Differential Equations Quiz: Systems In Matrix Form

What this quiz covers

This quiz focuses on Systems In Matrix Form, giving you a quick way to practice the rules, question types, and explanations that matter most for Differential Equations.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $\frac{dx}{dt} = x + 2y + t^2$, $\frac{dy}{dt} = y + z$, $\frac{dz}{dt} = 3x + 2z$
2. $\frac{dx}{dt} = x + 2y$, $\frac{dy}{dt} = y + z + t^2$, $\frac{dz}{dt} = 3x + 2z + \sin t$
3. $\frac{dx}{dt} = x + 2y + t^2$, $\frac{dy}{dt} = y + z$, $\frac{dz}{dt} = 3x + 2z + \sin t$ (correct answer)
4. $\frac{dx}{dt} = x + 2y$, $\frac{dy}{dt} = y + z$, $\frac{dz}{dt} = 3x + 2z + \sin t$

Explanation: When you encounter a matrix differential equation like $\mathbf{X}' = A\mathbf{X} + \mathbf{b}(t)$, you're looking at a compact way to represent a system of coupled differential equations. To convert it back to scalar form, you need to perform the matrix multiplication and addition component by component. Let's work through this systematically. If $\mathbf{X} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}$, then $\mathbf{X}' = \begin{pmatrix} x' \\ y' \\ z' \end{pmatrix}$. The right side becomes: $\begin{pmatrix} 1 & 2 & 0 \\ 0 & 1 & 1 \\ 3 & 0 & 2 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} + \begin{pmatrix} t^2 \\ 0 \\ \sin t \end{pmatrix} = \begin{pmatrix} x + 2y \\ y + z \\ 3x + 2z \end{pmatrix} + \begin{pmatrix} t^2 \\ 0 \\ \sin t \end{pmatrix} = \begin{pmatrix} x + 2y + t^2 \\ y + z \\ 3x + 2z + \sin t \end{pmatrix}$ This gives us: $x' = x + 2y + t^2$, $y' = y + z$, $z' = 3x + 2z + \sin t$, which is answer C. Answer A incorrectly omits $\sin t$ from the third equation. Answer B misplaces the forcing terms, putting $t^2$ in the second equation instead of the first. Answer D omits $t^2$ from the first equation entirely. Study tip: Always multiply the matrix and vector first, then add the forcing vector component-wise. Each row of the coefficient matrix corresponds to one scalar equation, and each component of the forcing vector gets added to its corresponding equation position.

Question 2

1. $\begin{pmatrix} 0 & 1 & 0 & 0 \\ -\frac{k_1+k_2}{m_1} & 0 & \frac{k_2}{m_1} & 0 \\ 0 & 0 & 0 & 1 \\ \frac{k_2}{m_2} & 0 & -\frac{k_2+k_3}{m_2} & 0 \end{pmatrix}$ (correct answer)
2. $\begin{pmatrix} 0 & 1 & 0 & 0 \\ -\frac{k_1-k_2}{m_1} & 0 & \frac{k_2}{m_1} & 0 \\ 0 & 0 & 0 & 1 \\ \frac{k_2}{m_2} & 0 & -\frac{k_2+k_3}{m_2} & 0 \end{pmatrix}$
3. $\begin{pmatrix} 0 & 1 & 0 & 0 \\ -\frac{k_1+k_2}{m_1} & 0 & \frac{k_2}{m_1} & 0 \\ 0 & 0 & 0 & 1 \\ \frac{k_2}{m_2} & 0 & -\frac{k_2-k_3}{m_2} & 0 \end{pmatrix}$
4. $\begin{pmatrix} 0 & 1 & 0 & 0 \\ -\frac{k_1+k_2}{m_1} & 0 & -\frac{k_2}{m_1} & 0 \\ 0 & 0 & 0 & 1 \\ \frac{k_2}{m_2} & 0 & -\frac{k_2+k_3}{m_2} & 0 \end{pmatrix}$

