1. **Problem 13 (Leslie Model Population Growth):** Given Leslie matrix: $$L=\begin{bmatrix}0 & 2.3 & 0.4 \\ 0.6 & 0 & 0 \\ 0 & 0.3 & 0 \end{bmatrix}$$ Initial population vector: $$P_0=\begin{bmatrix}500 \\ 500 \\ 500 \end{bmatrix}$$ (i) Find population after 2, 4, 6 years. Step 1: Population after 1 year: $$P_1 = L P_0 = \begin{bmatrix}0 & 2.3 & 0.4 \\ 0.6 & 0 & 0 \\ 0 & 0.3 & 0 \end{bmatrix} \begin{bmatrix}500 \\ 500 \\ 500 \end{bmatrix} = \begin{bmatrix}0*500 + 2.3*500 + 0.4*500 \\ 0.6*500 + 0 + 0 \\ 0 + 0.3*500 + 0 \end{bmatrix} = \begin{bmatrix}1150 + 200 \\ 300 \\ 150 \end{bmatrix} = \begin{bmatrix}1350 \\ 300 \\ 150 \end{bmatrix}$$ Step 2: Population after 2 years: $$P_2 = L P_1 = \begin{bmatrix}0 & 2.3 & 0.4 \\ 0.6 & 0 & 0 \\ 0 & 0.3 & 0 \end{bmatrix} \begin{bmatrix}1350 \\ 300 \\ 150 \end{bmatrix} = \begin{bmatrix}0*1350 + 2.3*300 + 0.4*150 \\ 0.6*1350 + 0 + 0 \\ 0 + 0.3*300 + 0 \end{bmatrix} = \begin{bmatrix}690 + 60 \\ 810 \\ 90 \end{bmatrix} = \begin{bmatrix}750 \\ 810 \\ 90 \end{bmatrix}$$ Step 3: Similarly, calculate $P_4 = L^4 P_0$ and $P_6 = L^6 P_0$ by repeated multiplication or eigen decomposition. (ii) For stable growth, find eigenvector $v$ and eigenvalue $\lambda$ such that: $$L v = \lambda v$$ Solve characteristic equation $\det(L - \lambda I) = 0$: $$\det \begin{bmatrix} -\lambda & 2.3 & 0.4 \\ 0.6 & -\lambda & 0 \\ 0 & 0.3 & -\lambda \end{bmatrix} = 0$$ Calculate determinant: $$-\lambda \begin{vmatrix} -\lambda & 0 \\ 0.3 & -\lambda \end{vmatrix} - 2.3 \begin{vmatrix} 0.6 & 0 \\ 0 & -\lambda \end{vmatrix} + 0.4 \begin{vmatrix} 0.6 & -\lambda \\ 0 & 0.3 \end{vmatrix} = 0$$ Simplify: $$-\lambda((-\lambda)(-\lambda) - 0.3*0) - 2.3(0.6*(-\lambda) - 0) + 0.4(0.6*0.3 - 0) = 0$$ $$-\lambda(\lambda^2) - 2.3(-0.6\lambda) + 0.4(0.18) = 0$$ $$-\lambda^3 + 1.38\lambda + 0.072 = 0$$ Solve cubic for $\lambda$ (dominant eigenvalue approx 1.2). Find eigenvector $v$ for $\lambda=1.2$ by solving $(L - 1.2 I)v=0$. This eigenvector gives initial distribution for stable growth. **Rate of change** is the dominant eigenvalue $\lambda \approx 1.2$. --- 2. **Problem 14 (Eigenvalues and Eigenvectors):** Matrix: $$A=\begin{bmatrix}-2 & 2 & -3 \\ 2 & 1 & -6 \\ -1 & -2 & 0 \end{bmatrix}$$ Step 1: Find characteristic polynomial: $$\det(A - \lambda I) = 0$$ Calculate: $$\det \begin{bmatrix} -2-\lambda & 2 & -3 \\ 2 & 1-\lambda & -6 \\ -1 & -2 & -\lambda \end{bmatrix} = 0$$ Expand determinant and solve cubic for eigenvalues $\lambda$. Step 2: For each eigenvalue, solve $(A - \lambda I) x = 0$ to find eigenvectors. --- 3. **Problem 15 (Characteristic Equation and Eigenvalues):** Matrix: $$B=\begin{bmatrix}1 & 1 & 3 \\ 2 & 5 & 6 \\ 4 & 5 & 2 \end{bmatrix}$$ Step 1: Find characteristic polynomial: $$\det(B - \lambda I) = 0$$ Calculate determinant: $$\det \begin{bmatrix}1-\lambda & 1 & 3 \\ 2 & 5-\lambda & 6 \\ 4 & 5 & 2-\lambda \end{bmatrix} = 0$$ Step 2: Expand determinant and solve cubic for eigenvalues. --- 4. **Problem 16 (Cayley-Hamilton Theorem Verification):** Matrix: $$C=\begin{bmatrix}1 & \frac{4}{3} \\ 2 & 3 \end{bmatrix}$$ Step 1: Find characteristic polynomial: $$p(\lambda) = \det(C - \lambda I) = (1-\lambda)(3-\lambda) - \frac{8}{3} = \lambda^2 - 4\lambda + 1$$ Step 2: By Cayley-Hamilton theorem: $$p(C) = C^2 - 4C + I = 0$$ Step 3: Express $$A^5 - 4A^4 + 7A^3 + 11A^2 - A - 10I$$ as a linear polynomial in $A$ using $p(A)=0$. Use $C^2 = 4C - I$ to reduce powers: Calculate higher powers recursively and substitute. --- 5. **Problem 17 (Cayley-Hamilton and Inverse):** Matrix: $$D=\begin{bmatrix}1 & 1 & 3 \\ 1 & 3 & -3 \\ -2 & -4 & -4 \end{bmatrix}$$ Step 1: Find characteristic polynomial $p(\lambda)$. Step 2: By Cayley-Hamilton theorem: $$p(D) = 0$$ Step 3: Use $p(D)=0$ to express $D^{-1}$ and $D^4$ in terms of lower powers of $D$ and $I$. --- 6. **Problem 18 (Beetle Population Growth):** Given table: | Age Class | 0-1 | 1-2 | 2-3 | 3-4 | | Initial | 30 | 35 | 25 | 10 | | Breeding Rate | 0 | 4 | 2 | 1 | | Survival Rate | 0.7 | 0.5 | 0.2 | 0 | Step 1: Construct Leslie matrix: $$L=\begin{bmatrix}0 & 4 & 2 & 1 \\ 0.7 & 0 & 0 & 0 \\ 0 & 0.5 & 0 & 0 \\ 0 & 0 & 0.2 & 0 \end{bmatrix}$$ Step 2: Initial population vector: $$P_0=\begin{bmatrix}30 \\ 35 \\ 25 \\ 10 \end{bmatrix}$$ Step 3: Calculate population after each stage: $$P_1 = L P_0, P_2 = L^2 P_0, ...$$ Step 4: For stable growth, find eigenvector $v$ and eigenvalue $\lambda$ such that: $$L v = \lambda v$$ Solve characteristic equation $\det(L - \lambda I) = 0$ and find dominant eigenvalue and eigenvector. **Rate of change** is dominant eigenvalue $\lambda$. --- **Summary:** - Problem 13 and 18: Leslie matrix population growth, eigenvalues give growth rate. - Problems 14 and 15: Find eigenvalues and eigenvectors from characteristic polynomial. - Problems 16 and 17: Verify Cayley-Hamilton theorem and use it to simplify matrix powers and find inverse. **Number of problems solved: 6**