Linear Algebra Quiz: Linear Algebra In Des
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Linear Algebra In DesQuestion 1 of 20

Consider the system x′=Ax\mathbf{x}' = A\mathbf{x} where AA has eigenvalues λ1=3\lambda_1 = 3 and λ2=−1\lambda_2 = -1 with corresponding eigenvectors v1=(21)\mathbf{v}_1 = \begin{pmatrix} 2 \\ 1 \end{pmatrix} and v2=(1−1)\mathbf{v}_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix} . Which initial condition will cause the solution to approach the origin as t→∞t \to \infty?

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Linear Algebra Quiz: Linear Algebra In Des

Practice Linear Algebra In Des in Linear Algebra with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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Question 1

Consider the system x′=Ax\mathbf{x}' = A\mathbf{x} where AA has eigenvalues λ1=3\lambda_1 = 3 and λ2=−1\lambda_2 = -1 with corresponding eigenvectors v1=(21)\mathbf{v}_1 = \begin{pmatrix} 2 \\ 1 \end{pmatrix} and v2=(1−1)\mathbf{v}_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix} . Which initial condition will cause the solution to approach the origin as t→∞t \to \infty?

  2. (correct answer)
Explanation: The general solution is x(t)=c1e3tv1+c2e−tv2\mathbf{x}(t) = c_1 e^{3t}\mathbf{v}_1 + c_2 e^{-t}\mathbf{v}_2. For the solution to approach the origin, we need the coefficient of the unstable mode (e3te^{3t}) to be zero, so c1=0c_1 = 0. This happens when the initial condition lies on the stable eigenspace spanned by $$\mathbf{v}_2 = \begin{pmatrix} 1 \ -1 \end{pmatrix}

Question 2

The system y⃗′=Ay⃗\vec{y}' = A\vec{y} has an equilibrium point at the origin. The eigenvalues of the coefficient matrix AA are λ1=−3\lambda_1 = -3 and λ2=1\lambda_2 = 1. How would the behavior of solutions near the origin be classified?

  1. A stable node, where all trajectories approach the origin.
  2. An unstable node, where all trajectories move away from the origin.
  3. A saddle point, where some trajectories approach and others move away from the origin. (correct answer)
  4. A stable spiral, where all trajectories spiral towards the origin.
Explanation: The stability of the equilibrium point at the origin is determined by the eigenvalues of AA. Since the eigenvalues are real and have opposite signs (−3<0-3 < 0 and 1>01 > 0), the origin is a saddle point. Trajectories corresponding to the negative eigenvalue approach the origin, while trajectories corresponding to the positive eigenvalue move away. A stable node occurs when both eigenvalues are negative. An unstable node occurs when both are positive. A spiral occurs for complex eigenvalues.

Question 3

The coefficient matrix AA for a real system x⃗′=Ax⃗\vec{x}' = A\vec{x} has a complex eigenvalue λ=−2+3i\lambda = -2 + 3i. Which statement accurately describes the behavior of the system's solutions near the origin?

  1. Solutions oscillate and grow in magnitude, spiraling away from the origin.
  2. Solutions oscillate and decay in magnitude, spiraling towards the origin. (correct answer)
  3. Solutions move along straight lines in a saddle point pattern.
  4. Solutions follow closed, stable elliptical orbits around the origin.
Explanation: For a complex eigenvalue λ=a+bi\lambda = a + bi, the real part aa governs the growth or decay of the solution's magnitude, and the imaginary part bb governs the oscillation. Here, a=−2a = -2 is negative, so the magnitude decays exponentially (e−2te^{-2t}). The imaginary part b=3b=3 is non-zero, causing oscillation. The combination of decay and oscillation results in trajectories that spiral towards the origin, which is classified as a stable spiral.

Question 4

The system x⃗′=Ax⃗\vec{x}' = A\vec{x} is governed by a matrix AA which has a repeated eigenvalue λ=2\lambda=2 with algebraic multiplicity 2 but geometric multiplicity 1. If v⃗\vec{v} is the unique (up to scaling) eigenvector, what is the correct form of the general solution?

