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Math 308 Week in Review 10 — Solutions

Differential Equations — Matrices, Systems, Eigenvalues · Spring 2026 • Sections 7.1, 7.2 & 7.3

Directions: Full worked solutions for Week in Review 10 (Differential Equations — Matrices, Systems, Eigenvalues). Click Solution under any problem to reveal the worked answer. Use the Contents panel or the Prev / Up / Next buttons to move between sections, and Search to find terms on this page.

Reference

Matrix Reference (Section 7.2)

Table of definitions
OperationDefinition
Addition \( (A+B)_{ij}=a_{ij}+b_{ij} \) (same dimensions required)
Scalar multiply \( (cA)_{ij}=c\,a_{ij} \)
Product \( (AB)_{ij}=\sum _{k=1}^{n}a_{ik}\,b_{kj} \)   ( \( A \) is \( m\times n \) , \( B \) is \( n\times p \) )
Trace \( \operatorname{tr}(A)=a_{11}+a_{22}+\cdots +a_{nn} \)
\( 2\times 2 \) determinant \( \begin{vmatrix} a & b \\ c & d \end{vmatrix}=ad-bc \)
\( 3\times 3 \) determinantCofactor expansion along any row or column

Eigenvalue Reference (Section 7.3)

Table of definitions
StepFormula / Procedure
Definition \( A\mathbf{v}=\lambda \mathbf{v} \) with \( \mathbf{v}\ne 0 \)
Characteristic equation \( det(A-\lambda I)=0 \)
\( 2\times 2 \) shortcut \( {\lambda }^{2}-\operatorname{tr}(A)\,\lambda +det(A)=0 \)
EigenvectorNonzero solution of \( (A-{\lambda }_{i}I)\,\mathbf{v}=0 \)
Linear dependence (two vectors)Dependent \( \,\iff \, \) one is a scalar multiple of the other (or one is zero)
Complex eigenvaluesIf \( \lambda =a+bi \) has eigenvector \( \mathbf{v} \) , then \( \overline{\lambda }=a-bi \) has eigenvector \( \overline{\mathbf{v}} \)
Symmetric matricesAll eigenvalues are real; eigenvectors for distinct eigenvalues are orthogonal

Section 7.2: Review of Matrices

1.

Perform the indicated matrix operations.

(a) \( \begin{bmatrix} 5 & -2 & 3 \\ 1 & 4 & -3 \end{bmatrix}+\begin{bmatrix} 2 & 4 & -1 \\ -3 & 2 & 5 \end{bmatrix} \)

(b) \( \begin{bmatrix} 6 & 4 & -3 \\ 5 & -2 & 2 \\ 1 & 3 & -5 \end{bmatrix}-\begin{bmatrix} 2 & -1 & 2 \\ 3 & -2 & 4 \\ 5 & 2 & -3 \end{bmatrix} \)

(c) \( 3\begin{bmatrix} 5 & -4 \\ 2 & 7 \end{bmatrix} \)

Part (a). Add entrywise (both matrices are \( 2\times 3 \) ):

\[ \begin{bmatrix} 5 & -2 & 3 \\ 1 & 4 & -3 \end{bmatrix}+\begin{bmatrix} 2 & 4 & -1 \\ -3 & 2 & 5 \end{bmatrix}=\begin{bmatrix} 5+2 & -2+4 & 3-1 \\ 1-3 & 4+2 & -3+5 \end{bmatrix}=\begin{bmatrix} 7 & 2 & 2 \\ -2 & 6 & 2 \end{bmatrix} \]

Part (b). Subtract entrywise:

\[ \begin{bmatrix} 6 & 4 & -3 \\ 5 & -2 & 2 \\ 1 & 3 & -5 \end{bmatrix}-\begin{bmatrix} 2 & -1 & 2 \\ 3 & -2 & 4 \\ 5 & 2 & -3 \end{bmatrix}=\begin{bmatrix} 4 & 5 & -5 \\ 2 & 0 & -2 \\ -4 & 1 & -2 \end{bmatrix} \]

Part (c). Multiply every entry by \( 3 \) :

\[ 3\begin{bmatrix} 5 & -4 \\ 2 & 7 \end{bmatrix}=\begin{bmatrix} 15 & -12 \\ 6 & 21 \end{bmatrix} \]
\[ \boxed{(a)\,\begin{bmatrix} 7 & 2 & 2 \\ -2 & 6 & 2 \end{bmatrix},\,(b)\,\begin{bmatrix} 4 & 5 & -5 \\ 2 & 0 & -2 \\ -4 & 1 & -2 \end{bmatrix},\,(c)\,\begin{bmatrix} 15 & -12 \\ 6 & 21 \end{bmatrix}} \]

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2.

Compute each matrix product. State the dimensions of the result before computing entries.

