Linear Algebra in DEs - Linear Algebra

1. The general solution to a $2 \times 2$ system $\vec{x}' = A\vec{x}$ is $\vec{x}(t) = c_1 e^{-3t} \begin{pmatrix} 1 \\ 0 \end{pmatrix} + c_2 e^{-3t} \begin{pmatrix} 0 \\ 1 \end{pmatrix}$. Which statement about the matrix $A$ must be true?
   Answer: $A = \begin{pmatrix} -3 & 0 \\ 0 & -3 \end{pmatrix}$. When you encounter a system of differential equations $\vec{x}' = A\vec{x}$ with a given general solution, you're working backwards from the solution to determine properties of the matrix $A$. The key insight is recognizing what the solution structure tells you about eigenvalues and eigenvectors. The given solution $\vec{x}(t) = c_1 e^{-3t} \begin{pmatrix} 1 \\ 0 \end{pmatrix} + c_2 e^{-3t} \begin{pmatrix} 0 \\ 1 \end{pmatrix}$ shows that both terms have the same exponential factor $e^{-3t}$, indicating a repeated eigenvalue $\lambda = -3$. The vectors $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ and $\begin{pmatrix} 0 \\ 1 \end{pmatrix}$ 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 $A$ is diagonalizable with $A = \begin{pmatrix} -3 & 0 \\ 0 & -3 \end{pmatrix}$, making D correct. 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 $A$.
2. Consider the matrix $A = \begin{pmatrix} 0 & 1 \\ -k & 0 \end{pmatrix}$ where $k > 0$. For what value of $k$ will the corresponding differential system $\mathbf{x}' = A\mathbf{x}$ have solutions that complete exactly one full revolution around the origin in time $t = \pi$?
   Answer: $k = 4$ The eigenvalues are $\lambda = \pm i\sqrt{k}$ (purely imaginary), creating a center with circular/elliptical orbits. The angular frequency is $\omega = \sqrt{k}$. For one complete revolution in time $\pi$, we need $\omega \cdot \pi = 2\pi$, so $\sqrt{k} \cdot \pi = 2\pi$, which gives $\sqrt{k} = 2$, thus $k = 4$.
3. A real $2 \times 2$ matrix $A$ has a complex eigenpair $\lambda = 2+5i$ and $\vec{v} = \begin{pmatrix} 1 \\ 2-i \end{pmatrix}$. The complex solution $\vec{z}(t) = e^{(2+5i)t}\vec{v}$ 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 $\vec{x}' = A\vec{x}$?
   Answer: $e^{2t}\begin{pmatrix} \cos(5t) \\ 2\cos(5t) + \sin(5t) \end{pmatrix}$ We use Euler's formula $e^{i\theta} = \cos\theta + i\sin\theta$ to expand the complex solution: $\vec{z}(t) = e^{2t}(\cos(5t) + i\sin(5t)) \begin{pmatrix} 1 \\ 2-i \end{pmatrix} = e^{2t} \begin{pmatrix} \cos(5t) + i\sin(5t) \\ (2-i)(\cos(5t) + i\sin(5t)) \end{pmatrix}$. Expanding the bottom component gives $2\cos(5t) + 2i\sin(5t) - i\cos(5t) + \sin(5t) = (2\cos(5t) + \sin(5t)) + i(2\sin(5t) - \cos(5t))$. The real part of the full vector solution is $\text{Re}(\vec{z}(t)) = e^{2t}\begin{pmatrix} \cos(5t) \\ 2\cos(5t) + \sin(5t) \end{pmatrix}$, 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 $\lambda$. Choice D is an equivalent but rearranged form of choice A; however, choice A is the direct result of taking the real part.
4. Consider the system $\begin{pmatrix} x_1' \\ x_2' \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \end{pmatrix}$. If the initial condition is $\mathbf{x}(0) = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$, which expression represents $x_1(t)$?