Explanation: Converting the second-order system requires $x_1' = x_1'$ and $x_1'' = \frac{-k_1x_1 + k_2(x_2-x_1)}{m_1} = \frac{-(k_1+k_2)x_1 + k_2x_2}{m_1}$. Similarly, $x_2'' = \frac{-k_2(x_2-x_1) - k_3x_2}{m_2} = \frac{k_2x_1 - (k_2+k_3)x_2}{m_2}$. Choice B incorrectly has $k_1-k_2$ instead of $k_1+k_2$. Choice C has $k_2-k_3$ instead of $k_2+k_3$. Choice D has the wrong sign for the coupling term $\frac{k_2}{m_1}$.

Question 3

1. $\frac{d}{dt}\begin{pmatrix} y_1 \\ y_2 \\ y_3 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & -3 & 2 \end{pmatrix}\begin{pmatrix} y_1 \\ y_2 \\ y_3 \end{pmatrix} + \begin{pmatrix} 0 \\ 0 \\ e^{2t} \end{pmatrix}$
2. $\frac{d}{dt}\begin{pmatrix} y_1 \\ y_2 \\ y_3 \end{pmatrix} = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -1 & 3 & -2 \end{pmatrix}\begin{pmatrix} y_1 \\ y_2 \\ y_3 \end{pmatrix} + \begin{pmatrix} 0 \\ 0 \\ e^{2t} \end{pmatrix}$
3. $\frac{d}{dt}\begin{pmatrix} y_1 \\ y_2 \\ y_3 \end{pmatrix} = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & -3 & 2 \end{pmatrix}\begin{pmatrix} y_1 \\ y_2 \\ y_3 \end{pmatrix} + \begin{pmatrix} 0 \\ e^{2t} \\ 0 \end{pmatrix}$
4. $\frac{d}{dt}\begin{pmatrix} y_1 \\ y_2 \\ y_3 \end{pmatrix} = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & -3 & 2 \end{pmatrix}\begin{pmatrix} y_1 \\ y_2 \\ y_3 \end{pmatrix} + \begin{pmatrix} 0 \\ 0 \\ e^{2t} \end{pmatrix}$ (correct answer)

Explanation: When you encounter a higher-order differential equation like this third-order equation, converting it to a first-order system using substitution is a fundamental technique. The key is systematically defining new variables for each derivative and then expressing their relationships. Starting with the substitution $y_1 = y$, $y_2 = y'$, $y_3 = y''$, you need to find the derivatives of each new variable. From the definitions: $y_1' = y' = y_2$ and $y_2' = y'' = y_3$. For $y_3' = y'''$, you use the original equation. Since $y''' - 2y'' + 3y' - y = e^{2t}$, solving for $y'''$ gives $y''' = 2y'' - 3y' + y + e^{2t} = 2y_3 - 3y_2 + y_1 + e^{2t}$. This creates the system: $y_1' = y_2$, $y_2' = y_3$, $y_3' = y_1 - 3y_2 + 2y_3 + e^{2t}$. In matrix form, this becomes answer choice D. Choice A is wrong because the coefficient matrix has the identity structure in the wrong positions—the first two rows should have the "shift" pattern $[0,1,0]$ and $[0,0,1]$, not $[1,0,0]$ and $[0,1,0]$. Choice B has the correct matrix structure but incorrect signs in the bottom row: it shows $[-1,3,-2]$ instead of $[1,-3,2]$. Choice C places the forcing term $e^{2t}$ in the wrong position—it should be in the third component, not the second. Strategy tip: Always work systematically through the substitution, carefully tracking signs when solving for the highest derivative. The coefficient matrix follows a predictable pattern with the challenging part being the bottom row.