  1. x⃗(t)=c1e2tv⃗+c2te2tv⃗\vec{x}(t) = c_1 e^{2t}\vec{v} + c_2 t e^{2t}\vec{v}
  2. x⃗(t)=c1e2tv⃗+c2e2t(tv⃗+u⃗)\vec{x}(t) = c_1 e^{2t}\vec{v} + c_2 e^{2t}(t\vec{v} + \vec{u}), where (A−2I)u⃗=v⃗(A-2I)\vec{u} = \vec{v}. (correct answer)
  3. x⃗(t)=c1e2tv⃗1+c2e2tv⃗2\vec{x}(t) = c_1 e^{2t}\vec{v}_1 + c_2 e^{2t}\vec{v}_2, where v⃗1\vec{v}_1 and v⃗2\vec{v}_2 are linearly independent.
  4. x⃗(t)=c1e2tcos⁡(t)v⃗+c2e2tsin⁡(t)u⃗\vec{x}(t) = c_1 e^{2t}\cos(t)\vec{v} + c_2 e^{2t}\sin(t)\vec{u}, for some vector u⃗\vec{u}.
Explanation: When an eigenvalue λ\lambda has algebraic multiplicity greater than its geometric multiplicity (a defective eigenvalue), there are not enough eigenvectors to form a full basis of solutions. The general solution is constructed using the eigenvector v⃗\vec{v} and a generalized eigenvector u⃗\vec{u}, which is found by solving (A−λI)u⃗=v⃗(A-\lambda I)\vec{u} = \vec{v}. The second linearly independent solution has the form eλt(tv⃗+u⃗)e^{\lambda t}(t\vec{v} + \vec{u}). Choice A is a common mistake that omits the generalized eigenvector u⃗\vec{u}. Choice C is incorrect because only one linearly independent eigenvector exists. Choice D describes the form for complex eigenvalues.

Question 5

A 2×22 \times 2 matrix AA for the system x⃗′=Ax⃗\vec{x}' = A\vec{x} has eigenvalues λ1=0\lambda_1 = 0 and λ2=−5\lambda_2 = -5. Let v⃗1\vec{v}_1 be the eigenvector corresponding to λ1\lambda_1. What does the zero eigenvalue imply about the solutions?

  1. All solutions decay to the origin as t→∞t \to \infty.
  2. The system possesses an entire line of equilibrium points. (correct answer)
  3. All non-trivial solutions grow without bound as t→∞t \to \infty.
  4. All solutions are periodic and form closed orbits.
Explanation: An eigenvalue of λ1=0\lambda_1 = 0 means that for its corresponding eigenvector v⃗1\vec{v}_1, we have Av⃗1=0v⃗1=0⃗A\vec{v}_1 = 0\vec{v}_1 = \vec{0}. Consider any point on the line spanned by v⃗1\vec{v}_1, say x⃗=cv⃗1\vec{x} = c\vec{v}_1. The derivative at this point is x⃗′=A(cv⃗1)=c(Av⃗1)=0⃗\vec{x}' = A(c\vec{v}_1) = c(A\vec{v}_1) = \vec{0}. A zero derivative means the point is an equilibrium (or fixed) point. Thus, the entire line of points cv⃗1c\vec{v}_1 consists of equilibria. The general solution is x⃗(t)=c1e0tv⃗1+c2e−5tv⃗2=c1v⃗1+c2e−5tv⃗2\vec{x}(t) = c_1e^{0t}\vec{v}_1 + c_2e^{-5t}\vec{v}_2 = c_1\vec{v}_1 + c_2e^{-5t}\vec{v}_2. As t→∞t \to \infty, solutions approach a point on this line.

Question 6

A system of linear differential equations is given by x⃗′=Ax⃗\vec{x}' = A\vec{x}. A proposed non-trivial solution is of the form x⃗(t)=e2tv⃗\vec{x}(t) = e^{2t}\vec{v}. What condition must the matrix AA and the vector v⃗\vec{v} satisfy for this to be a valid solution?

  1. Av⃗=2v⃗A\vec{v} = 2\vec{v} and v⃗\vec{v} must be a non-zero vector. (correct answer)
  2. Av⃗=0⃗A\vec{v} = \vec{0} and v⃗\vec{v} can be any vector.
  3. The determinant of AA must be 22.
  4. A=2IA = 2I, where II is the identity matrix.
Explanation: To check if x⃗(t)=e2tv⃗\vec{x}(t) = e^{2t}\vec{v} is a solution, we substitute it into the differential equation. The derivative is x⃗′(t)=2e2tv⃗\vec{x}'(t) = 2e^{2t}\vec{v}. The right side is Ax⃗(t)=A(e2tv⃗)=e2t(Av⃗)A\vec{x}(t) = A(e^{2t}\vec{v}) = e^{2t}(A\vec{v}). Equating the two gives 2e2tv⃗=e2t(Av⃗)2e^{2t}\vec{v} = e^{2t}(A\vec{v}). Since e2te^{2t} is never zero, we can divide by it to obtain 2v⃗=Av⃗2\vec{v} = A\vec{v}. This is the definition of an eigenvector v⃗\vec{v} with eigenvalue λ=2\lambda=2. For the solution to be non-trivial, the eigenvector v⃗\vec{v} must be non-zero.

