue λ=2\lambda = 2. For a system with a repeated eigenvalue that has only one linearly independent eigenvector, the general solution takes the form x⃗(t)=c1v⃗eλt+c2(tv⃗+w⃗)eλt\vec{x}(t) = c_1\vec{v}e^{\lambda t} + c_2(t\vec{v} + \vec{w})e^{\lambda t}, where v⃗\vec{v} is an eigenvector and w⃗\vec{w} is a generalized eigenvector. Option D correctly represents this form with λ=2\lambda=2.