1. **Problem 1a:** Express staffing numbers as a matrix. Given: $$S=\begin{bmatrix}17 & 38 & 8 \\ 21 & 34 & 5 \end{bmatrix}$$ This matrix has 2 rows (Female, Male) and 3 columns (Full-time Rangers, Part-time Rangers, Admin staff). 2. **Problem 1b:** Find total Female and Male staff using row sums. For Female: $$F=17+38+8=63$$ For Male: $$M=21+34+5=60$$ Hence the column matrix: $$\begin{bmatrix}63 \\ 60 \end{bmatrix}$$ 3. **Problem 1f i and ii:** Transition matrix for Happy/Unhappy workers. The transition matrix T: $$T=\begin{bmatrix}0.85 & 0.3\\0.15 & 0.7\end{bmatrix}$$ Rows: Today (Happy, Unhappy) Columns: Tomorrow (Happy, Unhappy) 4. **Problem 1f iii:** Initial state matrix (E0) with 98 Happy and 25 Unhappy workers: $$E_0=\begin{bmatrix}98 \\ 25\end{bmatrix}$$ 5. **Problem 1f iv:** Number expected next day: $$E_1 = T \times E_0 = \begin{bmatrix}0.85 & 0.3\\0.15 & 0.7\end{bmatrix} \times \begin{bmatrix}98 \\ 25\end{bmatrix} = \begin{bmatrix}0.85\times98 + 0.3\times25 \\ 0.15\times98 + 0.7\times25\end{bmatrix} = \begin{bmatrix}91 \\ 32\end{bmatrix}$$ Rounded values: 91 Happy, 32 Unhappy 6. **Problem 1f v:** Four days later, multiply by $T^4$: Using given values: $$E_4 = \begin{bmatrix}83 \\ 40\end{bmatrix}$$ 7. **Problem 1f vi:** Steady state vector $E_s = \begin{bmatrix}h \\ u\end{bmatrix}$ satisfies: $$E_s = T E_s$$ So solve: $$\begin{cases}h = 0.85h + 0.3u \\ u = 0.15h + 0.7u\end{cases}$$ With constraint $$h + u = 123$$. Solving yields: $$h \approx 82, \quad u \approx 41$$ (rounded) 8. **Problem 2a:** Female frogs in age groups: One year old: 850, Two years: 940, Three years: 530, Four years: 280 Initial state matrix: $$S_0 = \begin{bmatrix}850 \\ 940 \\ 530 \\ 280\end{bmatrix}$$ 9. **Problem 2b & c:** Leslie matrix $L$ (4x4) from birth and survival rates: $$L = \begin{bmatrix}3.6 & 2.9 & 1.7 & 0.6 \\ 0.43 & 0 & 0 & 0 \\ 0 & 0.31 & 0 & 0 \\ 0 & 0 & 0.15 & 0 \end{bmatrix}$$ 10. **Problem 2d i:** Predict number of female frogs after 1 year: $$L \times S_0 = \begin{bmatrix}3.6 & 2.9 & 1.7 & 0.6 \\ 0.43 & 0 & 0 & 0 \\ 0 & 0.31 & 0 & 0 \\ 0 & 0 & 0.15 & 0 \end{bmatrix} \times \begin{bmatrix}850 \\ 940 \\ 530 \\ 280\end{bmatrix} = \begin{bmatrix}6855 \\ 366 \\ 291 \\ 42 \end{bmatrix}$$ 11. **Problem 2d ii:** Population after 3 years (given): $$S_3 = \begin{bmatrix}406089 \\ 44408 \\ 3500 \\ 0 \end{bmatrix}$$ Total: $$406089 + 44408 + 3500 + 0=453997$$ 12. **Problem 3a:** One-step dominance matrix $T$ shows which species dominates which according to given matrix: Species order: Gr (Grass), M (Mouse), R (Rabbit), G (Grasshopper), S (Snake), L (Lizard), H (Hawk) $$T=\begin{bmatrix}0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 1 & 0 & 1 & 1 & 0 \end{bmatrix}$$ 13. **Problem 3b:** Two-step dominance matrix $T^2$: $$T^2=\begin{bmatrix}0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 3 & 0 & 0 & 0 \end{bmatrix}$$ 14. **Problem 3 Rank Order:** Summing $T + T^2$ results in dominance totals: - Hawk (H) dominates 10 species. - Snake (S) and Lizard (L) dominate 2 species each. - Grasshopper (G), Mouse (M), Rabbit (R) dominate 1 species each. - Grass (Gr) dominates 0. Final rank order from highest to lowest dominance is: $$\text{Hawk} > \text{Snake, Lizard} > \text{Grasshopper, Mouse, Rabbit} > \text{Grass}$$