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IB Mathematics AHL 1.15 Eigenvalues and eigenvectors AI HL Paper 2- Exam Style Questions- New Syllabus

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Question

A geneticist uses a Markov chain to model the changes in a specific gene as a cell divides. The system is defined by the relation \(\binom{X_{n+1}}{Z_{n+1}} = \mathbf{M} \binom{X_n}{Z_n}\), where \(X_n\) represents the probability of the gene being in its normal state after \(n\) divisions, \(Z_n\) is the probability of it being in a mutated state, and the transition matrix is \(\mathbf{M} = \begin{pmatrix}0.94 & b \\ 0.06 & 0.98\end{pmatrix}\).
(a) (i) State the value of \(b\).
(ii) Interpret the meaning of \(b\) within the context of this genetic model.
(b) Calculate the eigenvalues of the matrix \(\mathbf{M}\).
(c) Determine the eigenvectors corresponding to each eigenvalue found in part (b).
It is given that the gene is in its normal state at the start (\(n = 0\)).
(d) Calculate the probability that the gene is in its normal state:
(i) after exactly \(5\) cell divisions.
(ii) in the long term as \(n \to \infty\).

Most-appropriate topic codes:

• AHL 1.15: Eigenvalues and eigenvectors — parts (b) and (c)
• AHL 4.19: Markov chains, transition matrices, and steady state probabilities — parts (a) and (d)
▶️ Answer/Explanation
Detailed solution

(a)
(i) In a stochastic (transition) matrix, each column must sum to \(1\).
\(b + 0.98 = 1 \Rightarrow \mathbf{b = 0.02}\).
(ii) \(b\) represents the probability of the gene reverting from the mutated state to the normal state during a cell division.

(b)
Solve the characteristic equation \(\det(\mathbf{M} – \lambda\mathbf{I}) = 0\):
\(\det\begin{pmatrix}0.94 – \lambda & 0.02 \\ 0.06 & 0.98 – \lambda\end{pmatrix} = 0\)
\(\lambda^2 – 1.92\lambda + 0.92 = 0\)
\(\lambda = 1\) and \(\lambda = 0.92\) (or \(\frac{23}{25}\)).

(c)
For \(\lambda = 1\): \(\begin{pmatrix}-0.06 & 0.02 \\ 0.06 & -0.02\end{pmatrix}\binom{x}{y} = 0 \Rightarrow -3x + y = 0 \Rightarrow y = 3x\). Eigenvector is \(\binom{1}{3}\).
For \(\lambda = 0.92\): \(\begin{pmatrix}0.02 & 0.02 \\ 0.06 & 0.06\end{pmatrix}\binom{x}{y} = 0 \Rightarrow x + y = 0 \Rightarrow y = -x\). Eigenvector is \(\binom{1}{-1}\).

(d)
Initial state \(\mathbf{s}_0 = \binom{1}{0}\).
(i) \(\mathbf{s}_5 = \mathbf{M}^5 \binom{1}{0}\). Using technology to find \(\mathbf{M}^5 \approx \begin{pmatrix}0.7443 & 0.0852 \\ 0.2557 & 0.9148\end{pmatrix}\).
Probability of normal state \(\approx \mathbf{0.744}\).
(ii) The long-term state is the steady-state vector. Normalize the eigenvector for \(\lambda=1\):
Sum of components \(= 1 + 3 = 4\).
Steady state \(= \binom{1/4}{3/4} = \binom{0.25}{0.75}\).
Long-term probability of normal state = \(0.25\).

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