Matrices Class Log

Grade 12 – Mathematics HL (A&I) - Matrices Class Log

Date: Tuesday 3rd of February 2026

Duration: 45 minutes
Overview
  • Focus: Differential equations (coupled systems)

  • Homework Correction: Kognity assignment


Differential Equations

Exact Solutions of Coupled Systems of Differential Equations
  • Theorem: The solutions of the coupled linear system dxdt=Mx\frac{d\mathbf{x}}{dt} = \mathbf{M}\mathbf{x} where x=C(x y)D\mathbf{x} = C \begin{pmatrix} x \ y \end{pmatrix} D can be expressed as: x=Aeλ<em>1v</em>1t+Beλ<em>2v</em>2t\mathbf{x} = \mathbf{A} e^{\lambda<em>1 \mathbf{v</em>1} t} + \mathbf{B} e^{\lambda<em>2 \mathbf{v</em>2} t}

    • Where:

    • λ<em>1\lambda<em>1 and λ</em>2\lambda</em>2 are the eigenvalues of the matrix M\mathbf{M} associated with the eigenvectors v<em>1\mathbf{v<em>1} and v</em>2\mathbf{v</em>2}

    • Constants A,BR\mathbf{A}, \mathbf{B} \in \mathbb{R} depend on initial conditions

  • Examination Note: Exact solutions will only be calculated for real distinct eigenvalues.


Phase Portraits

Definition and Description
  • Phase Portrait: A representative set of solutions of the coupled system of differential equations
    dxdt=Mx\frac{d\mathbf{x}}{dt} = \mathbf{M}\mathbf{x}

  • Parametric Curves:

    • Curves plotted on the Cartesian plane.

    • Illustrate the path of each particular solution over time ($t$).

  • Specific Example:

    • Lucia's Mathematical Exploration involved plotting phase trajectories (solution curves) in a phase plane representing the effector cell and tumour cell populations.

  • Equilibrium Point: The system stabilizes where both populations remain constant.

    • For equilibrium point (12,16)(12, 16), it holds that:

    • dEdt=0\frac{dE}{dt} = 0

    • dTdt=0\frac{dT}{dt} = 0

  • Components of Phase Portraits:

    • Consists of equilibrium points and a selection of phase trajectories, demonstrating how variable values change over time.


Properties of Eigenvalues and Eigenvectors

Behaviour of the Coupled System
  • Distinct Eigenvalues: When the matrix representing the coupled system has distinct eigenvalues, several observations can be made:

    • Lines through the origin along the direction of eigenvectors indicate gradients for long-term behaviour as t+t \rightarrow +\infty and tt \rightarrow -\infty.

    • Positive and Negative Eigenvalues:

    • Negative Eigenvalue ($\lambda$):

      • Solutions are parallel to the eigenvector v\mathbf{v} and are directed towards the origin as tt increases.

    • Positive Eigenvalue ($\lambda$):

      • Solutions are parallel to v\mathbf{v} and move away from the origin as tt increases.

    • Mixed Signs:

      • With one positive and one negative eigenvalue, the solution demonstrates a combined behaviour:

      • For large negative tt, the curve aligns with the negative eigenvalue’s solution, moving towards the origin.

      • For large positive tt, it aligns with the positive eigenvalue’s solution, moving away from the origin.


Example of a Coupled System of Differential Equations
  • System:
    dxdt=xdydt=y\frac{dx}{dt} = x \qquad \frac{dy}{dt} = -y
    This can be expressed as:
    (dxdt dydt)=(1amp;0 0amp;1)(x y)\begin{pmatrix} \frac{dx}{dt} \ \frac{dy}{dt} \end{pmatrix} = \begin{pmatrix} 1 &amp; 0 \ 0 &amp; -1 \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix}

  • Eigenvalues:

    • λ<em>1=1\lambda<em>1 = 1 and λ</em>2=1\lambda</em>2 = -1

  • Eigenvectors:

    • (1 0)\begin{pmatrix} 1 \ 0 \end{pmatrix} and (0 1)\begin{pmatrix} 0 \ 1 \end{pmatrix}

  • General Solution:
    (x y)=Aet(1 0)+Bet(0 1)\begin{pmatrix} x \ y \end{pmatrix} = \mathbf{A} e^{t} \begin{pmatrix} 1 \ 0 \end{pmatrix} + \mathbf{B} e^{-t} \begin{pmatrix} 0 \ 1 \end{pmatrix}, where tRt \in \mathbb{R}.