(a) \( \begin{bmatrix} 4 & -2 \\ 3 & 5 \end{bmatrix}\begin{bmatrix} 2 & -1 & 3 \\ 4 & 6 & 1 \end{bmatrix} \)

(b) \( \begin{bmatrix} 5 & 3 & -1 \\ 7 & 2 & -3 \end{bmatrix}\begin{bmatrix} 4 & 1 \\ 2 & 5 \\ -3 & 4 \end{bmatrix} \)

(c) \( \begin{bmatrix} 6 & -3 & 2 \\ 1 & 4 & 5 \\ -2 & 3 & 7 \end{bmatrix}\begin{bmatrix} 2 \\ -1 \\ 4 \end{bmatrix} \)

Part (a). \( (2\times 2)(2\times 3)\to 2\times 3 \) . Each entry is the dot product of a row of the left matrix with a column of the right matrix.

\[ \begin{bmatrix} 4 & -2 \\ 3 & 5 \end{bmatrix}\begin{bmatrix} 2 & -1 & 3 \\ 4 & 6 & 1 \end{bmatrix}=\begin{bmatrix} 4(2)+(-2)(4) & 4(-1)+(-2)(6) & 4(3)+(-2)(1) \\ 3(2)+5(4) & 3(-1)+5(6) & 3(3)+5(1) \end{bmatrix}=\begin{bmatrix} 0 & -16 & 10 \\ 26 & 27 & 14 \end{bmatrix} \]

Part (b). \( (2\times 3)(3\times 2)\to 2\times 2 \) .

\[ \begin{bmatrix} 5 & 3 & -1 \\ 7 & 2 & -3 \end{bmatrix}\begin{bmatrix} 4 & 1 \\ 2 & 5 \\ -3 & 4 \end{bmatrix}=\begin{bmatrix} 20+6+3 & 5+15-4 \\ 28+4+9 & 7+10-12 \end{bmatrix}=\begin{bmatrix} 29 & 16 \\ 41 & 5 \end{bmatrix} \]

Part (c). \( (3\times 3)(3\times 1)\to 3\times 1 \) .

\[ \begin{bmatrix} 6 & -3 & 2 \\ 1 & 4 & 5 \\ -2 & 3 & 7 \end{bmatrix}\begin{bmatrix} 2 \\ -1 \\ 4 \end{bmatrix}=\begin{bmatrix} 12+3+8 \\ 2-4+20 \\ -4-3+28 \end{bmatrix}=\begin{bmatrix} 23 \\ 18 \\ 21 \end{bmatrix} \]
\[ \boxed{(a)\,\begin{bmatrix} 0 & -16 & 10 \\ 26 & 27 & 14 \end{bmatrix},\,(b)\,\begin{bmatrix} 29 & 16 \\ 41 & 5 \end{bmatrix},\,(c)\,\begin{bmatrix} 23 \\ 18 \\ 21 \end{bmatrix}} \]

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3.

Compute the trace of each matrix.

(a) \( \operatorname{tr}\,\left( \begin{bmatrix} 6 & -3 \\ 4 & -8 \end{bmatrix} \right) \)

(b) \( \operatorname{tr}\,\left( \begin{bmatrix} 4 & -1 & 5 \\ 7 & 3 & -2 \\ 1 & 6 & -5 \end{bmatrix} \right) \)

Part (a). Diagonal entries are \( 6 \) and \( -8 \) :

\[ \operatorname{tr}\,\begin{bmatrix} 6 & -3 \\ 4 & -8 \end{bmatrix}=6+(-8)=-2 \]

Part (b). Diagonal entries are \( 4,\,3,\,-5 \) :

\[ \operatorname{tr}\,\begin{bmatrix} 4 & -1 & 5 \\ 7 & 3 & -2 \\ 1 & 6 & -5 \end{bmatrix}=4+3+(-5)=2 \]
\[ \boxed{(a)\,\operatorname{tr}=-2,\,(b)\,\operatorname{tr}=2} \]

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4.

Compute each determinant. For the \( 3\times 3 \) case, expand along whichever row or column has the most zeros (or the simplest entries).

(a) \( \begin{vmatrix} 4 & 3 \\ -2 & 6 \end{vmatrix} \)

(b) \( \begin{vmatrix} -5 & 4 \\ 3 & -7 \end{vmatrix} \)

(c) \( \begin{vmatrix} 2 & -1 & 4 \\ 3 & 5 & -2 \\ 1 & 4 & 3 \end{vmatrix} \)

Part (a). Use \( ad-bc \) :

\[ \begin{vmatrix} 4 & 3 \\ -2 & 6 \end{vmatrix}=(4)(6)-(3)(-2)=24+6=30 \]

Part (b).

\[ \begin{vmatrix} -5 & 4 \\ 3 & -7 \end{vmatrix}=(-5)(-7)-(4)(3)=35-12=23 \]

Part (c). Expand along row 1:

\[ \begin{vmatrix} 2 & -1 & 4 \\ 3 & 5 & -2 \\ 1 & 4 & 3 \end{vmatrix}=2\begin{vmatrix} 5 & -2 \\ 4 & 3 \end{vmatrix}-(-1)\begin{vmatrix} 3 & -2 \\ 1 & 3 \end{vmatrix}+4\begin{vmatrix} 3 & 5 \\ 1 & 4 \end{vmatrix} \]

Evaluate each \( 2\times 2 \) minor:

\[ =2(15-(-8))+1(9-(-2))+4(12-5)=2(23)+1(11)+4(7)=46+11+28=85 \]
\[ \boxed{(a)\,30,\,(b)\,23,\,(c)\,85} \]

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5.

Multiply out the following matrix-vector product.

\[ \begin{bmatrix} 5 & -2 & 4 \\ 3 & 6 & -1 \\ 7 & 2 & 5 \end{bmatrix}\begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} \]

Express your answer as a column vector whose entries are linear combinations of \( x_{1},x_{2},x_{3} \) .

