Transition matrices, state matrices and equilibrium: VCE General Mathematics Unit 4 Matrices

(a) The leading diagonal holds the proportions who stay in the same activity, and here every one is $0$: no child does the same activity in two consecutive weeks. (1 mark; $67\%$.)

(b)(i) Week 1 is $S_1 = \begin{bmatrix} 27 \\ 0 \\ 0 \end{bmatrix}$, so Week 10 is nine steps later: $S_{10} = K^9 S_1 = \begin{bmatrix} 10.0856\ldots \\ 9.9648\ldots \\ 6.9495\ldots \end{bmatrix}$. Percentage cooking $= \frac{10.0856}{27} \times 100 = 37.4\%$. (1 mark; only $9\%$. Rounding 10.0856 to 10 before dividing gives $37.0\%$; the examiners stressed rounding only at the final step.)

(b)(ii) Week 3 is $S_3 = K^2 S_1 = \begin{bmatrix} 15.66 \\ 7.776 \\ 3.564 \end{bmatrix}$, so 7.776 children are expected to do gardening. The proportion moving from $G$ to $M$ is $k_{32} = 0.24$: $7.776 \times 0.24 = 1.866 \approx 2$ children. (2 marks; $17\%$ full marks.)

Source: VCAA 2025 General Mathematics Examination 2, Question 13, and the 2025 examination report.

(a) $S_{2024} = MS_{2023}$, so step backwards with the inverse: $S_{2023} = M^{-1}S_{2024} = \begin{bmatrix} 16 \\ 20 \\ 25 \\ 0 \end{bmatrix}$. (1 mark; $53\%$.) Check: $M \begin{bmatrix} 16 & 20 & 25 & 0 \end{bmatrix}^T = \begin{bmatrix} 4 & 15 & 15 & 27 \end{bmatrix}^T$.

(b) $S_{2025} = MS_{2024} + B = \begin{bmatrix} 13 \\ 11.25 \\ 20.875 \\ 42.875 \end{bmatrix}$ and $S_{2026} = MS_{2025} + B = \begin{bmatrix} 15.25 \\ 15.9375 \\ 19.1188 \\ 64.6938 \end{bmatrix}$.

Children enrolled are those in $N$, $T$ and $P$: $15.25 + 15.94 + 19.12 = 50.3$, so 50 children. (1 mark; only $13\%$. Many students answered 115 by also counting the children who had left.)

Source: VCAA 2025 General Mathematics Examination 2, Question 12, and the 2025 examination report.

(a) From $W$: $0.95$ stay, $0.05$ leave. From $L$: everyone stays left (they never return), so that column is $0$ and $1$.

(1 mark; $34\%$. $0.5$ was often written for $0.05$, and the $1$ was left out or put in the first row.)

(b) Each show $0.05 \times 180 = 9$ workers leave, so 9 new workers must be added to $W$: $B = \begin{bmatrix} 9 \\ 0 \end{bmatrix}$. (1 mark; $39\%$. The examiners also accepted $\begin{bmatrix} 9 \\ -9 \end{bmatrix}$.) Check: $TS_0 + B = \begin{bmatrix} 171 + 9 \\ 9 + 0 \end{bmatrix} = \begin{bmatrix} 180 \\ 9 \end{bmatrix}$.

Source: VCAA 2023 General Mathematics Examination 2, Question 11, and the 2023 examination report.

Column $N$ is $\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}$: every bird at $N$ stays at $N$ forever. Birds at $M$ and $O$ keep leaking to $N$ ($\frac{1}{3}$ of $M$ and $\frac{1}{10}$ of $O$ each day) and none ever come back, so in the long term all birds end up at $N$ and none remain at $M$ or $O$. Of the options, only A is true. (A large power of $T$, such as $T^{60}$, has row $N$ equal to $1\;1\;1$.)

Only $39\%$ answered correctly. $T$ is not regular (its powers always keep the zeros in column $N$), which is why the long-term state does not depend on sharing between all three locations.

Source: VCAA 2023 General Mathematics Examination 1, Question 27, and the 2023 examination report.

$S_{\text{Tue}} = TS_{\text{Mon}} = \begin{bmatrix} 132 \\ 134 \\ 140 \\ 54 \end{bmatrix}$ and $S_{\text{Wed}} = TS_{\text{Tue}} = \begin{bmatrix} 143.6 \\ 132.6 \\ 127 \\ 56.8 \end{bmatrix}$.