Question 7

Consider the system (x1′x2′)=(2102)(x1x2)\begin{pmatrix} x_1' \\ x_2' \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \end{pmatrix}. If the initial condition is x(0)=(11)\mathbf{x}(0) = \begin{pmatrix} 1 \\ 1 \end{pmatrix} , which expression represents x1(t)x_1(t)?

  1. x1(t)=e2t+te2tx_1(t) = e^{2t} + te^{2t}
  2. x1(t)=e2tx_1(t) = e^{2t}
  3. x1(t)=e2t+2te2tx_1(t) = e^{2t} + 2te^{2t}
  4. x1(t)=(1+t)e2tx_1(t) = (1 + t)e^{2t} (correct answer)
Explanation: The matrix has repeated eigenvalue λ=2\lambda = 2 with only one eigenvector (10)\begin{pmatrix} 1 \\ 0 \end{pmatrix}. The general solution involves e2te^{2t} and te2tte^{2t} terms. The fundamental matrix is Φ(t)=(e2tte2t0e2t)\Phi(t) = \begin{pmatrix} e^{2t} & te^{2t} \\ 0 & e^{2t} \end{pmatrix} . With x(0)=(11)\mathbf{x}(0) = \begin{pmatrix} 1 \\ 1 \end{pmatrix} , we get x1(t)=e2t⋅1+te2t⋅1=(1+t)e2tx_1(t) = e^{2t} \cdot 1 + te^{2t} \cdot 1 = (1 + t)e^{2t}.

Question 8

A population model is given by (x′y′)=(0.10.20.3−0.1)(xy)\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} 0.1 & 0.2 \\ 0.3 & -0.1 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} where x(t)x(t) and y(t)y(t) represent two interacting species. The eigenvalues are approximately λ1≈0.236\lambda_1 \approx 0.236 and λ2≈−0.236\lambda_2 \approx -0.236. What does this suggest about the long-term behavior?

  1. Both populations will grow exponentially without bound due to the positive diagonal entries
  2. The populations will oscillate indefinitely around their equilibrium values due to nearly equal magnitude eigenvalues
  3. Both populations will decay to zero since the sum of eigenvalues equals the trace which is zero
  4. One linear combination of the populations grows while another decays, leading to eventual dominance of one mode (correct answer)
Explanation: When analyzing linear dynamical systems like population models, the eigenvalues of the coefficient matrix determine the long-term behavior. Each eigenvalue corresponds to a fundamental mode of the system, and the sign of the real part tells you whether that mode grows or decays over time. With eigenvalues λ1≈0.236>0\lambda_1 \approx 0.236 > 0 and λ2≈−0.236<0\lambda_2 \approx -0.236 < 0, you have one growing mode and one decaying mode. The general solution is a linear combination of these modes: eλ1tv1+eλ2tv2e^{\lambda_1 t}\mathbf{v_1} + e^{\lambda_2 t}\mathbf{v_2}, where v1\mathbf{v_1} and v2\mathbf{v_2} are the corresponding eigenvectors. As t→∞t \to \infty, the positive eigenvalue dominates since e0.236te^{0.236t} grows while e−0.236t→0e^{-0.236t} \to 0. This means D is correct—one linear combination grows while another decays. A incorrectly focuses on diagonal entries rather than eigenvalues. The individual matrix entries don't directly determine stability; eigenvalues do. B mistakes the similar magnitudes for oscillatory behavior, but these are real eigenvalues, not complex ones. Complex eigenvalues with nonzero imaginary parts create oscillations. C misapplies the trace concept. While the trace (sum of eigenvalues) being zero means the determinant of eAte^{At} stays constant, it doesn't mean both populations decay—one grows and one shrinks. Study tip: For linear systems, eigenvalue signs determine fate: positive means growth, negative means decay, and complex means oscillation. Don't get distracted by individual matrix entries.