Key idea. The product \( A\mathbf{x} \) is a linear combination of the columns of \( A \) , with the entries of \( \mathbf{x} \) as the weights. If \( A=[\,{\mathbf{a}}_{1}\ {\mathbf{a}}_{2}\ {\mathbf{a}}_{3}\,] \) , then

\[ A\mathbf{x}=x_{1}\,{\mathbf{a}}_{1}+x_{2}\,{\mathbf{a}}_{2}+x_{3}\,{\mathbf{a}}_{3}. \]

This is the viewpoint that matters for Chapter 7: it tells us the range of \( A \) is the span of its columns, and it is the idea behind writing solutions of \( {\mathbf{x}}^{'}=A\mathbf{x} \) as combinations of eigenvectors.

Apply it. The three columns of \( A \) are

\[ {\mathbf{a}}_{1}=\begin{bmatrix} 5 \\ 3 \\ 7 \end{bmatrix},\,{\mathbf{a}}_{2}=\begin{bmatrix} -2 \\ 6 \\ 2 \end{bmatrix},\,{\mathbf{a}}_{3}=\begin{bmatrix} 4 \\ -1 \\ 5 \end{bmatrix}. \]

Therefore

\[ A\mathbf{x}=x_{1}\begin{bmatrix} 5 \\ 3 \\ 7 \end{bmatrix}+x_{2}\begin{bmatrix} -2 \\ 6 \\ 2 \end{bmatrix}+x_{3}\begin{bmatrix} 4 \\ -1 \\ 5 \end{bmatrix}. \]

Adding the three weighted columns entry-by-entry gives the column vector form:

\[ A\mathbf{x}=\begin{bmatrix} 5x_{1}-2x_{2}+4x_{3} \\ 3x_{1}+6x_{2}-x_{3} \\ 7x_{1}+2x_{2}+5x_{3} \end{bmatrix}. \]
\[ \boxed{A\mathbf{x}=x_{1}\begin{bmatrix} 5 \\ 3 \\ 7 \end{bmatrix}+x_{2}\begin{bmatrix} -2 \\ 6 \\ 2 \end{bmatrix}+x_{3}\begin{bmatrix} 4 \\ -1 \\ 5 \end{bmatrix}=\begin{bmatrix} 5x_{1}-2x_{2}+4x_{3} \\ 3x_{1}+6x_{2}-x_{3} \\ 7x_{1}+2x_{2}+5x_{3} \end{bmatrix}} \]

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6.

Rewrite each system of linear equations in the matrix-vector form \( A\mathbf{x}=\mathbf{b} \) . Be sure to identify \( A \) , \( \mathbf{x} \) , and \( \mathbf{b} \) explicitly. In part (c), reorder the equations and unknowns as needed so that the coefficient matrix is in the natural \( x_{1},x_{2},x_{3} \) column order.

(a) \( \begin{matrix} 4x_{1}-2x_{2} & =6 \\ 3x_{1}+5x_{2} & =-1 \end{matrix} \)

(b) \( \begin{matrix} 5x_{1}+2x_{2}-3x_{3} & =4 \\ 7x_{2}+4x_{3} & =1 \\ -2x_{3} & =6 \end{matrix} \)

(c) \( \begin{matrix} x_{3}-2x_{1} & =5 \\ 4x_{2}+x_{1} & =3 \\ 3x_{3}-x_{2}+2x_{1} & =8 \end{matrix} \)

Part (a). Read coefficients off the two equations:

\[ A=\begin{bmatrix} 4 & -2 \\ 3 & 5 \end{bmatrix},\,\mathbf{x}=\begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix},\,\mathbf{b}=\begin{bmatrix} 6 \\ -1 \end{bmatrix} \]

Part (b). Insert zero coefficients for any missing variables. The system is upper triangular.

\[ A=\begin{bmatrix} 5 & 2 & -3 \\ 0 & 7 & 4 \\ 0 & 0 & -2 \end{bmatrix},\,\mathbf{x}=\begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix},\,\mathbf{b}=\begin{bmatrix} 4 \\ 1 \\ 6 \end{bmatrix} \]

Part (c). First rearrange each equation so that \( x_{1},x_{2},x_{3} \) appear in order on the left:

\[ \begin{matrix} -2x_{1}\phantom{+0x_{2}}+x_{3} & =5 \\ \phantom{-}x_{1}+4x_{2}\phantom{+0x_{3}} & =3 \\ \phantom{-}2x_{1}-x_{2}+3x_{3} & =8 \end{matrix} \]

Now read off the matrix form:

\[ A=\begin{bmatrix} -2 & 0 & 1 \\ 1 & 4 & 0 \\ 2 & -1 & 3 \end{bmatrix},\,\mathbf{x}=\begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix},\,\mathbf{b}=\begin{bmatrix} 5 \\ 3 \\ 8 \end{bmatrix} \]

In every part, \( A\mathbf{x}=\mathbf{b} \) with the matrices above.

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Section 7.1: Introduction to Systems of First-Order ODEs

7.

For each system below, determine whether it is linear or nonlinear. If linear, also state whether it is homogeneous or nonhomogeneous, and write it in matrix-vector form \( {\mathbf{x}}^{'}=Q(t)\mathbf{x}+\mathbf{g}(t) \) .