   Answer: $x_1(t) = (1 + t)e^{2t}$ The matrix has repeated eigenvalue $\lambda = 2$ with only one eigenvector $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$. The general solution involves $e^{2t}$ and $te^{2t}$ terms. The fundamental matrix is $\Phi(t) = \begin{pmatrix} e^{2t} & te^{2t} \\ 0 & e^{2t} \end{pmatrix}$. With $\mathbf{x}(0) = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$, we get $x_1(t) = e^{2t} \cdot 1 + te^{2t} \cdot 1 = (1 + t)e^{2t}$.
5. Consider the system of differential equations $\frac{d\mathbf{x}}{dt} = A\mathbf{x}$ where $A = \begin{pmatrix} 3 & -2 \\ 4 & -1 \end{pmatrix}$. If the eigenvalues of $A$ are $\lambda_1 = 1$ and $\lambda_2 = 1$, what can be concluded about the nature of the solutions near the origin?
   Answer: The origin is an unstable improper node since $A$ has only one linearly independent eigenvector When $A$ has repeated eigenvalues $\lambda = 1 > 0$, we must check if there are two linearly independent eigenvectors. Computing $(A - I)\mathbf{v} = \mathbf{0}$: $\begin{pmatrix} 2 & -2 \\ 4 & -2 \end{pmatrix}\mathbf{v} = \mathbf{0}$. This gives only one linearly independent eigenvector, so $A$ is not diagonalizable. The Jordan form will have a non-zero off-diagonal entry, creating an improper node. Since $\lambda = 1 > 0$, solutions grow exponentially, making it unstable.
6. A population model is given by $\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} 0.1 & 0.2 \\ 0.3 & -0.1 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}$ where $x(t)$ and $y(t)$ represent two interacting species. The eigenvalues are approximately $\lambda_1 \approx 0.236$ and $\lambda_2 \approx -0.236$. What does this suggest about the long-term behavior?
   Answer: One linear combination of the populations grows while another decays, leading to eventual dominance of one mode 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 $\lambda_1 \approx 0.236 > 0$ and $\lambda_2 \approx -0.236 < 0$, you have one growing mode and one decaying mode. The general solution is a linear combination of these modes: $e^{\lambda_1 t}\mathbf{v_1} + e^{\lambda_2 t}\mathbf{v_2}$, where $\mathbf{v_1}$ and $\mathbf{v_2}$ are the corresponding eigenvectors. As $t \to \infty$, the positive eigenvalue dominates since $e^{0.236t}$ grows while $e^{-0.236t} \to 0$. 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 $e^{At}$ 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.
7. Two competing species have populations $x(t)$ and $y(t)$ modeled by the linear system $\vec{p}' = A\vec{p}$, where $\vec{p} = \begin{pmatrix} x \\ y \end{pmatrix}$. The matrix $A$ has eigenvalues $\lambda_1 = 0.5$ and $\lambda_2 = -0.2$, with corresponding eigenvectors $\vec{v}_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$ and $\vec{v}_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}$. If the initial populations are $x(0)=300$ and $y(0)=100$, what is the long-term behavior of the population ratio $y/x$ as $t \to \infty$?
   Answer: The ratio approaches $1$. The general solution is $\vec{p}(t) = c_1 e^{0.5t} \vec{v}_1 + c_2 e^{-0.2t} \vec{v}_2$. At $t=0$, we have $c_1\begin{pmatrix} 1 \\ 1 \end{pmatrix} + c_2\begin{pmatrix} 1 \\ -1 \end{pmatrix} = \begin{pmatrix} 300 \\ 100 \end{pmatrix}$, which gives $c_1=200, c_2=100$. So $x(t) = 200e^{0.5t} + 100e^{-0.2t}$ and $y(t) = 200e^{0.5t} - 100e^{-0.2t}$. As $t \to \infty$, the $e^{0.5t}$ term dominates. The ratio becomes $\frac{y(t)}{x(t)} \approx \frac{200e^{0.5t}}{200e^{0.5t}} = 1$. Conceptually, the solution vector becomes parallel to the eigenvector associated with the largest positive eigenvalue, which is $\vec{v}_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$. The ratio of the components of this vector is $1/1=1$.