Question 9

Consider the matrix A=(01−k0)A = \begin{pmatrix} 0 & 1 \\ -k & 0 \end{pmatrix} where k>0k > 0. For what value of kk will the corresponding differential system x′=Ax\mathbf{x}' = A\mathbf{x} have solutions that complete exactly one full revolution around the origin in time t=πt = \pi?

  1. k=1k = 1
  2. k=2k = 2
  3. k=4k = 4 (correct answer)
  4. k=πk = \pi
Explanation: The eigenvalues are λ=±ik\lambda = \pm i\sqrt{k} (purely imaginary), creating a center with circular/elliptical orbits. The angular frequency is ω=k\omega = \sqrt{k}. For one complete revolution in time π\pi, we need ω⋅π=2π\omega \cdot \pi = 2\pi, so k⋅π=2π\sqrt{k} \cdot \pi = 2\pi, which gives k=2\sqrt{k} = 2, thus k=4k = 4.

Question 10

For the system x′=Ax\mathbf{x}' = A\mathbf{x} with A=(abcd)A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} , suppose det⁡(A)=−6\det(A) = -6 and tr(A)=1\text{tr}(A) = 1. What type of critical point occurs at the origin?

  1. Stable node, since the trace is positive and determinant is negative
  2. Unstable node, since both eigenvalues have the same sign as the trace
  3. Saddle point, since the determinant is negative indicating eigenvalues of opposite signs (correct answer)
  4. Spiral point, since the discriminant Δ=tr2−4det⁡>0\Delta = \text{tr}^2 - 4\det > 0 creates complex eigenvalues
Explanation: For eigenvalues λ1,λ2\lambda_1, \lambda_2: λ1+λ2=tr(A)=1\lambda_1 + \lambda_2 = \text{tr}(A) = 1 and λ1λ2=det⁡(A)=−6\lambda_1 \lambda_2 = \det(A) = -6. Since the product is negative, the eigenvalues have opposite signs. This always creates a saddle point regardless of the trace value. A negative determinant is the key indicator of a saddle.

Question 11

For the linear system x′=Ax\mathbf{x}' = A\mathbf{x} where AA has eigenvalues λ1=−2+3i\lambda_1 = -2 + 3i and λ2=−2−3i\lambda_2 = -2 - 3i, which statement best describes the long-term behavior of solutions?

  1. Solutions spiral outward from the origin with increasing amplitude due to the positive real parts
  2. Solutions spiral inward toward the origin with decreasing amplitude due to the negative real parts (correct answer)
  3. Solutions oscillate in closed elliptical orbits around the origin due to the purely imaginary components
  4. Solutions exhibit damped oscillations that approach a limit cycle rather than the origin itself
Explanation: Complex eigenvalues λ=α±βi\lambda = \alpha \pm \beta i create spiral behavior. The real part α=−2<0\alpha = -2 < 0 determines stability: negative real parts cause exponential decay, so solutions spiral inward. The imaginary parts ±3i\pm 3i create the oscillatory/spiral motion with angular frequency related to ∣β∣=3|\beta| = 3. This makes the origin a stable spiral point.

Question 12

A spring-mass system with damping leads to the equation mx¨+cx˙+kx=0m\ddot{x} + c\dot{x} + kx = 0. When written as a first-order system y′=Ay\mathbf{y}' = A\mathbf{y} with $$\mathbf{y} = \begin{pmatrix} x \ \dot{x} \end{pmatrix}

  1. c2=4mkc^2 = 4mk (correct answer)
  2. c2=mkc^2 = mk
  3. c=2mkc = 2\sqrt{mk}
  4. c=4mkc = \sqrt{4mk}
Explanation: The system matrix is A=(01−k/m−c/m)A = \begin{pmatrix} 0 & 1 \\ -k/m & -c/m \end{pmatrix} . The characteristic equation is λ2+(c/m)λ+k/m=0\lambda^2 + (c/m)\lambda + k/m = 0. For repeated real roots (critical damping), the discriminant must be zero: (c/m)2−4(k/m)=0(c/m)^2 - 4(k/m) = 0, which gives c2/m2=4k/mc^2/m^2 = 4k/m, so c2=4mkc^2 = 4mk. Note that choices C and D are equivalent to A, but A is the standard form.