(a) \( \begin{matrix} x_{1}^{'}+2x_{2} & =4t \\ x_{1}\,x_{2}^{'}-x_{2} & =e^{t} \end{matrix} \)

(b) \( \begin{matrix} 2t\,x_{1}^{'}-x_{2}+4tx_{3} & =\sin (t) \\ x_{3}^{'}+3x_{3} & =0 \\ x_{2}^{'}-\cos (t)\,x_{3} & =0 \end{matrix} \)

(c) \( \begin{matrix} x_{1}^{'}+x_{2}\,x_{3} & =5 \\ x_{2}^{'}+\sin (x_{1}) & =2t \\ 4x_{3}^{'}-(t^{2}+1)\,x_{1} & =0 \end{matrix} \)

(d) \( \begin{matrix} e^{t}\,x_{1}^{'}-x_{3} & =0 \\ x_{3}^{'}-4t\,x_{2} & =3\ln (t)\,x_{3} \\ x_{2}^{'}+5x_{3} & =0 \end{matrix} \)

Linearity test. A system is linear when each equation is a linear combination of \( x_{1},\ldots ,x_{n} \) and their first derivatives, with coefficients depending only on \( t \) . Products of unknowns, products of unknowns with derivatives, powers, and transcendental functions of unknowns make it nonlinear.

(a) Nonlinear. The second equation contains \( x_{1}\,x_{2}^{'} \) .

(b) Linear nonhomogeneous. Solve each equation for the appropriate derivative:

  • Eq. 2: \( x_{3}^{'}=-3x_{3} \) .
  • Eq. 3: \( x_{2}^{'}=\cos (t)\,x_{3} \) .
  • Eq. 1 (divide by \( 2t \) ): \( x_{1}^{'}=\dfrac{1}{2t}\,x_{2}-2x_{3}+\dfrac{\sin (t)}{2t} \) .
\[ {\mathbf{x}}^{'}=\begin{bmatrix} 0 & \dfrac{1}{2t} & -2 \\ 0 & 0 & \cos (t) \\ 0 & 0 & -3 \end{bmatrix}\mathbf{x}+\begin{bmatrix} \sin (t)/(2t) \\ 0 \\ 0 \end{bmatrix},\,t\ne 0. \]

(c) Nonlinear. Equation 1 contains \( x_{2}x_{3} \) and equation 2 contains \( \sin (x_{1}) \) .

(d) Linear homogeneous. Solve each equation for the appropriate derivative:

  • Eq. 1: \( x_{1}^{'}=e^{-t}\,x_{3} \) .
  • Eq. 3: \( x_{2}^{'}=-5\,x_{3} \) .
  • Eq. 2: \( x_{3}^{'}=4t\,x_{2}+3\ln (t)\,x_{3} \) .
\[ {\mathbf{x}}^{'}=\begin{bmatrix} 0 & 0 & e^{-t} \\ 0 & 0 & -5 \\ 0 & 4t & 3\ln (t) \end{bmatrix}\mathbf{x},\,t>0. \]

(a) nonlinear • (b) linear nonhomogeneous • (c) nonlinear • (d) linear homogeneous.

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8.

Two tanks are connected by pipes. Tank 1 holds \( 25\,\text{L} \) and Tank 2 holds \( 30\,\text{L} \) (volumes remain constant throughout).

  • Salt water containing \( 3\,\text{g/L} \) of salt flows into Tank 1 at a rate of \( 6\,\text{L/min} \) .
  • Salt water containing \( 5\,\text{g/L} \) of salt flows into Tank 2 at a rate of \( 4\,\text{L/min} \) .
  • Salt water flows out of Tank 1 at a rate of \( 9\,\text{L/min} \) , of which \( 6\,\text{L/min} \) enters Tank 2 and the remainder leaves the system.
  • Salt water flows out of Tank 2 at a rate of \( 10\,\text{L/min} \) , of which \( 3\,\text{L/min} \) enters Tank 1 and the remainder leaves the system.

Initially, Tank 1 contains \( 80\,\text{g} \) of salt and Tank 2 contains \( 50\,\text{g} \) of salt. Let \( Q_{1}(t) \) and \( Q_{2}(t) \) denote the amounts of salt (in grams) in Tanks 1 and 2 at time \( t \) . Write down an initial value problem (a system of two first-order ODEs together with initial conditions) that models the salt content. Express your final answer in matrix-vector form.

Step 1: Always check volume balance first. For the concentration formulas \( Q_{i}/V_{i} \) to stay valid, each tank's total inflow rate must equal its total outflow rate.

  • Tank 1: in \( =6\,\text{L/min} \) (external) \( +\,3\,\text{L/min} \) (from T2) \( =9\,\text{L/min} \) ; out \( =9\,\text{L/min} \) . Balanced. ✓
  • Tank 2: in \( =4\,\text{L/min} \) (external) \( +\,6\,\text{L/min} \) (from T1) \( =10\,\text{L/min} \) ; out \( =10\,\text{L/min} \) (of which \( 3\,\text{L/min} \) returns to Tank 1 and \( 7\,\text{L/min} \) leaves the system). Balanced. ✓

Both volumes are constant, so we may use the simple concentration formulas \( Q_{1}/25 \) and \( Q_{2}/30 \) throughout.