Question 4

1. $\lambda^2 + 4\lambda + 13$
2. $\lambda^2 - 4\lambda + 13$ (correct answer)
3. $\lambda^2 - 4\lambda + 3$
4. $\lambda^2 - 4\lambda - 13$

Explanation: When you encounter a linear system of differential equations, you're typically working toward finding eigenvalues, which requires computing the characteristic polynomial $\det(\mathbf{A} - \lambda \mathbf{I})$. First, identify the coefficient matrix $\mathbf{A}$ from the given system. The equations $\frac{dx}{dt} = 3x - 2y$ and $\frac{dy}{dt} = 5x + y$ give us: $\mathbf{A} = \begin{pmatrix} 3 & -2 \\ 5 & 1 \end{pmatrix}$ Next, form $\mathbf{A} - \lambda \mathbf{I}$: $\mathbf{A} - \lambda \mathbf{I} = \begin{pmatrix} 3-\lambda & -2 \\ 5 & 1-\lambda \end{pmatrix}$ Calculate the determinant: $\det(\mathbf{A} - \lambda \mathbf{I}) = (3-\lambda)(1-\lambda) - (-2)(5) = (3-\lambda)(1-\lambda) + 10$ Expanding: $3 - 3\lambda - \lambda + \lambda^2 + 10 = \lambda^2 - 4\lambda + 13$ This matches answer choice B. Now for the wrong answers: Choice A ($\lambda^2 + 4\lambda + 13$) has the wrong sign on the middle term—this happens if you incorrectly expand $(3-\lambda)(1-\lambda)$ or mix up the trace calculation. Choice C ($\lambda^2 - 4\lambda + 3$) gets the linear term right but miscalculates the constant term, likely forgetting to add the off-diagonal product $(-2)(5) = -10$. Choice D ($\lambda^2 - 4\lambda - 13$) makes sign errors in both the expansion and the off-diagonal calculation. Study tip: Remember that for a 2×2 matrix, the characteristic polynomial is always $\lambda^2 - \text{trace}(\mathbf{A})\lambda + \det(\mathbf{A})$. Here, trace = $3+1=4$ and determinant = $3(1)-(-2)(5)=13$, giving $\lambda^2-4\lambda+13$.

Question 5

1. $\frac{d}{dt}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 3 & -2 \\ 1 & 4 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} + \begin{pmatrix} t \\ -e^t \end{pmatrix}$ (correct answer)
2. $\frac{d}{dt}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 3 & 2 \\ 1 & 4 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} + \begin{pmatrix} t \\ -e^t \end{pmatrix}$
3. $\frac{d}{dt}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 3 & -2 \\ 1 & 4 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} + \begin{pmatrix} t \\ e^t \end{pmatrix}$
4. $\frac{d}{dt}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 3 & -2 \\ -1 & 4 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} + \begin{pmatrix} t \\ -e^t \end{pmatrix}$

Explanation: The correct matrix form is A. The coefficient matrix has entries from the coefficients of x and y: first row [3, -2] from 3x - 2y, second row [1, 4] from x + 4y. The forcing vector contains the non-homogeneous terms [t, -e^t]. Choice B incorrectly has +2 instead of -2. Choice C has +e^t instead of -e^t. Choice D has -1 instead of +1 in the coefficient matrix.

Question 6

1. $\mathbf{P}$ has the form $\begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ -c & -b & -a \end{pmatrix}$ and $\mathbf{q}(t) = \begin{pmatrix} 0 \\ 0 \\ f(t) \end{pmatrix}$
2. $\mathbf{P}$ has the form $\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -c & -b & -a \end{pmatrix}$ and $\mathbf{q}(t) = \begin{pmatrix} 0 \\ 0 \\ f(t) \end{pmatrix}$ (correct answer)
3. $\mathbf{P}$ has the form $\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -c & -b & -a \end{pmatrix}$ and $\mathbf{q}(t) = \begin{pmatrix} f(t) \\ 0 \\ 0 \end{pmatrix}$
4. $\mathbf{P}$ has the form $\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ c & b & a \end{pmatrix}$ and $\mathbf{q}(t) = \begin{pmatrix} 0 \\ 0 \\ f(t) \end{pmatrix}$