Question 13

A second-order linear differential equation y′′+py′+qy=0y'' + py' + qy = 0 can be written as a first-order system x′=Ax\mathbf{x}' = A\mathbf{x} where x=(yy′)\mathbf{x} = \begin{pmatrix} y \\ y' \end{pmatrix} . If the characteristic equation has roots r1=−1r_1 = -1 and r2=−3r_2 = -3, what are the eigenvalues of the coefficient matrix AA?

  1. λ1=1,λ2=3\lambda_1 = 1, \lambda_2 = 3
  2. λ1=−1,λ2=−3\lambda_1 = -1, \lambda_2 = -3 (correct answer)
  3. λ1=−1,λ2=3\lambda_1 = -1, \lambda_2 = 3
  4. λ1=1,λ2=−3\lambda_1 = 1, \lambda_2 = -3
Explanation: When converting y′′+py′+qy=0y'' + py' + qy = 0 to the system (yy′)′=(01−q−p)(yy′)\begin{pmatrix} y \\ y' \end{pmatrix}' = \begin{pmatrix} 0 & 1 \\ -q & -p \end{pmatrix}\begin{pmatrix} y \\ y' \end{pmatrix}, the eigenvalues of the coefficient matrix are exactly the roots of the characteristic equation of the original second-order equation. This is because det⁡(A−λI)=λ2+pλ+q\det(A - \lambda I) = \lambda^2 + p\lambda + q, which matches the characteristic equation r2+pr+q=0r^2 + pr + q = 0.

Question 14

Two competing species have populations x(t)x(t) and y(t)y(t) modeled by the linear system p⃗′=Ap⃗\vec{p}' = A\vec{p}, where p⃗=(xy)\vec{p} = \begin{pmatrix} x \\ y \end{pmatrix}. The matrix AA has eigenvalues λ1=0.5\lambda_1 = 0.5 and λ2=−0.2\lambda_2 = -0.2, with corresponding eigenvectors v⃗1=(11)\vec{v}_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix} and v⃗2=(1−1)\vec{v}_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}. If the initial populations are x(0)=300x(0)=300 and y(0)=100y(0)=100, what is the long-term behavior of the population ratio y/xy/x as t→∞t \to \infty?

  1. The ratio approaches 1/31/3.
  2. The ratio approaches −1-1.
  3. The ratio approaches 00.
  4. The ratio approaches 11. (correct answer)
Explanation: The general solution is p⃗(t)=c1e0.5tv⃗1+c2e−0.2tv⃗2\vec{p}(t) = c_1 e^{0.5t} \vec{v}_1 + c_2 e^{-0.2t} \vec{v}_2. At t=0t=0, we have c1(11)+c2(1−1)=(300100)c_1\begin{pmatrix} 1 \\ 1 \end{pmatrix} + c_2\begin{pmatrix} 1 \\ -1 \end{pmatrix} = \begin{pmatrix} 300 \\ 100 \end{pmatrix}, which gives c1=200,c2=100c_1=200, c_2=100. So x(t)=200e0.5t+100e−0.2tx(t) = 200e^{0.5t} + 100e^{-0.2t} and y(t)=200e0.5t−100e−0.2ty(t) = 200e^{0.5t} - 100e^{-0.2t}. As t→∞t \to \infty, the e0.5te^{0.5t} term dominates. The ratio becomes y(t)x(t)≈200e0.5t200e0.5t=1\frac{y(t)}{x(t)} \approx \frac{200e^{0.5t}}{200e^{0.5t}} = 1. Conceptually, the solution vector becomes parallel to the eigenvector associated with the largest positive eigenvalue, which is v⃗1=(11)\vec{v}_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}. The ratio of the components of this vector is 1/1=11/1=1.

Question 15

A real 2×22 \times 2 matrix AA has a complex eigenpair λ=2+5i\lambda = 2+5i and v⃗=(12−i)\vec{v} = \begin{pmatrix} 1 \\ 2-i \end{pmatrix}. The complex solution z⃗(t)=e(2+5i)tv⃗\vec{z}(t) = e^{(2+5i)t}\vec{v} can be separated into its real and imaginary parts to find two linearly independent real solutions. Which of the following is a valid real-valued solution to the system x⃗′=Ax⃗\vec{x}' = A\vec{x}?