Step 2: Write each rate equation as (rate in) \( - \) (rate out). For each salt stream, the mass rate is (concentration) \( \,\times \, \) (volumetric flow rate).

Tank 1 derivative. Salt enters from the external pipe and from Tank 2; it leaves through the single \( 9\,\text{L/min} \) outflow.

\[ Q_{1}^{'}=\underbrace{\underbrace{(3\,\text{g/L})(6\,\text{L/min})}}+\underbrace{\underbrace{\left( \dfrac{Q_{2}}{30} \right)(3\,\text{L/min})}}-\underbrace{\underbrace{\left( \dfrac{Q_{1}}{25} \right)(9\,\text{L/min})}}=18+\dfrac{Q_{2}}{10}-\dfrac{9Q_{1}}{25} \]

Tank 2 derivative. Salt enters from the external pipe and from Tank 1; it leaves through the total \( 10\,\text{L/min} \) outflow (we do not need to split it into the " \( 3 \) to T1" and " \( 7 \) out" parts — the concentration leaving is the same regardless of destination).

\[ Q_{2}^{'}=\underbrace{\underbrace{(5\,\text{g/L})(4\,\text{L/min})}}+\underbrace{\underbrace{\left( \dfrac{Q_{1}}{25} \right)(6\,\text{L/min})}}-\underbrace{\underbrace{\left( \dfrac{Q_{2}}{30} \right)(10\,\text{L/min})}}=20+\dfrac{6Q_{1}}{25}-\dfrac{Q_{2}}{3} \]

Step 3: Collect in matrix-vector form. Read the coefficients of \( Q_{1} \) and \( Q_{2} \) off each equation to fill in the \( 2\times 2 \) matrix \( Q \) , and the constants into the forcing vector \( \mathbf{g} \) .

\[ \begin{bmatrix} Q_{1}^{'} \\ Q_{2}^{'} \end{bmatrix}=\begin{bmatrix} -\dfrac{9}{25} & \dfrac{1}{10} \\ \dfrac{6}{25} & -\dfrac{1}{3} \end{bmatrix}\begin{bmatrix} Q_{1} \\ Q_{2} \end{bmatrix}+\begin{bmatrix} 18 \\ 20 \end{bmatrix} \]

Append the initial conditions \( Q_{1}(0)=80\,\text{g} \) and \( Q_{2}(0)=50\,\text{g} \) to complete the IVP.

\[ \boxed{{\mathbf{Q}}^{'}(t)=\begin{bmatrix} -9/25 & 1/10 \\ 6/25 & -1/3 \end{bmatrix}\mathbf{Q}(t)+\begin{bmatrix} 18 \\ 20 \end{bmatrix},\,\mathbf{Q}(0)=\begin{bmatrix} 80 \\ 50 \end{bmatrix}} \]

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9.

At time \( t=0 \) , a population of \( 200 \) deer lives in a forest, and \( 12 \) cougars live in the same area. Make the following modeling assumptions:

  • In the absence of cougars, the deer population would grow at a rate proportional to itself, with positive growth constant \( a \) .
  • The cougars kill deer at a rate proportional to the product of the two populations, with positive predation constant \( b \) .
  • In the absence of deer, the cougar population would die out at a rate proportional to itself, with positive death constant \( c \) .
  • The cougar population grows at a rate proportional to the number of deer killed, with positive conversion constant \( d \) .
  • No deer or cougars enter or leave the area from outside, and there are no other sources of birth or death for either species.

Let \( D(t) \) denote the deer population and \( C(t) \) denote the cougar population at time \( t\ge 0 \) . Write down an initial value problem (a system of two first-order ODEs together with initial conditions) that models this situation. Then state whether your system is linear or nonlinear, and justify your answer.

Step 1: Translate each assumption into a term.

  • Deer grow at a rate proportional to \( D \) : contributes \( +aD \) to \( D^{'} \) .
  • Cougars kill deer at a rate proportional to \( DC \) : contributes \( -bDC \) to \( D^{'} \) .
  • Cougars die at a rate proportional to \( C \) : contributes \( -cC \) to \( C^{'} \) .
  • Cougar births proportional to kills: contributes \( +dDC \) to \( C^{'} \) .

Step 2: Assemble the system (Lotka–Volterra).

\[ \begin{cases} D^{'}(t)=aD-bDC \\ C^{'}(t)=-cC+dDC \end{cases},\,D(0)=200,\ C(0)=12 \]

Step 3: Classify. Both equations contain the product \( DC \) of the two unknowns. This violates linearity (a linear system would allow only sums of the form \( \alpha (t)\,D+\beta (t)\,C \) on the right). The system is therefore nonlinear. Because of this nonlinearity, it cannot be written in the matrix form \( {\mathbf{x}}^{'}=Q(t)\mathbf{x}+\mathbf{g}(t) \) .

\[ \boxed{\begin{cases} D^{'}=aD-bDC \\ C^{'}=-cC+dDC \end{cases},\,D(0)=200,\ C(0)=12\,\text{(nonlinear)}} \]

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10.

Rewrite the following second-order differential equation as an equivalent system of two first-order equations. Then express your system in matrix-vector form.

\[ y^{''}+4y^{'}-3y=\sin (2t) \]

Step 1: Introduce new variables. Let \( x_{1}=y \) and \( x_{2}=y^{'} \) . Then automatically \( x_{1}^{'}=x_{2} \) .

