IB Mathematics AHL 1.15 Eigenvalues and eigenvectors AI HL Paper 3- Exam Style Questions- New Syllabus

(a)
The transition matrix columns represent “from” states (Alpha, Beta, Gamma) and rows represent “to” states. Each column sums to 1 (100%).
\(T = \begin{pmatrix} 0.91 & 0.04 & 0.06 \\ 0.05 & 0.95 & 0.02 \\ 0.04 & 0.01 & 0.92 \end{pmatrix}\)

(b)
Over 5 years means 5 transitions. We need \(T^5\). The (1,3) entry gives the proportion moving from Gamma to Alpha after 5 years.
Using technology: \(T^5 \approx \begin{pmatrix} 0.659 & 0.156 & 0.219 \\ 0.196 & 0.792 & 0.101 \\ 0.146 & 0.0514 & 0.680 \end{pmatrix}\)
The (1,3) entry is 0.219 → 21.9%.
\(\boxed{21.9\%}\)

(c)
The steady state vector \(\mathbf{v}\) satisfies \(T\mathbf{v} = \mathbf{v}\) with components summing to 1.
Solving \((T-I)\mathbf{v} = \mathbf{0}\) with \(v_1+v_2+v_3=1\):
Using technology or solving: \(\mathbf{v} \approx \begin{pmatrix} 0.342 \\ 0.432 \\ 0.225 \end{pmatrix}\)
\(\boxed{\begin{pmatrix} 0.342 \\ 0.432 \\ 0.225 \end{pmatrix}}\)

(d)
From steady state vector, long-term proportion of customers at Alpha is 34.2%.
\(\boxed{34.2\%}\)

(e)
Let new transition from Gamma to Alpha be \(x\) (instead of 0.06). Then Gamma to Gamma becomes \(0.92 – (x-0.06) = 0.98 – x\).
New \(T’ = \begin{pmatrix} 0.91 & 0.04 & x \\ 0.05 & 0.95 & 0.02 \\ 0.04 & 0.01 & 0.98-x \end{pmatrix}\)
We want steady-state Alpha component \(\geq 0.40\).
Testing integers: at \(x=0.13\) (13%), steady state ≈ \(\begin{pmatrix} 0.403 \\ 0.459 \\ 0.138 \end{pmatrix}\).
At \(x=0.12\) (12%), steady state ≈ \(\begin{pmatrix} 0.395 \\ 0.456 \\ 0.149 \end{pmatrix}\) (under 40%).
Minimum integer percentage is \(\boxed{13\%}\).

(f)
From (e), new steady state gives Beta 45.9% vs original 43.2%. Since 0.459 > 0.432, Beta’s market share increases.
\(\boxed{\text{Yes, Beta benefits as its long-term market share increases from 43.2% to 45.9%.}}\)

(g)
Real-world factors: stores may change marketing strategies, new competitors may enter, economic conditions change, customer preferences evolve.
\(\boxed{\text{Changes in store marketing strategies or economic conditions could alter customer behavior year to year.}}\)

(h)
Arithmetic sequence: \(u_n = u_1 + (n-1)d\)
\(u_{10} = 600 + (10-1) \times 30 = 600 + 270 = 870\)
\(\boxed{870}\)

(i)
Week 5 customers: \(600 + 4 \times 30 = 720\) (Model 1 up to week 5)
Week 6 under Model 2: \(720 + 30 \times 1.12 = 720 + 33.6 = 753.6\)
\(\boxed{753.6}\)

(j)
From week 5: 720 customers.
Week 6–10 is 5 weeks of geometric growth in new customers: first new increment = 33.6, ratio = 1.12
Sum of geometric series: \(S_5 = 33.6 \times \frac{1.12^5 – 1}{1.12 – 1} \approx 33.6 \times 6.3528 \approx 213.455\)
Total at week 10: \(720 + 213.455 \approx 933.455\)
\(\boxed{933}\) (or 933.46)

(k)
Method 1:
Model 1: \(A_n = 600 + 30(n-1)\)
Model 2: For \(n \geq 6\), \(G_n = 720 + 30 \times 1.12 \times \frac{1.12^{n-5} – 1}{0.12}\)
Solve \(G_n – A_n \geq 500\)
Using technology: \(n \approx 17.505\) → week 18.
Method 2:
Create lists:
Week 17: Model 1 = 1080, Model 2 ≈ 1530.87 (difference ≈ 450.87)
Week 18: Model 1 = 1110, Model 2 ≈ 1661.77 (difference ≈ 551.77) ≥ 500
First week with difference ≥ 500 is \(\boxed{\text{week 18}}\).