  1. e2t(cos⁡(5t)2cos⁡(5t)+sin⁡(5t))e^{2t}\begin{pmatrix} \cos(5t) \\ 2\cos(5t) + \sin(5t) \end{pmatrix} (correct answer)
  2. e2t(cos⁡(5t)2cos⁡(5t)−sin⁡(5t))e^{2t}\begin{pmatrix} \cos(5t) \\ 2\cos(5t) - \sin(5t) \end{pmatrix}
  3. e5t(cos⁡(2t)cos⁡(2t)−sin⁡(2t))e^{5t}\begin{pmatrix} \cos(2t) \\ \cos(2t) - \sin(2t) \end{pmatrix}
  4. e2t(12)cos⁡(5t)−e2t(0−1)sin⁡(5t)e^{2t}\begin{pmatrix} 1 \\ 2 \end{pmatrix}\cos(5t) - e^{2t}\begin{pmatrix} 0 \\ -1 \end{pmatrix}\sin(5t)
Explanation: We use Euler's formula eiθ=cos⁡θ+isin⁡θe^{i\theta} = \cos\theta + i\sin\theta to expand the complex solution: z⃗(t)=e2t(cos⁡(5t)+isin⁡(5t))(12−i)=e2t(cos⁡(5t)+isin⁡(5t)(2−i)(cos⁡(5t)+isin⁡(5t)))\vec{z}(t) = e^{2t}(\cos(5t) + i\sin(5t)) \begin{pmatrix} 1 \\ 2-i \end{pmatrix} = e^{2t} \begin{pmatrix} \cos(5t) + i\sin(5t) \\ (2-i)(\cos(5t) + i\sin(5t)) \end{pmatrix}. Expanding the bottom component gives 2cos⁡(5t)+2isin⁡(5t)−icos⁡(5t)+sin⁡(5t)=(2cos⁡(5t)+sin⁡(5t))+i(2sin⁡(5t)−cos⁡(5t))2\cos(5t) + 2i\sin(5t) - i\cos(5t) + \sin(5t) = (2\cos(5t) + \sin(5t)) + i(2\sin(5t) - \cos(5t)). The real part of the full vector solution is Re(z⃗(t))=e2t(cos⁡(5t)2cos⁡(5t)+sin⁡(5t))\text{Re}(\vec{z}(t)) = e^{2t}\begin{pmatrix} \cos(5t) \\ 2\cos(5t) + \sin(5t) \end{pmatrix}, which is a valid real solution. This matches choice A. Choice B has a sign error in the expansion. Choice C swaps the real and imaginary parts of λ\lambda. Choice D is an equivalent but rearranged form of choice A; however, choice A is the direct result of taking the real part.

Question 16

A system of differential equations is described by x⃗′=Ax⃗\vec{x}' = A\vec{x}. The 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}. Which of the following represents the general solution x⃗(t)\vec{x}(t)?

  1. x⃗(t)=c1e−t(−21)+c2e−4t(11)\vec{x}(t) = c_1 e^{-t} \begin{pmatrix} -2 \\ 1 \end{pmatrix} + c_2 e^{-4t} \begin{pmatrix} 1 \\ 1 \end{pmatrix}
  2. x⃗(t)=c1et(11)+c2e4t(−21)\vec{x}(t) = c_1 e^{t} \begin{pmatrix} 1 \\ 1 \end{pmatrix} + c_2 e^{4t} \begin{pmatrix} -2 \\ 1 \end{pmatrix}
  3. x⃗(t)=c1e−t(11)+c2e−4t(−21)\vec{x}(t) = c_1 e^{-t} \begin{pmatrix} 1 \\ 1 \end{pmatrix} + c_2 e^{-4t} \begin{pmatrix} -2 \\ 1 \end{pmatrix} (correct answer)
  4. x⃗(t)=(c1e−t+c2e−4t)(−12)\vec{x}(t) = (c_1 e^{-t} + c_2 e^{-4t}) \begin{pmatrix} -1 \\ 2 \end{pmatrix}
Explanation: The general solution of a linear system x⃗′=Ax⃗\vec{x}' = A\vec{x} is a linear combination of solutions of the form eλtv⃗e^{\lambda t}\vec{v} for each eigenpair (λ,v⃗)(\lambda, \vec{v}). Therefore, the general solution is x⃗(t)=c1eλ1tv⃗1+c2eλ2tv⃗2\vec{x}(t) = c_1 e^{\lambda_1 t} \vec{v}_1 + c_2 e^{\lambda_2 t} \vec{v}_2. Substituting the given values yields x⃗(t)=c1e−t(11)+c2e−4t(−21)\vec{x}(t) = c_1 e^{-t} \begin{pmatrix} 1 \\ 1 \end{pmatrix} + c_2 e^{-4t} \begin{pmatrix} -2 \\ 1 \end{pmatrix}. Choice A incorrectly swaps the eigenvectors. Choice B has incorrect signs in the exponents. Choice D demonstrates a misunderstanding of how to combine the basis solutions.