Step 2: Solve the ODE for the highest derivative.

\[ y^{''}=-4y^{'}+3y+\sin (2t)=-4x_{2}+3x_{1}+\sin (2t) \]

So \( x_{2}^{'}=y^{''}=3x_{1}-4x_{2}+\sin (2t) \) .

Step 3: Assemble the system in matrix-vector form.

\[ \begin{bmatrix} x_{1}^{'} \\ x_{2}^{'} \end{bmatrix}=\begin{bmatrix} 0 & 1 \\ 3 & -4 \end{bmatrix}\begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix}+\begin{bmatrix} 0 \\ \sin (2t) \end{bmatrix} \]
\[ \boxed{{\mathbf{x}}^{'}=\begin{bmatrix} 0 & 1 \\ 3 & -4 \end{bmatrix}\mathbf{x}+\begin{bmatrix} 0 \\ \sin (2t) \end{bmatrix}} \]

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11.

Rewrite the following initial value problem as an equivalent first-order system together with initial conditions for the new variables.

\[ y^{‴}+t^{2}\,y^{''}-\cos (t)\,y^{'}+2y=5t^{3},\,y(0)=1,\ y^{'}(0)=-2,\ y^{''}(0)=4 \]

Step 1: Define the new variables. Set \( x_{1}=y,\ x_{2}=y^{'},\ x_{3}=y^{''} \) . Then \( x_{1}^{'}=x_{2} \) and \( x_{2}^{'}=x_{3} \) automatically.

Step 2: Solve for \( y^{‴} \) and substitute.

\[ y^{‴}=-t^{2}y^{''}+\cos (t)\,y^{'}-2y+5t^{3}=-t^{2}x_{3}+\cos (t)\,x_{2}-2x_{1}+5t^{3} \]

Therefore \( x_{3}^{'}=-2x_{1}+\cos (t)\,x_{2}-t^{2}x_{3}+5t^{3} \) .

Step 3: Matrix form and initial conditions.

\[ \begin{bmatrix} x_{1}^{'} \\ x_{2}^{'} \\ x_{3}^{'} \end{bmatrix}=\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -2 & \cos (t) & -t^{2} \end{bmatrix}\begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix}+\begin{bmatrix} 0 \\ 0 \\ 5t^{3} \end{bmatrix} \]

Initial conditions (translated from \( y,y^{'},y^{''} \) to \( x_{1},x_{2},x_{3} \) ):

\[ x_{1}(0)=1,\,x_{2}(0)=-2,\,x_{3}(0)=4 \]
\[ \boxed{{\mathbf{x}}^{'}=\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -2 & \cos (t) & -t^{2} \end{bmatrix}\mathbf{x}+\begin{bmatrix} 0 \\ 0 \\ 5t^{3} \end{bmatrix},\,\mathbf{x}(0)=\begin{bmatrix} 1 \\ -2 \\ 4 \end{bmatrix}} \]

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12.

Rewrite the following initial value problem as an equivalent first-order system with appropriate initial conditions.

\[ z^{(4)}+2t\,z^{‴}-z=e^{t}+\sin (t),\,z(2)=1,\ z^{'}(2)=4,\ z^{''}(2)=-3,\ z^{‴}(2)=5 \]

Step 1: Define the new variables. For a fourth-order ODE we need four: \( x_{1}=z,\ x_{2}=z^{'},\ x_{3}=z^{''},\ x_{4}=z^{‴} \) . Then \( x_{1}^{'}=x_{2} \) , \( x_{2}^{'}=x_{3} \) , \( x_{3}^{'}=x_{4} \) automatically.

Step 2: Solve for \( z^{(4)} \) .

\[ z^{(4)}=-2t\,z^{‴}+z+e^{t}+\sin (t)=-2t\,x_{4}+x_{1}+e^{t}+\sin (t) \]

So \( x_{4}^{'}=x_{1}-2t\,x_{4}+e^{t}+\sin (t) \) .

Step 3: Matrix form and initial conditions (at \( t_{0}=2 \) ).

\[ {\mathbf{x}}^{'}=\begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & -2t \end{bmatrix}\mathbf{x}+\begin{bmatrix} 0 \\ 0 \\ 0 \\ e^{t}+\sin (t) \end{bmatrix} \]
\[ \mathbf{x}(2)=\begin{bmatrix} 1 \\ 4 \\ -3 \\ 5 \end{bmatrix} \]

System, forcing vector, and initial condition as above.

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Section 7.3: Linear Independence, Eigenvalues, and Eigenvectors

13.

For each pair of vectors, decide whether they are linearly dependent or linearly independent. Justify each answer briefly.

(a) \( \left\{ \begin{bmatrix} 3 \\ -6 \end{bmatrix},\ \begin{bmatrix} -2 \\ 4 \end{bmatrix} \right\} \)

(b) \( \left\{ \begin{bmatrix} 4 \\ 3 \end{bmatrix},\ \begin{bmatrix} -2 \\ 5 \end{bmatrix} \right\} \)

(c) \( \left\{ \begin{bmatrix} 2 \\ 7 \end{bmatrix},\ \begin{bmatrix} 0 \\ 0 \end{bmatrix} \right\} \)

(d) \( \left\{ \begin{bmatrix} 5 \\ -3 \\ 2 \end{bmatrix},\ \begin{bmatrix} 4 \\ 6 \\ -1 \end{bmatrix} \right\} \)

Shortcut for pairs: two nonzero vectors are linearly dependent if and only if one is a scalar multiple of the other; if either vector is zero, the pair is automatically dependent.

