Problem Statement
Two competing retail chains have store growth rates modelled by the coupled linear differential equations:\n\n$$\begin{pmatrix} \frac{dx}{dt} \\ \frac{dy}{dt} \end{pmatrix} = \begin{pmatrix} 1 & 3 \\ 2 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}$$
**(a)** [4 marks] Find the eigenvalues $\lambda_1$ and $\lambda_2$ of the coefficient matrix.
**(b)** [3 marks] Find the normalized eigenvectors corresponding to each eigenvalue.
**(c)** [2 marks] Classify the equilibrium point at the origin $(0, 0)$ and state whether it is stable or unstable.
Verified Solution & Marking Scheme
Part (a) [M1]
Setting $\det(A - \lambda I) = 0 \implies (1 - \lambda)(2 - \lambda) - 6 = 0$
Part (a) [A1]
$\lambda^2 - 3\lambda + 2 - 6 = 0 \implies \lambda^2 - 3\lambda - 4 = 0$
Part (a) [A1]
$(\lambda - 4)(\lambda + 1) = 0$
Part (a) [A1]
$\lambda_1 = 4, \quad \lambda_2 = -1$
Part (b) [M1]
For $\lambda_1 = 4$: $\begin{pmatrix} -3 & 3 \\ 2 & -2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \mathbf{0} \implies -3x + 3y = 0 \implies x = y$
Part (b) [A1]
$\mathbf{v}_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$ (or normalized $\frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 1 \end{pmatrix}$)
Part (b) [M1]
For $\lambda_2 = -1$: $\begin{pmatrix} 2 & 3 \\ 2 & 3 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \mathbf{0} \implies 2x + 3y = 0 \implies y = -\frac{2}{3}x$
Part (b) [A1]
$\mathbf{v}_2 = \begin{pmatrix} 3 \\ -2 \end{pmatrix}$
Part (c) [R1]
Since one eigenvalue is positive ($\lambda_1 = 4 > 0$) and one is negative ($\lambda_2 = -1 < 0$)
Part (c) [A1]
The equilibrium point is an unstable saddle point.