IB Mathematics AHL 4.19 Transition matrices AI HL Paper 1 - Exam Style Questions- New Syllabus
▶️ Answer/Explanation
(a)
To find the eigenvector \(\begin{pmatrix} x \\ y \end{pmatrix}\), we solve \((\mathbf{T} – \mathbf{I})\mathbf{v} = \mathbf{0}\):
\(\begin{pmatrix} p-1 & 1-q \\ 1-p & q-1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}\).
From the first row: \((p-1)x + (1-q)y = 0\).
Rearranging gives \((1-p)x = (1-q)y\), which implies \(\frac{x}{y} = \frac{1-q}{1-p}\).
A corresponding eigenvector is \(\begin{pmatrix} 1-q \\ 1-p \end{pmatrix}\).
(b)
The steady-state proportion in state \(A\) is the value of \(x\) when the components of the eigenvector are normalized to sum to \(1\).
Proportion in \(A = \frac{x}{x+y} = \frac{1-q}{(1-q)+(1-p)} = \frac{1-q}{2-p-q}\).
(c)
Substituting \(p=0.3\) and \(q=0.6\) into the expression from part (b):
\(\frac{1-0.6}{2-0.3-0.6} = \frac{0.4}{1.1} = \frac{4}{11}\).
Proportion \(\approx 0.364\).