8. A system of linear differential equations is given by $\vec{x}' = A\vec{x}$. A proposed non-trivial solution is of the form $\vec{x}(t) = e^{2t}\vec{v}$. What condition must the matrix $A$ and the vector $\vec{v}$ satisfy for this to be a valid solution?
   Answer: $A\vec{v} = 2\vec{v}$ and $\vec{v}$ must be a non-zero vector. To check if $\vec{x}(t) = e^{2t}\vec{v}$ is a solution, we substitute it into the differential equation. The derivative is $\vec{x}'(t) = 2e^{2t}\vec{v}$. The right side is $A\vec{x}(t) = A(e^{2t}\vec{v}) = e^{2t}(A\vec{v})$. Equating the two gives $2e^{2t}\vec{v} = e^{2t}(A\vec{v})$. Since $e^{2t}$ is never zero, we can divide by it to obtain $2\vec{v} = A\vec{v}$. This is the definition of an eigenvector $\vec{v}$ with eigenvalue $\lambda=2$. For the solution to be non-trivial, the eigenvector $\vec{v}$ must be non-zero.
9. The matrix $A = \begin{pmatrix} \alpha & \beta \\ -\beta & \alpha \end{pmatrix}$ appears in a differential system $\mathbf{x}' = A\mathbf{x}$. For what relationship between $\alpha$ and $\beta$ will the origin be a center (closed orbits)?
   Answer: $\alpha = 0$ and $\beta \neq 0$ The eigenvalues of this matrix are $\lambda = \alpha \pm i\beta$. For a center, we need purely imaginary eigenvalues, which occurs when the real part $\alpha = 0$. We also need $\beta \neq 0$ to ensure non-trivial dynamics. This gives eigenvalues $\pm i\beta$, creating closed orbits around the origin.
10. The system $\vec{x}' = A\vec{x}$ is governed by a matrix $A$ which has a repeated eigenvalue $\lambda=2$ with algebraic multiplicity 2 but geometric multiplicity 1. If $\vec{v}$ is the unique (up to scaling) eigenvector, what is the correct form of the general solution?
    Answer: $\vec{x}(t) = c_1 e^{2t}\vec{v} + c_2 e^{2t}(t\vec{v} + \vec{u})$, where $(A-2I)\vec{u} = \vec{v}$. When an eigenvalue $\lambda$ 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 $\vec{v}$ and a generalized eigenvector $\vec{u}$, which is found by solving $(A-\lambda I)\vec{u} = \vec{v}$. The second linearly independent solution has the form $e^{\lambda t}(t\vec{v} + \vec{u})$. Choice A is a common mistake that omits the generalized eigenvector $\vec{u}$. Choice C is incorrect because only one linearly independent eigenvector exists. Choice D describes the form for complex eigenvalues.
11. A second-order linear differential equation $y'' + py' + qy = 0$ can be written as a first-order system $\mathbf{x}' = A\mathbf{x}$ where $\mathbf{x} = \begin{pmatrix} y \\ y' \end{pmatrix}$. If the characteristic equation has roots $r_1 = -1$ and $r_2 = -3$, what are the eigenvalues of the coefficient matrix $A$?
    Answer: $\lambda_1 = -1, \lambda_2 = -3$ When converting $y'' + py' + qy = 0$ to the system $\begin{pmatrix} y \\ y' \end{pmatrix}' = \begin{pmatrix} 0 & 1 \\ -q & -p \end{pmatrix}\begin{pmatrix} y \\ y' \end{pmatrix}$, the eigenvalues of the coefficient matrix are exactly the roots of the characteristic equation of the original second-order equation. This is because $\det(A - \lambda I) = \lambda^2 + p\lambda + q$, which matches the characteristic equation $r^2 + pr + q = 0$.
12. The coefficient matrix $A$ for a real system $\vec{x}' = A\vec{x}$ has a complex eigenvalue $\lambda = -2 + 3i$. Which statement accurately describes the behavior of the system's solutions near the origin?
    Answer: Solutions oscillate and decay in magnitude, spiraling towards the origin. For a complex eigenvalue $\lambda = a + bi$, the real part $a$ governs the growth or decay of the solution's magnitude, and the imaginary part $b$ governs the oscillation. Here, $a = -2$ is negative, so the magnitude decays exponentially ($e^{-2t}$). The imaginary part $b=3$ 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.
