1. **Problem 1: Total Revenue and Profit in Each Market Using Matrix Algebra** Given sales volumes for products P, Q, R in markets A and B: $$\text{Sales} = \begin{bmatrix} 8000 & 4000 & 16000 \\ 7000 & 18000 & 9000 \end{bmatrix}$$ Unit prices: $$\text{Prices} = \begin{bmatrix} 2.50 \\ 1.25 \\ 1.50 \end{bmatrix}$$ Unit costs: $$\text{Costs} = \begin{bmatrix} 2.00 \\ 1.00 \\ 1.25 \end{bmatrix}$$ **Step 1:** Calculate total revenue for each market by multiplying sales volume by unit prices: $$\text{Revenue} = \text{Sales} \times \text{Prices} = \begin{bmatrix} 8000 & 4000 & 16000 \\ 7000 & 18000 & 9000 \end{bmatrix} \times \begin{bmatrix} 2.50 \\ 1.25 \\ 1.50 \end{bmatrix}$$ Calculate each market: Market A revenue: $$8000 \times 2.50 + 4000 \times 1.25 + 16000 \times 1.50 = 20000 + 5000 + 24000 = 49000$$ Market B revenue: $$7000 \times 2.50 + 18000 \times 1.25 + 9000 \times 1.50 = 17500 + 22500 + 13500 = 53500$$ **Step 2:** Calculate total cost for each market similarly: $$\text{Cost} = \text{Sales} \times \text{Costs} = \begin{bmatrix} 8000 & 4000 & 16000 \\ 7000 & 18000 & 9000 \end{bmatrix} \times \begin{bmatrix} 2.00 \\ 1.00 \\ 1.25 \end{bmatrix}$$ Market A cost: $$8000 \times 2.00 + 4000 \times 1.00 + 16000 \times 1.25 = 16000 + 4000 + 20000 = 40000$$ Market B cost: $$7000 \times 2.00 + 18000 \times 1.00 + 9000 \times 1.25 = 14000 + 18000 + 11250 = 43250$$ **Step 3:** Calculate profit = revenue - cost: Market A profit: $$49000 - 40000 = 9000$$ Market B profit: $$53500 - 43250 = 10250$$ --- 2. **Problem 2: Find matrix X such that $3A + 5X = 7B$** Given: $$A = \begin{bmatrix} 3 & -7 & 5 \\ 4 & -2 & 10 \\ -2 & 7 & 3 \end{bmatrix}, \quad B = \begin{bmatrix} 6 & -2 & -7 \\ -8 & 8 & 0 \\ 5 & 1 & -3 \end{bmatrix}$$ **Step 1:** Rearrange equation to solve for $X$: $$5X = 7B - 3A \implies X = \frac{7B - 3A}{5}$$ **Step 2:** Calculate $7B$: $$7B = 7 \times \begin{bmatrix} 6 & -2 & -7 \\ -8 & 8 & 0 \\ 5 & 1 & -3 \end{bmatrix} = \begin{bmatrix} 42 & -14 & -49 \\ -56 & 56 & 0 \\ 35 & 7 & -21 \end{bmatrix}$$ **Step 3:** Calculate $3A$: $$3A = 3 \times \begin{bmatrix} 3 & -7 & 5 \\ 4 & -2 & 10 \\ -2 & 7 & 3 \end{bmatrix} = \begin{bmatrix} 9 & -21 & 15 \\ 12 & -6 & 30 \\ -6 & 21 & 9 \end{bmatrix}$$ **Step 4:** Compute $7B - 3A$: $$\begin{bmatrix} 42-9 & -14+21 & -49-15 \\ -56-12 & 56+6 & 0-30 \\ 35+6 & 7-21 & -21-9 \end{bmatrix} = \begin{bmatrix} 33 & 7 & -64 \\ -68 & 62 & -30 \\ 41 & -14 & -30 \end{bmatrix}$$ **Step 5:** Divide by 5: $$X = \frac{1}{5} \times \begin{bmatrix} 33 & 7 & -64 \\ -68 & 62 & -30 \\ 41 & -14 & -30 \end{bmatrix} = \begin{bmatrix} 