Explanation: When you encounter a higher-order differential equation that needs to be converted to a system, you're using a standard technique that transforms one nth-order equation into n first-order equations by introducing new variables. To convert $y''' + ay'' + by' + cy = f(t)$ into a first-order system, define $y_1 = y$, $y_2 = y'$, and $y_3 = y''$. This gives us the relationships: $y_1' = y_2$, $y_2' = y_3$, and from the original equation, $y_3' = y''' = -ay'' - by' - cy + f(t) = -ay_3 - by_2 - cy_1 + f(t)$. Writing this as $\frac{d\mathbf{y}}{dt} = \mathbf{P}\mathbf{y} + \mathbf{q}(t)$ where $\mathbf{y} = \begin{pmatrix} y_1 \\ y_2 \\ y_3 \end{pmatrix}$, we get the system with matrix $\mathbf{P} = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -c & -b & -a \end{pmatrix}$ and forcing vector $\mathbf{q}(t) = \begin{pmatrix} 0 \\ 0 \\ f(t) \end{pmatrix}$. This confirms answer B is correct. Answer A has the wrong structure in the first two rows—the 1's are misplaced, disrupting the derivative relationships. Answer C places the forcing term $f(t)$ in the wrong position; it should appear in the equation for $y_3'$, not $y_1'$. Answer D has the correct matrix structure but wrong signs—the coefficients in the bottom row should be negative, not positive. Remember: the companion matrix always has 1's on the superdiagonal, zeros elsewhere in the first $n-1$ rows, and the negative coefficients in the last row. The forcing term appears in the position corresponding to the highest derivative.

Question 7

1. $a = 2$, $c = 4$, $g(t) = \cos t$ (correct answer)
2. $a = 2$, $c = -3$, $g(t) = \cos t$
3. $a = 4$, $c = 2$, $g(t) = \sin t$
4. $a = 2$, $c = 4$, $g(t) = \sin t$

Explanation: From the matrix form $\mathbf{A} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} 2 & -3 \\ 4 & 1 \end{pmatrix}$, we read $a = 2$ and $c = 4$. The forcing vector $\mathbf{F}(t) = \begin{pmatrix} f(t) \\ g(t) \end{pmatrix} = \begin{pmatrix} \sin t \\ \cos t \end{pmatrix}$ gives $g(t) = \cos t$. Choice B confuses $c$ with $b = -3$. Choice C swaps $a$ and $c$ and confuses $g(t)$ with $f(t)$. Choice D has the correct $a$ and $c$ but wrong $g(t)$.

Question 8

1. $\frac{d}{dt}\begin{pmatrix} u \\ v \\ w \end{pmatrix} = \begin{pmatrix} 0 & -1 & 1 \\ 2 & -1 & 0 \\ 0 & -2 & -1 \end{pmatrix}\begin{pmatrix} u \\ v \\ w \end{pmatrix}$
2. $\frac{d}{dt}\begin{pmatrix} u \\ v \\ w \end{pmatrix} = \begin{pmatrix} 2 & -1 & 1 \\ 2 & -3 & 0 \\ 0 & 2 & -1 \end{pmatrix}\begin{pmatrix} u \\ v \\ w \end{pmatrix}$
3. $\frac{d}{dt}\begin{pmatrix} u \\ v \\ w \end{pmatrix} = \begin{pmatrix} 0 & -1 & 1 \\ 2 & -3 & 0 \\ 0 & 2 & -1 \end{pmatrix}\begin{pmatrix} u \\ v \\ w \end{pmatrix}$
4. $\frac{d}{dt}\begin{pmatrix} u \\ v \\ w \end{pmatrix} = \begin{pmatrix} 0 & -1 & 1 \\ 2 & -1 & 0 \\ 0 & 2 & -1 \end{pmatrix}\begin{pmatrix} u \\ v \\ w \end{pmatrix}$ (correct answer)