13. Consider the initial value problem $\vec{x}' = \begin{pmatrix} 1 & -2 \\ 3 & -4 \end{pmatrix} \vec{x}$ with $\vec{x}(0) = \begin{pmatrix} 3 \\ 2 \end{pmatrix}$. What is the solution vector $\vec{x}(t)$?
    Answer: $\vec{x}(t) = \begin{pmatrix} 5e^{-t} - 2e^{-2t} \\ 5e^{-t} - 3e^{-2t} \end{pmatrix}$ First, find the eigenvalues of $A = \begin{pmatrix} 1 & -2 \\ 3 & -4 \end{pmatrix}$. The characteristic equation is $(1-\lambda)(-4-\lambda) - (-2)(3) = \lambda^2 + 3\lambda + 2 = (\lambda+1)(\lambda+2) = 0$, so $\lambda_1 = -1$ and $\lambda_2 = -2$. The corresponding eigenvectors are $\vec{v}_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$ and $\vec{v}_2 = \begin{pmatrix} 2 \\ 3 \end{pmatrix}$. The general solution is $\vec{x}(t) = c_1 e^{-t} \vec{v}_1 + c_2 e^{-2t} \vec{v}_2$. Using the initial condition, $\vec{x}(0) = c_1\begin{pmatrix} 1 \\ 1 \end{pmatrix} + c_2\begin{pmatrix} 2 \\ 3 \end{pmatrix} = \begin{pmatrix} 3 \\ 2 \end{pmatrix}$. This gives the system $c_1+2c_2=3$ and $c_1+3c_2=2$, which solves to $c_1=5$ and $c_2=-1$. Thus, the solution is $\vec{x}(t) = 5e^{-t}\begin{pmatrix} 1 \\ 1 \end{pmatrix} - e^{-2t}\begin{pmatrix} 2 \\ 3 \end{pmatrix} = \begin{pmatrix} 5e^{-t} - 2e^{-2t} \\ 5e^{-t} - 3e^{-2t} \end{pmatrix}$.
14. For the linear system $\mathbf{x}' = A\mathbf{x}$ where $A$ has eigenvalues $\lambda_1 = -2 + 3i$ and $\lambda_2 = -2 - 3i$, which statement best describes the long-term behavior of solutions?
    Answer: Solutions spiral inward toward the origin with decreasing amplitude due to the negative real parts Complex eigenvalues $\lambda = \alpha \pm \beta i$ create spiral behavior. The real part $\alpha = -2 < 0$ determines stability: negative real parts cause exponential decay, so solutions spiral inward. The imaginary parts $\pm 3i$ create the oscillatory/spiral motion with angular frequency related to $|\beta| = 3$. This makes the origin a stable spiral point.
15. Consider the system $\mathbf{x}' = A\mathbf{x}$ where $A$ has eigenvalues $\lambda_1 = 3$ and $\lambda_2 = -1$ with corresponding eigenvectors $\mathbf{v}_1 = \begin{pmatrix} 2 \\ 1 \end{pmatrix}$ and $\mathbf{v}_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}$. Which initial condition will cause the solution to approach the origin as $t \to \infty$?
    Answer: $\mathbf{x}(0) = \begin{pmatrix} 1 \\ -1 \end{pmatrix}$ The general solution is $\mathbf{x}(t) = c_1 e^{3t}\mathbf{v}_1 + c_2 e^{-t}\mathbf{v}_2$. For the solution to approach the origin, we need the coefficient of the unstable mode ($e^{3t}$) to be zero, so $c_1 = 0$. This happens when the initial condition lies on the stable eigenspace spanned by $\mathbf{v}_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}$. Choice B is exactly this eigenvector.
16. A system of differential equations is described by $\vec{x}' = A\vec{x}$. The matrix $A$ has eigenvalues $\lambda_1 = -1$ and $\lambda_2 = -4$ with corresponding eigenvectors $\vec{v}_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$ and $\vec{v}_2 = \begin{pmatrix} -2 \\ 1 \end{pmatrix}$. Which of the following represents the general solution $\vec{x}(t)$?