  • Long-Term Behaviour:

    • For large negative tt, solutions approach the origin along the yy-axis.

    • For large positive tt, solutions diverge along the xx-axis.

  • Equilibrium Point:

    • A point where both derivatives equal zero:

    • dxdt=0\frac{dx}{dt} = 0 and dydt=0\frac{dy}{dt} = 0

  • Stability Conditions:

    • Stable Equilibrium Point:

    • All nearby points converge to this point (conditions met when both eigenvalues are negative).

    • Unstable Equilibrium Point:

    • All nearby points diverge from this point (conditions met when both eigenvalues are positive).

    • Saddle Point:

    • One positive and one negative eigenvalue indicates that trajectories approach the origin and subsequently turn away, leading to instability.


Further Example of a Coupled System of Differential Equations
  • Coupled System:
    dxdtamp;=x+9y dydtamp;=6x4y\begin{aligned} \frac{dx}{dt} &amp;= -x + 9y \ \frac{dy}{dt} &amp;= 6x - 4y \end{aligned}

  • Matrix Representation:
    (dxdt dydt)=(1amp;9 6amp;4)(x y)\begin{pmatrix} \frac{dx}{dt} \ \frac{dy}{dt} \end{pmatrix} = \begin{pmatrix} -1 &amp; 9 \ 6 &amp; -4 \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix}

  • Eigenvalues:

    • λ<em>1=1\lambda<em>1 = 1 and λ</em>2=2\lambda</em>2 = -2

  • Eigenvectors:

    • (3 2)\begin{pmatrix} 3 \ 2 \end{pmatrix} and (1 1)\begin{pmatrix} 1 \ -1 \end{pmatrix}

  • General Solution:
    (x y)=Aet(3 2)+Be2t(1 1),  tR\begin{pmatrix} x \ y \end{pmatrix} = \mathbf{A} e^{t} \begin{pmatrix} 3 \ 2 \end{pmatrix} + \mathbf{B} e^{-2t} \begin{pmatrix} 1 \ -1 \end{pmatrix}, \; t \in \mathbb{R}

  • Behaviour as tt Approaches Infinity:

    • For large negative tt, curves follow (1 1)\begin{pmatrix} 1 \ -1 \end{pmatrix} (towards origin).

    • For large positive tt, curves follow (3 2)\begin{pmatrix} 3 \ 2 \end{pmatrix} (away from origin).

  • Equilibrium Point Analysis:

    • The origin (0,0)(0, 0) is classified as a saddle point.


Additional Exercise and Resources
  • Exercise 20.4.1:

    • Complete questions (a) to (d) on pages 881-882.

  • Applet for Phase Portrait Verification:

    • Utilize the applet available at:
      GeoGebra Applet to verify the phase portrait.

  • Slope Field Diagram:

    • A visual representation of the coupled system of differential equations defined as:
      dxdtamp;=y+x×(x2+y2)×sin(πx2+y2) dydtamp;=x+y×(x2+y2)×sin(πx2+y2)\begin{aligned} \frac{dx}{dt} &amp;= -y + x \times (x^2 + y^2) \times \sin\left(\frac{\pi}{x^2 + y^2}\right) \ \frac{dy}{dt} &amp;= x + y \times (x^2 + y^2) \times \sin\left(\frac{\pi}{x^2 + y^2}\right) \end{aligned}

    • Reference for visual representation:

      Slope Field Diagram