1. **Problem 1.1.1: Calculate equilibrium price and quantity using matrices.** Given demand: $Q_d = 20 - 2P$ Given supply: $Q_s = -10 + 2P$ At equilibrium, $Q_d = Q_s = Q$. Rewrite equations: $Q = 20 - 2P$ $Q = -10 + 2P$ Set equal: $20 - 2P = -10 + 2P$ Rearranged: $20 + 10 = 2P + 2P$ $30 = 4P$ $P = \frac{30}{4} = 7.5$ Substitute $P=7.5$ into demand: $Q = 20 - 2(7.5) = 20 - 15 = 5$ Using matrices: Write system as: $\begin{cases} Q + 2P = 20 \\ Q - 2P = -10 \end{cases}$ Matrix form: $\begin{bmatrix}1 & 2 \\ 1 & -2\end{bmatrix} \begin{bmatrix}Q \\ P\end{bmatrix} = \begin{bmatrix}20 \\ -10\end{bmatrix}$ Calculate determinant: $\Delta = 1 \times (-2) - 1 \times 2 = -2 - 2 = -4$ Find inverse matrix: $\frac{1}{\Delta} \begin{bmatrix}-2 & -2 \\ -1 & 1\end{bmatrix} = \frac{1}{-4} \begin{bmatrix}-2 & -2 \\ -1 & 1\end{bmatrix}$ Multiply inverse by constants: $\begin{bmatrix}Q \\ P\end{bmatrix} = \frac{1}{-4} \begin{bmatrix}-2 & -2 \\ -1 & 1\end{bmatrix} \begin{bmatrix}20 \\ -10\end{bmatrix} = \frac{1}{-4} \begin{bmatrix}(-2)(20) + (-2)(-10) \\ (-1)(20) + (1)(-10)\end{bmatrix} = \frac{1}{-4} \begin{bmatrix}-40 + 20 \\ -20 - 10\end{bmatrix} = \frac{1}{-4} \begin{bmatrix}-20 \\ -30\end{bmatrix} = \begin{bmatrix}5 \\ 7.5\end{bmatrix}$ So, equilibrium quantity $Q=5$ and price $P=7.5$. 2. **Problem 1.1.2: Determine consumer surplus and producer surplus at equilibrium.** Consumer surplus (CS) is area between demand curve and price line up to equilibrium quantity: Demand intercept price when $Q=0$: $0 = 20 - 2P \Rightarrow P = 10$ CS formula: $CS = \frac{1}{2} \times (\text{base}) \times (\text{height}) = \frac{1}{2} \times 5 \times (10 - 7.5) = \frac{1}{2} \times 5 \times 2.5 = 6.25$ Producer surplus (PS) is area between supply curve and price line up to equilibrium quantity: Supply intercept price when $Q=0$: $0 = -10 + 2P \Rightarrow P = 5$ PS formula: $PS = \frac{1}{2} \times 5 \times (7.5 - 5) = \frac{1}{2} \times 5 \times 2.5 = 6.25$ 3. **Problem 1.1.3: Determine P and Q intercepts.** Demand: - $P$-intercept: set $Q=0$, solve $0=20 - 2P \Rightarrow P=10$ - $Q$-intercept: set $P=0$, $Q=20$ Supply: - $P$-intercept: set $Q=0$, solve $0=-10 + 2P \Rightarrow P=5$ - $Q$-intercept: set $P=0$, $Q=-10$ 4. **Problem 1.1.4: Determine slopes of demand and supply functions.** Demand slope from $Q_d = 20 - 2P$: Rewrite as $Q_d = -2P + 20$, slope = $-2$ Supply slope from $Q_s = -10 + 2P$: Rewrite as $Q_s = 2P - 10$, slope = $2$ 5. **Problem 2.1.1: Factorize $K^2L^2 - 25Q^2$.** Recognize difference of squares: $K^2L^2 - (5Q)^2 = (KL - 5Q)(KL + 5Q)$ 6. **Problem 2.1.2: Factorize $Q^2 + 4QK + 4K^2$.** Recognize perfect square trinomial: $Q^2 + 4QK + 4K^2 = (Q + 2K)^2$ 7. **Problem 2.1.3: Factorize $\delta L^{-3} + (1 - \delta) L^{-2}$.** Factor out $L^{-3}$: $L^{-3}(\delta + (1 - \delta)L)$ 8. **Problem 2.2.1: Solve $\frac{x^{-3}}{4} + 2 = 3x$.** Rewrite: $\frac{1}{4x^3} + 2 = 3x$ Multiply both sides by $4x^3$: $1 + 8x^3 = 12x^4$ Rearranged: $12x^4 - 8x^3 - 1 = 0$ This quartic can be solved numerically or by substitution methods. **Final answers:** - Equilibrium price $P=7.5$, quantity $Q=5$. - Consumer surplus = 6.25, Producer surplus = 6.25. - Demand intercepts: $P=10$, $Q=20$; Supply intercepts: $P=5$, $Q=-10$. - Slopes: Demand = $-2$, Supply = $2$. - Factorizations: - $K^2L^2 - 25Q^2 = (KL - 5Q)(KL + 5Q)$ - $Q^2 + 4QK + 4K^2 = (Q + 2K)^2$ - $\delta L^{-3} + (1 - \delta) L^{-2} = L^{-3}(\delta + (1 - \delta)L)$ - Equation $\frac{x^{-3}}{4} + 2 = 3x$ leads to quartic $12x^4 - 8x^3 - 1 = 0$ to solve for $x$.