    Answer: $\vec{x}(t) = c_1 e^{-t} \begin{pmatrix} 1 \\ 1 \end{pmatrix} + c_2 e^{-4t} \begin{pmatrix} -2 \\ 1 \end{pmatrix}$ The general solution of a linear system $\vec{x}' = A\vec{x}$ is a linear combination of solutions of the form $e^{\lambda t}\vec{v}$ for each eigenpair $(\lambda, \vec{v})$. Therefore, the general solution is $\vec{x}(t) = c_1 e^{\lambda_1 t} \vec{v}_1 + c_2 e^{\lambda_2 t} \vec{v}_2$. Substituting the given values yields $\vec{x}(t) = c_1 e^{-t} \begin{pmatrix} 1 \\ 1 \end{pmatrix} + c_2 e^{-4t} \begin{pmatrix} -2 \\ 1 \end{pmatrix}$. 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.
17. A $2 \times 2$ matrix $A$ for the system $\vec{x}' = A\vec{x}$ has eigenvalues $\lambda_1 = 0$ and $\lambda_2 = -5$. Let $\vec{v}_1$ be the eigenvector corresponding to $\lambda_1$. What does the zero eigenvalue imply about the solutions?
    Answer: The system possesses an entire line of equilibrium points. An eigenvalue of $\lambda_1 = 0$ means that for its corresponding eigenvector $\vec{v}_1$, we have $A\vec{v}_1 = 0\vec{v}_1 = \vec{0}$. Consider any point on the line spanned by $\vec{v}_1$, say $\vec{x} = c\vec{v}_1$. The derivative at this point is $\vec{x}' = A(c\vec{v}_1) = c(A\vec{v}_1) = \vec{0}$. A zero derivative means the point is an equilibrium (or fixed) point. Thus, the entire line of points $c\vec{v}_1$ consists of equilibria. The general solution is $\vec{x}(t) = c_1e^{0t}\vec{v}_1 + c_2e^{-5t}\vec{v}_2 = c_1\vec{v}_1 + c_2e^{-5t}\vec{v}_2$. As $t \to \infty$, solutions approach a point on this line.
18. For the system $\mathbf{x}' = A\mathbf{x}$ with $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, suppose $\det(A) = -6$ and $\text{tr}(A) = 1$. What type of critical point occurs at the origin?
    Answer: Saddle point, since the determinant is negative indicating eigenvalues of opposite signs For eigenvalues $\lambda_1, \lambda_2$: $\lambda_1 + \lambda_2 = \text{tr}(A) = 1$ and $\lambda_1 \lambda_2 = \det(A) = -6$. 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.
19. The system $\vec{y}' = A\vec{y}$ has an equilibrium point at the origin. The eigenvalues of the coefficient matrix $A$ are $\lambda_1 = -3$ and $\lambda_2 = 1$. How would the behavior of solutions near the origin be classified?
    Answer: A saddle point, where some trajectories approach and others move away from the origin. The stability of the equilibrium point at the origin is determined by the eigenvalues of $A$. Since the eigenvalues are real and have opposite signs ($-3 < 0$ and $1 > 0$), 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.
20. A spring-mass system with damping leads to the equation $m\ddot{x} + c\dot{x} + kx = 0$. When written as a first-order system $\mathbf{y}' = A\mathbf{y}$ with $\mathbf{y} = \begin{pmatrix} x \\ \dot{x} \end{pmatrix}$, under what condition on the parameters will the system be critically damped (repeated real eigenvalues)?
    Answer: $c^2 = 4mk$ The system matrix is $A = \begin{pmatrix} 0 & 1 \\ -k/m & -c/m \end{pmatrix}$. The characteristic equation is $\lambda^2 + (c/m)\lambda + k/m = 0$. For repeated real roots (critical damping), the discriminant must be zero: $(c/m)^2 - 4(k/m) = 0$, which gives $c^2/m^2 = 4k/m$, so $c^2 = 4mk$. Note that choices C and D are equivalent to A, but A is the standard form.

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