Part (a). Check if \( \begin{bmatrix} -2 \\ 4 \end{bmatrix}=c\begin{bmatrix} 3 \\ -6 \end{bmatrix} \) . From the first component \( -2=3c\Rightarrow c=-2/3 \) . Check the second: \( -6\cdot (-2/3)=4 \) . ✓ So the second vector is \( -\dfrac{2}{3} \) times the first. Dependent.

Part (b). Check if \( \begin{bmatrix} -2 \\ 5 \end{bmatrix}=c\begin{bmatrix} 4 \\ 3 \end{bmatrix} \) . First component gives \( c=-1/2 \) , but then the second would be \( 3\cdot (-1/2)=-3/2\ne 5 \) . Not a scalar multiple. (Equivalently, the \( 2\times 2 \) determinant is \( 4(5)-3(-2)=26\ne 0 \) .) Independent.

Part (c). One of the vectors is the zero vector. Choosing \( c_{1}=0 \) and \( c_{2}=1 \) gives \( 0\cdot \begin{bmatrix} 2 \\ 7 \end{bmatrix}+1\cdot \begin{bmatrix} 0 \\ 0 \end{bmatrix}=0 \) with not-all-zero coefficients. Dependent.

Part (d). Check if \( \begin{bmatrix} 4 \\ 6 \\ -1 \end{bmatrix}=c\begin{bmatrix} 5 \\ -3 \\ 2 \end{bmatrix} \) . First component gives \( c=4/5 \) , but then the second would be \( -3\cdot (4/5)=-12/5\ne 6 \) . Not a scalar multiple. Independent.

(a) dependent • (b) independent • (c) dependent • (d) independent.

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14.

Find all eigenvalues and a corresponding eigenvector for each eigenvalue of the matrix below.

\[ A=\begin{bmatrix} 5 & -2 \\ 4 & -1 \end{bmatrix} \]

Step 1: Characteristic equation. For a \( 2\times 2 \) matrix, \( {\lambda }^{2}-\operatorname{tr}(A)\,\lambda +det(A)=0 \) . Here \( \operatorname{tr}(A)=5+(-1)=4 \) and \( det(A)=(5)(-1)-(-2)(4)=-5+8=3 \) :

\[ {\lambda }^{2}-4\lambda +3=0\ \implies \ (\lambda -1)(\lambda -3)=0 \]

Eigenvalues: \( {\lambda }_{1}=1,\ {\lambda }_{2}=3 \) .

Step 2: Eigenvector for \( {\lambda }_{1}=1 \) . Solve \( (A-I)\mathbf{v}=0 \) :

\[ A-I=\begin{bmatrix} 4 & -2 \\ 4 & -2 \end{bmatrix}\ \implies \ 4\mathbf{v}_{1}-2\mathbf{v}_{2}=0\ \implies \ \mathbf{v}_{2}=2\mathbf{v}_{1} \]

Take \( \mathbf{v}_{1}=1 \) : \( {\mathbf{v}}^{(1)}=\begin{bmatrix} 1 \\ 2 \end{bmatrix} \) . Quick check: \( A{\mathbf{v}}^{(1)}=\begin{bmatrix} 5-4 \\ 4-2 \end{bmatrix}=\begin{bmatrix} 1 \\ 2 \end{bmatrix}=1\cdot {\mathbf{v}}^{(1)} \) . ✓

Step 3: Eigenvector for \( {\lambda }_{2}=3 \) . Solve \( (A-3I)\mathbf{v}=0 \) :

\[ A-3I=\begin{bmatrix} 2 & -2 \\ 4 & -4 \end{bmatrix}\ \implies \ 2\mathbf{v}_{1}-2\mathbf{v}_{2}=0\ \implies \ \mathbf{v}_{1}=\mathbf{v}_{2} \]

Take \( \mathbf{v}_{2}=1 \) : \( {\mathbf{v}}^{(2)}=\begin{bmatrix} 1 \\ 1 \end{bmatrix} \) . Check: \( A{\mathbf{v}}^{(2)}=\begin{bmatrix} 5-2 \\ 4-1 \end{bmatrix}=\begin{bmatrix} 3 \\ 3 \end{bmatrix}=3\cdot {\mathbf{v}}^{(2)} \) . ✓

\[ \boxed{{\lambda }_{1}=1,\ {\mathbf{v}}^{(1)}=\begin{bmatrix} 1 \\ 2 \end{bmatrix};\,{\lambda }_{2}=3,\ {\mathbf{v}}^{(2)}=\begin{bmatrix} 1 \\ 1 \end{bmatrix}} \]

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15.

Find all eigenvalues and a corresponding eigenvector for each eigenvalue of the matrix below.

\[ A=\begin{bmatrix} 2 & -5 \\ 1 & 0 \end{bmatrix} \]

Observation: The eigenvalues here turn out to be complex conjugates. When this happens, the corresponding eigenvectors are also complex conjugates of each other — so once you find one eigenvector, you get the other for free by conjugation.