Question 17

Consider the system of differential equations dxdt=Ax\frac{d\mathbf{x}}{dt} = A\mathbf{x} where A=(3−24−1)A = \begin{pmatrix} 3 & -2 \\ 4 & -1 \end{pmatrix} . If the eigenvalues of AA are λ1=1\lambda_1 = 1 and λ2=1\lambda_2 = 1, what can be concluded about the nature of the solutions near the origin?

  1. The origin is an unstable node since both eigenvalues are positive and equal
  2. The origin is a stable node since the repeated eigenvalue indicates convergent behavior
  3. The origin is an unstable improper node since AA has only one linearly independent eigenvector (correct answer)
  4. The origin is a center since the repeated eigenvalue creates oscillatory solutions around the equilibrium
Explanation: When AA has repeated eigenvalues λ=1>0\lambda = 1 > 0, we must check if there are two linearly independent eigenvectors. Computing (A−I)v=0(A - I)\mathbf{v} = \mathbf{0}: (2−24−2)v=0 \begin{pmatrix} 2 & -2 \\ 4 & -2 \end{pmatrix}\mathbf{v} = \mathbf{0}. This gives only one linearly independent eigenvector, so AA is not diagonalizable. The Jordan form will have a non-zero off-diagonal entry, creating an improper node. Since λ=1>0\lambda = 1 > 0, solutions grow exponentially, making it unstable.

Question 18

Consider the initial value problem x⃗′=(1−23−4)x⃗\vec{x}' = \begin{pmatrix} 1 & -2 \\ 3 & -4 \end{pmatrix} \vec{x} with x⃗(0)=(32)\vec{x}(0) = \begin{pmatrix} 3 \\ 2 \end{pmatrix}. What is the solution vector x⃗(t)\vec{x}(t)?

  1. x⃗(t)=(e−t+2e−2te−t+3e−2t)\vec{x}(t) = \begin{pmatrix} e^{-t} + 2e^{-2t} \\ e^{-t} + 3e^{-2t} \end{pmatrix}
  2. x⃗(t)=(5e−t−2e−2t5e−t−3e−2t)\vec{x}(t) = \begin{pmatrix} 5e^{-t} - 2e^{-2t} \\ 5e^{-t} - 3e^{-2t} \end{pmatrix} (correct answer)
  3. x⃗(t)=(5et−2e2t5et−3e2t)\vec{x}(t) = \begin{pmatrix} 5e^{t} - 2e^{2t} \\ 5e^{t} - 3e^{2t} \end{pmatrix}
  4. x⃗(t)=(7e−t−4e−2t7e−t−6e−2t)\vec{x}(t) = \begin{pmatrix} 7e^{-t} - 4e^{-2t} \\ 7e^{-t} - 6e^{-2t} \end{pmatrix}
Explanation: First, find the eigenvalues of A=(1−23−4)A = \begin{pmatrix} 1 & -2 \\ 3 & -4 \end{pmatrix}. The characteristic equation is (1−λ)(−4−λ)−(−2)(3)=λ2+3λ+2=(λ+1)(λ+2)=0(1-\lambda)(-4-\lambda) - (-2)(3) = \lambda^2 + 3\lambda + 2 = (\lambda+1)(\lambda+2) = 0, so λ1=−1\lambda_1 = -1 and λ2=−2\lambda_2 = -2. The corresponding eigenvectors are v⃗1=(11)\vec{v}_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix} and v⃗2=(23)\vec{v}_2 = \begin{pmatrix} 2 \\ 3 \end{pmatrix}. The general solution is x⃗(t)=c1e−tv⃗1+c2e−2tv⃗2\vec{x}(t) = c_1 e^{-t} \vec{v}_1 + c_2 e^{-2t} \vec{v}_2. Using the initial condition, x⃗(0)=c1(11)+c2(23)=(32)\vec{x}(0) = c_1\begin{pmatrix} 1 \\ 1 \end{pmatrix} + c_2\begin{pmatrix} 2 \\ 3 \end{pmatrix} = \begin{pmatrix} 3 \\ 2 \end{pmatrix}. This gives the system c1+2c2=3c_1+2c_2=3 and c1+3c2=2c_1+3c_2=2, which solves to c1=5c_1=5 and c2=−1c_2=-1. Thus, the solution is x⃗(t)=5e−t(11)−e−2t(23)=(5e−t−2e−2t5e−t−3e−2t)\vec{x}(t) = 5e^{-t}\begin{pmatrix} 1 \\ 1 \end{pmatrix} - e^{-2t}\begin{pmatrix} 2 \\ 3 \end{pmatrix} = \begin{pmatrix} 5e^{-t} - 2e^{-2t} \\ 5e^{-t} - 3e^{-2t} \end{pmatrix}.