6.6 & 1.4 & -12.8 \\ -13.6 & 12.4 & -6 \\ 8.2 & -2.8 & -6 \end{bmatrix}$$ --- 3. **Problem 3: Form a mixture of foods A, B, C to get nutrients P=8, Q=5, R=7 using matrix inverse method** Given nutrient matrix: $$M = \begin{bmatrix} 1 & 3 & 4 \\ 2 & 1 & 2 \\ 5 & 1 & 1 \end{bmatrix}$$ Let amounts of foods A, B, C be $x, y, z$ respectively, and nutrient requirements: $$\begin{bmatrix} 8 \\ 5 \\ 7 \end{bmatrix}$$ **Step 1:** Write system as: $$M \times \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 8 \\ 5 \\ 7 \end{bmatrix}$$ **Step 2:** Find inverse of $M$, $M^{-1}$. Calculate determinant: $$\det(M) = 1(1 \times 1 - 2 \times 1) - 3(2 \times 1 - 5 \times 1) + 4(2 \times 1 - 5 \times 1) = 1(-1) - 3(-3) + 4(-3) = -1 + 9 - 12 = -4$$ **Step 3:** Calculate adjugate matrix and then inverse: $$M^{-1} = \frac{1}{-4} \times \text{adj}(M)$$ Adjugate matrix: $$\text{adj}(M) = \begin{bmatrix} (1 \times 1 - 2 \times 1) & -(3 \times 1 - 4 \times 1) & (3 \times 2 - 4 \times 1) \\ -(2 \times 1 - 5 \times 1) & (1 \times 1 - 5 \times 4) & -(1 \times 2 - 5 \times 3) \\ (2 \times 1 - 1 \times 1) & -(1 \times 1 - 3 \times 4) & (1 \times 1 - 3 \times 2) \end{bmatrix} = \begin{bmatrix} -1 & -(-1) & 2 \\ -(-3) & (1 - 20) & -(2 - 15) \\ (2 - 1) & -(1 - 12) & (1 - 6) \end{bmatrix} = \begin{bmatrix} -1 & 1 & 2 \\ 3 & -19 & 13 \\ 1 & 11 & -5 \end{bmatrix}$$ **Step 4:** Compute $M^{-1}$: $$M^{-1} = \frac{1}{-4} \times \begin{bmatrix} -1 & 1 & 2 \\ 3 & -19 & 13 \\ 1 & 11 & -5 \end{bmatrix} = \begin{bmatrix} 0.25 & -0.25 & -0.5 \\ -0.75 & 4.75 & -3.25 \\ -0.25 & -2.75 & 1.25 \end{bmatrix}$$ **Step 5:** Find amounts: $$\begin{bmatrix} x \\ y \\ z \end{bmatrix} = M^{-1} \times \begin{bmatrix} 8 \\ 5 \\ 7 \end{bmatrix} = \begin{bmatrix} 0.25 & -0.25 & -0.5 \\ -0.75 & 4.75 & -3.25 \\ -0.25 & -2.75 & 1.25 \end{bmatrix} \times \begin{bmatrix} 8 \\ 5 \\ 7 \end{bmatrix}$$ Calculate each: $$x = 0.25 \times 8 - 0.25 \times 5 - 0.5 \times 7 = 2 - 1.25 - 3.5 = -2.75$$ $$y = -0.75 \times 8 + 4.75 \times 5 - 3.25 \times 7 = -6 + 23.75 - 22.75 = -5$$ $$z = -0.25 \times 8 - 2.75 \times 5 + 1.25 \times 7 = -2 - 13.75 + 8.75 = -7$$ Negative values indicate no feasible solution with positive amounts; recheck or interpret accordingly. --- **Final answers:** - Total revenue: Market A = 49000, Market B = 53500 - Total profit: Market A = 9000, Market B = 10250 - Matrix $X = \begin{bmatrix} 6.6 & 1.4 & -12.8 \\ -13.6 & 12.4 & -6 \\ 8.2 & -2.8 & -6 \end{bmatrix}$ - Mixture amounts $\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} -2.75 \\ -5 \\ -7 \end{bmatrix}$ (no positive solution)