IB Maths AI HL 1.15 Eigenvalues and eigenvectors Question Bank

Practise IB Mathematics HL 1.15 by applying eigenvalues and eigenvectors methods to exam-style questions.

[Maximum number: 6]

A geneticist uses a Markov chain model to investigate changes in a specific gene in a cell as it divides. Every time the cell divides, the gene may mutate between its normal state and other states.
The model is of the form

\binom{X_{n+1}}{Z_{n+1}}=\boldsymbol{M}\binom{X_n}{Z_n}

where $X_n$ is the probability of the gene being in its normal state after dividing for the nth time, and $Z_n$ is the probability of it being in another state after dividing for the nth time, where $n\in\mathbb{N}$ .
Matrix $\boldsymbol{M}$ is found to be $\left(\begin{smallmatrix}0.94&b\\0.06&0.98\end{smallmatrix}\right)$ .

(a)

Find the eigenvalues of M.

[ 3 ]

$\det\left(\begin{smallmatrix}0.94-\lambda&0.02\\0.06&0.98-\lambda\end{smallmatrix}\right)=0$
(M1)
$(0.94-\lambda)(0.98-\lambda)-0.0012=0$ OR $\lambda^2-1.92\lambda+0.92=0$ $\lambda=1,\ 0.92\ (\frac{23}{25})$
Note: Award M1 for an attempt to find eigenvalues. Any indication that $\det(\boldsymbol{M}-\lambda\boldsymbol{I})=0$ has been used is sufficient.

(b)

Find the eigenvectors of M.

[ 3 ]

$\left(\begin{smallmatrix}0.94&0.02\\0.06&0.98\end{smallmatrix}\right)\binom{x}{y}=\binom{x}{y}$ OR $\left(\begin{smallmatrix}0.94&0.02\\0.06&0.98\end{smallmatrix}\right)\binom{x}{y}=0.92\binom{x}{y}$
Note: This M1 can be awarded for attempting to find either eigenvector.
0.02y-0.06x=0 OR 0.02y+0.02x=0
Eigenvectors $\binom13$ and $\binom1{-1}$ .
Note: Accept any multiple of the given eigenvectors.

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AHL 1.15 (HL)—Eigenvalues and eigenvectors question 1

Question (a)

Question (b)