Explanation: When analyzing nonlinear systems of differential equations, linearization helps you understand local behavior near equilibrium points. You linearize by computing the Jacobian matrix of partial derivatives and evaluating it at the equilibrium point. For this system, you need to find the Jacobian matrix of the right-hand sides: $J = \begin{pmatrix} \frac{\partial}{\partial x}(2x - xy + z) & \frac{\partial}{\partial y}(2x - xy + z) & \frac{\partial}{\partial z}(2x - xy + z) \\ \frac{\partial}{\partial x}(xy - 3y) & \frac{\partial}{\partial y}(xy - 3y) & \frac{\partial}{\partial z}(xy - 3y) \\ \frac{\partial}{\partial x}(-z + 2y) & \frac{\partial}{\partial y}(-z + 2y) & \frac{\partial}{\partial z}(-z + 2y) \end{pmatrix}$ Computing the partial derivatives: $J = \begin{pmatrix} 2 - y & -x & 1 \\ y & x - 3 & 0 \\ 0 & 2 & -1 \end{pmatrix}$ Evaluating at the equilibrium point $(1, 2, 4)$: $J(1,2,4) = \begin{pmatrix} 2 - 2 & -1 & 1 \\ 2 & 1 - 3 & 0 \\ 0 & 2 & -1 \end{pmatrix} = \begin{pmatrix} 0 & -1 & 1 \\ 2 & -2 & 0 \\ 0 & 2 & -1 \end{pmatrix}$ Answer D matches this calculation with the correct entry $-2$ in position $(2,2)$. Answer A has $-1$ instead of $-2$ in this position. Answer B incorrectly keeps the original constant terms like $2$ and $-3$ that should be removed during linearization. Answer C also has $-3$ instead of $-2$ in the $(2,2)$ position. Remember: linearization requires computing partial derivatives of each equation with respect to each variable, then substituting the equilibrium values. Don't confuse this with simply copying coefficients from the original equations.

Question 9

1. $\frac{d}{dt}\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ -4 & -3 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} + \begin{pmatrix} 0 \\ 4\cos(3t) \end{pmatrix}$
2. $\frac{d}{dt}\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ -8 & -6 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} + \begin{pmatrix} 0 \\ 4\cos(3t) \end{pmatrix}$
3. $\frac{d}{dt}\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ -4 & -3 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} + \begin{pmatrix} 0 \\ 2\cos(3t) \end{pmatrix}$ (correct answer)
4. $\frac{d}{dt}\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ -4 & -3 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} + \begin{pmatrix} 0 \\ 2\cos(3t) \end{pmatrix}$

Explanation: When converting a second-order differential equation to a first-order system, you're transforming one equation with higher derivatives into multiple equations with only first derivatives. This is essential for numerical methods and state-space analysis. Start by defining your state variables: $x_1 = x$ and $x_2 = \dot{x}$. This immediately gives you the first equation: $\dot{x_1} = x_2$. For the second equation, you need $\dot{x_2} = \ddot{x}$, so solve the original equation for $\ddot{x}$: $2\ddot{x} + 6\dot{x} + 8x = 4\cos(3t)$ $\ddot{x} = -3\dot{x} - 4x + 2\cos(3t)$ $\dot{x_2} = -4x_1 - 3x_2 + 2\cos(3t)$ This gives you the system in matrix form: $\frac{d}{dt}\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ -4 & -3 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} + \begin{pmatrix} 0 \\ 2\cos(3t) \end{pmatrix}$ Option A has the correct matrix but wrong forcing term (forgot to divide $F(t)$ by $m$). Option B forgot to divide all coefficients by $m = 2$, keeping $-8$ and $-6$ instead of $-4$ and $-3$. Option D has the wrong matrix structure with $1$ in the wrong position, breaking the fundamental relationship $\dot{x_1} = x_2$. Study tip: Always divide the entire equation by the coefficient of the highest derivative first, then carefully track how each term transforms. The top row of your matrix should always be $[0, 1, 0, ...]$ for these standard conversions.