Step 1: Characteristic equation. \( \operatorname{tr}(A)=2 \) , \( det(A)=(2)(0)-(-5)(1)=5 \) :

\[ {\lambda }^{2}-2\lambda +5=0\ \implies \ \lambda =\dfrac{2\pm \sqrt{4-20}}{2}=\dfrac{2\pm \sqrt{-16}}{2}=1\pm 2i \]

Step 2: Eigenvector for \( {\lambda }_{1}=1+2i \) . Solve \( (A-{\lambda }_{1}I)\mathbf{v}=0 \) :

\[ A-(1+2i)I=\begin{bmatrix} 1-2i & -5 \\ 1 & -1-2i \end{bmatrix} \]

The second row (which is usually simpler) gives \( \mathbf{v}_{1}+(-1-2i)\,\mathbf{v}_{2}=0 \) , i.e.\ \( \mathbf{v}_{1}=(1+2i)\mathbf{v}_{2} \) . Taking \( \mathbf{v}_{2}=1 \) :

\[ {\mathbf{v}}^{(1)}=\begin{bmatrix} 1+2i \\ 1 \end{bmatrix} \]

Row-1 consistency check: \( (1-2i)(1+2i)-5(1)=(1-(2i)^{2})-5=(1+4)-5=0 \) . ✓

Step 3: Eigenvector for \( {\lambda }_{2}=1-2i \) . Because \( A \) is real, the second eigenvalue/eigenvector pair is the conjugate of the first (the observation in the problem statement):

\[ {\mathbf{v}}^{(2)}={\mathbf{v}}^{(1)}^{―}=\begin{bmatrix} 1-2i \\ 1 \end{bmatrix} \]
\[ \boxed{{\lambda }_{1,2}=1\pm 2i,\,{\mathbf{v}}^{(1,2)}=\begin{bmatrix} 1\pm 2i \\ 1 \end{bmatrix}} \]

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16.

Find all eigenvalues and a corresponding eigenvector for each eigenvalue of the symmetric matrix below.

\[ A=\begin{bmatrix} 2 & \sqrt{5} \\ \sqrt{5} & -2 \end{bmatrix} \]

Observation: Symmetric matrices always have real eigenvalues, and eigenvectors corresponding to distinct eigenvalues are orthogonal. After you finish, verify orthogonality by computing the dot product of the two eigenvectors.

Step 1: Characteristic equation. \( \operatorname{tr}(A)=2+(-2)=0 \) , and \( det(A)=(2)(-2)-(\sqrt{5})(\sqrt{5})=-4-5=-9 \) :

\[ {\lambda }^{2}-0\cdot \lambda +(-9)=0\ \implies \ {\lambda }^{2}=9\ \implies \ \lambda =\pm 3 \]

(As expected for a symmetric matrix, both eigenvalues are real.)

Step 2: Eigenvector for \( {\lambda }_{1}=3 \) . Solve \( (A-3I)\mathbf{v}=0 \) :

\[ A-3I=\begin{bmatrix} -1 & \sqrt{5} \\ \sqrt{5} & -5 \end{bmatrix}\ \implies \ -\mathbf{v}_{1}+\sqrt{5}\,\mathbf{v}_{2}=0\ \implies \ \mathbf{v}_{1}=\sqrt{5}\,\mathbf{v}_{2} \]

Take \( \mathbf{v}_{2}=1 \) : \( {\mathbf{v}}^{(1)}=\begin{bmatrix} \sqrt{5} \\ 1 \end{bmatrix} \) . Check with row 2: \( \sqrt{5}(\sqrt{5})-5(1)=5-5=0 \) . ✓

Step 3: Eigenvector for \( {\lambda }_{2}=-3 \) . Solve \( (A+3I)\mathbf{v}=0 \) :

\[ A+3I=\begin{bmatrix} 5 & \sqrt{5} \\ \sqrt{5} & 1 \end{bmatrix}\ \implies \ 5\mathbf{v}_{1}+\sqrt{5}\,\mathbf{v}_{2}=0\ \implies \ \mathbf{v}_{2}=-\sqrt{5}\,\mathbf{v}_{1} \]

Take \( \mathbf{v}_{1}=1 \) : \( {\mathbf{v}}^{(2)}=\begin{bmatrix} 1 \\ -\sqrt{5} \end{bmatrix} \) .

Step 4: Verify orthogonality. Compute the dot product:

\[ {\mathbf{v}}^{(1)}\cdot {\mathbf{v}}^{(2)}=(\sqrt{5})(1)+(1)(-\sqrt{5})=\sqrt{5}-\sqrt{5}=0\ ✓ \]

The eigenvectors are indeed perpendicular, as guaranteed by the spectral theorem for symmetric matrices.

\[ \boxed{{\lambda }_{1}=3,\ {\mathbf{v}}^{(1)}=\begin{bmatrix} \sqrt{5} \\ 1 \end{bmatrix};\,{\lambda }_{2}=-3,\ {\mathbf{v}}^{(2)}=\begin{bmatrix} 1 \\ -\sqrt{5} \end{bmatrix};\,{\mathbf{v}}^{(1)}\cdot {\mathbf{v}}^{(2)}=0} \]

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