1. **Problem Statement:** Explain the concepts of Consumer Surplus and Producer Surplus with graphs and formulas. Explain solving simultaneous linear equations using Cramer's rule with an example. Explain matrix inversion with an example. 2. **Consumer Surplus and Producer Surplus:** - Consumer Surplus (CS) is the difference between what consumers are willing to pay and what they actually pay. - Producer Surplus (PS) is the difference between the price producers receive and the minimum price they are willing to accept. 3. **Formulas and Graphs:** - Demand curve: $Q_d = f(P)$, Supply curve: $Q_s = g(P)$. - Equilibrium price $P^*$ where $Q_d = Q_s$. - Consumer Surplus: $$CS = \int_0^{Q^*} (P_{max} - P^*) \, dQ$$ approximated as area of triangle $$\frac{1}{2} \times (Q^*) \times (P_{max} - P^*)$$. - Producer Surplus: $$PS = \int_0^{Q^*} (P^* - P_{min}) \, dQ$$ approximated as area of triangle $$\frac{1}{2} \times (Q^*) \times (P^* - P_{min})$$. 4. **Example for Cramer's Rule:** Solve the system: $$\begin{cases} 2x + 3y = 8 \\ 5x - y = 2 \end{cases}$$ - Write coefficient matrix $A = \begin{bmatrix} 2 & 3 \\ 5 & -1 \end{bmatrix}$ and constants vector $B = \begin{bmatrix} 8 \\ 2 \end{bmatrix}$. - Calculate determinant $$D = \det(A) = 2 \times (-1) - 3 \times 5 = -2 - 15 = -17$$. - Calculate $$D_x = \det\begin{bmatrix} 8 & 3 \\ 2 & -1 \end{bmatrix} = 8 \times (-1) - 3 \times 2 = -8 - 6 = -14$$. - Calculate $$D_y = \det\begin{bmatrix} 2 & 8 \\ 5 & 2 \end{bmatrix} = 2 \times 2 - 8 \times 5 = 4 - 40 = -36$$. - Solutions: $$x = \frac{D_x}{D} = \frac{-14}{-17} = \frac{14}{17}$$ $$y = \frac{D_y}{D} = \frac{-36}{-17} = \frac{36}{17}$$ 5. **Example of Matrix Inversion:** Find inverse of matrix $$A = \begin{bmatrix} 4 & 7 \\ 2 & 6 \end{bmatrix}$$. - Calculate determinant $$\det(A) = 4 \times 6 - 7 \times 2 = 24 - 14 = 10$$. - Inverse formula for 2x2 matrix: $$A^{-1} = \frac{1}{\det(A)} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} = \frac{1}{10} \begin{bmatrix} 6 & -7 \\ -2 & 4 \end{bmatrix}$$. - So, $$A^{-1} = \begin{bmatrix} 0.6 & -0.7 \\ -0.2 & 0.4 \end{bmatrix}$$. This completes the explanation and examples for the first question.