Question 19

The general solution to a 2×22 \times 2 system x⃗′=Ax⃗\vec{x}' = A\vec{x} is x⃗(t)=c1e−3t(10)+c2e−3t(01)\vec{x}(t) = c_1 e^{-3t} \begin{pmatrix} 1 \\ 0 \end{pmatrix} + c_2 e^{-3t} \begin{pmatrix} 0 \\ 1 \end{pmatrix}. Which statement about the matrix AA must be true?

  1. AA has a repeated eigenvalue λ=−3\lambda=-3 with geometric multiplicity 1.
  2. The matrix AA is not uniquely determined by the general solution.
  3. AA could be any matrix with trace −6-6 and determinant 99.
  4. A=(−300−3)A = \begin{pmatrix} -3 & 0 \\ 0 & -3 \end{pmatrix}. (correct answer)
Explanation: When you encounter a system of differential equations x⃗′=Ax⃗\vec{x}' = A\vec{x} with a given general solution, you're working backwards from the solution to determine properties of the matrix AA. The key insight is recognizing what the solution structure tells you about eigenvalues and eigenvectors. The given solution x⃗(t)=c1e−3t(10)+c2e−3t(01)\vec{x}(t) = c_1 e^{-3t} \begin{pmatrix} 1 \\ 0 \end{pmatrix} + c_2 e^{-3t} \begin{pmatrix} 0 \\ 1 \end{pmatrix} shows that both terms have the same exponential factor e−3te^{-3t}, indicating a repeated eigenvalue λ=−3\lambda = -3. The vectors (10)\begin{pmatrix} 1 \\ 0 \end{pmatrix} and (01)\begin{pmatrix} 0 \\ 1 \end{pmatrix} are the standard basis vectors, which serve as eigenvectors. Since we have two linearly independent eigenvectors for the repeated eigenvalue, the geometric multiplicity equals the algebraic multiplicity (both equal 2). This means AA is diagonalizable with $$A = \begin{pmatrix} -3 & 0 \ 0 & -3 \end{pmatrix} Option A is wrong because the geometric multiplicity is 2, not 1 (we have two independent eigenvectors). Option B is incorrect because the general solution uniquely determines $$A$$ when you have the complete eigenspace information. Option C is wrong because while $$A$$ does have trace $$-6$$ and determinant $$9$$, the eigenvector constraint severely limits the possibilities—only diagonal matrices with these eigenvalues work. **Study tip:** When working backwards from solutions to matrices, always identify the eigenvalues from exponential terms and eigenvectors from the coefficient vectors. The structure tells you everything about $$A$$.

Question 20

The matrix A=(αβ−βα)A = \begin{pmatrix} \alpha & \beta \\ -\beta & \alpha \end{pmatrix} appears in a differential system x′=Ax\mathbf{x}' = A\mathbf{x}. For what relationship between α\alpha and β\beta will the origin be a center (closed orbits)?

  1. α=0\alpha = 0 and β≠0\beta \neq 0 (correct answer)
  2. α>0\alpha > 0 and β>0\beta > 0
  3. α<0\alpha < 0 and β=0\beta = 0
  4. α=β\alpha = \beta and both are non-zero
Explanation: The eigenvalues of this matrix are λ=α±iβ\lambda = \alpha \pm i\beta. For a center, we need purely imaginary eigenvalues, which occurs when the real part α=0\alpha = 0. We also need β≠0\beta \neq 0 to ensure non-trivial dynamics. This gives eigenvalues ±iβ\pm i\beta, creating closed orbits around the origin.