Of the 140 on the oval on Tuesday, $0.3 \times 140 = 42$ stay on the oval on Wednesday. As a percentage of Wednesday's 127 oval students: $\frac{42}{127} \times 100 = 33\%$, option C.

Only $18\%$ answered correctly; $54\%$ chose B, $30\%$, by quoting the proportion $t_{33} = 0.3$ who stay on the oval. That is a percentage of Tuesday's oval students, but the question asks for a percentage of Wednesday's.

Source: VCAA 2023 General Mathematics Examination 1, Question 32, and the 2023 examination report.

(a) Column $M$ (from members): $0.7$ stay, $0.3$ leave. Column $N$ (from non-members): $0.4$ join, $0.6$ stay out.

Each column sums to 1. (1 mark)

(b) $t_{21}$ is row 2 ($N$ next month), column 1 ($M$ this month): the proportion ($30\%$) of this month's members who are not members next month. (1 mark)

(1 mark)

(1 mark) The total stays at 2000 because each column of $T$ sums to 1.

(a) $T^{50}S_0 = \begin{bmatrix} 1142.857\ldots \\ 857.142\ldots \end{bmatrix}$, and $T^{51}S_0$ is the same, so in the long run about 1143 members (and 857 non-members). (1 mark)

(b) $T$ is regular (all its elements are already positive), so the state settles to the same equilibrium from any starting state with the same total, 2000. Only the total matters. (1 mark)

(c) The exact equilibrium is $\frac{4}{7}$ and $\frac{3}{7}$ of 2000, since $0.3M = 0.4N$ (as many leave as join). $T\begin{bmatrix} 8000/7 \\ 6000/7 \end{bmatrix} = \begin{bmatrix} 0.7(8000/7) + 0.4(6000/7) \\ 0.3(8000/7) + 0.6(6000/7) \end{bmatrix} = \begin{bmatrix} 8000/7 \\ 6000/7 \end{bmatrix}$, unchanged. (1 mark)

$S_{\text{this}} = TS_{\text{last}}$, so $S_{\text{last}} = T^{-1}\begin{bmatrix} 800 \\ 1200 \end{bmatrix}$. $\det(T) = 0.42 - 0.12 = 0.3$ and $T^{-1} = \frac{1}{0.3}\begin{bmatrix} 0.6 & -0.4 \\ -0.3 & 0.7 \end{bmatrix}$. (1 mark)

$S_{\text{last}} = \frac{1}{0.3}\begin{bmatrix} 0.6(800) - 0.4(1200) \\ -0.3(800) + 0.7(1200) \end{bmatrix} = \frac{1}{0.3}\begin{bmatrix} 0 \\ 600 \end{bmatrix} = \begin{bmatrix} 0 \\ 2000 \end{bmatrix}$.

Last month the gym had no members: all 2000 people were non-members, and $40\%$ of them (800) joined. (1 mark)

(a) $S_0 = \begin{bmatrix} 2000 \\ 3000 \end{bmatrix}$, $S_{n+1} = \begin{bmatrix} 0.85 & 0.05 \\ 0.15 & 0.95 \end{bmatrix}S_n + \begin{bmatrix} 200 \\ -100 \end{bmatrix}$. (1 mark)

(b) $S_1 = \begin{bmatrix} 1700 + 150 + 200 \\ 300 + 2850 - 100 \end{bmatrix} = \begin{bmatrix} 2050 \\ 3050 \end{bmatrix}$, then $S_2 = \begin{bmatrix} 1742.5 + 152.5 + 200 \\ 307.5 + 2897.5 - 100 \end{bmatrix} = \begin{bmatrix} 2095 \\ 3105 \end{bmatrix}$. (1 mark)

(c) The total rises by 100 fish each year ($5000$, $5100$, $5200$). The transition matrix conserves the total because its columns sum to 1, so only $B$ changes it: $200 - 100 = 100$. (1 mark)

(a) No. Every power of $T$ keeps a $0$ in row 1, column 2 (anything in $B$ stays in $B$), so no power of $T$ has all positive elements. (1 mark)

(b) $B$ is an absorbing state: each step $40\%$ of whatever is in $A$ moves to $B$ and nothing returns. In the long run everything ends up in $B$ ($T^n$ approaches $\begin{bmatrix} 0 & 0 \\ 1 & 1 \end{bmatrix}$). (1 mark)