1. **Problem Statement:** Verify the quantities of wheat, corn, and rice using the matrix inverse method given the system of linear equations from the fertilizer, water, and labor constraints. 2. **Matrix Form:** The system is $AX = B$ where $$A = \begin{bmatrix}4 & 2 & 3 \\ 3 & 2 & 4 \\ 2 & 3 & 1\end{bmatrix}, \quad X = \begin{bmatrix}x_\text{wheat} \\ x_\text{corn} \\ x_\text{rice}\end{bmatrix}, \quad B = \begin{bmatrix}700 \\ 800 \\ 500\end{bmatrix}$$ 3. **Step 1: Find $A^{-1}$ (the inverse of matrix $A$).** 4. **Step 2: Calculate $X = A^{-1}B$.** 5. **Step 3: Compute the inverse and multiply:** The inverse matrix $A^{-1}$ is calculated (using standard methods or a calculator) as: $$A^{-1} = \frac{1}{\det(A)} \text{adj}(A)$$ Calculate determinant: $$\det(A) = 4(2 \times 1 - 4 \times 3) - 2(3 \times 1 - 4 \times 2) + 3(3 \times 3 - 2 \times 2) = 4(2 - 12) - 2(3 - 8) + 3(9 - 4) = 4(-10) - 2(-5) + 3(5) = -40 + 10 + 15 = -15$$ Calculate adjugate matrix: $$\text{adj}(A) = \begin{bmatrix} (2 \times 1 - 4 \times 3) & -(3 \times 1 - 4 \times 2) & (3 \times 3 - 2 \times 2) \\ -(2 \times 1 - 3 \times 3) & (4 \times 1 - 3 \times 3) & -(4 \times 3 - 2 \times 2) \\ (2 \times 4 - 3 \times 2) & -(4 \times 4 - 3 \times 2) & (4 \times 2 - 2 \times 3) \end{bmatrix}^T = \begin{bmatrix} -10 & 5 & 5 \\ 7 & -5 & -8 \\ 2 & -10 & 2 \end{bmatrix}^T = \begin{bmatrix} -10 & 7 & 2 \\ 5 & -5 & -10 \\ 5 & -8 & 2 \end{bmatrix}$$ 6. **Step 4: Multiply $A^{-1}$ by $B$:** $$X = \frac{1}{-15} \begin{bmatrix} -10 & 7 & 2 \\ 5 & -5 & -10 \\ 5 & -8 & 2 \end{bmatrix} \begin{bmatrix}700 \\ 800 \\ 500\end{bmatrix} = \frac{1}{-15} \begin{bmatrix} -10 \times 700 + 7 \times 800 + 2 \times 500 \\ 5 \times 700 - 5 \times 800 - 10 \times 500 \\ 5 \times 700 - 8 \times 800 + 2 \times 500 \end{bmatrix} = \frac{1}{-15} \begin{bmatrix} -7000 + 5600 + 1000 \\ 3500 - 4000 - 5000 \\ 3500 - 6400 + 1000 \end{bmatrix} = \frac{1}{-15} \begin{bmatrix} -400 \\ -3500 \\ -1900 \end{bmatrix}$$ 7. **Step 5: Simplify:** $$X = \begin{bmatrix} \frac{-400}{-15} \\ \frac{-3500}{-15} \\ \frac{-1900}{-15} \end{bmatrix} = \begin{bmatrix} 26.67 \\ 233.33 \\ 126.67 \end{bmatrix}$$ 8. **Interpretation:** The quantities are approximately: - Wheat: 26.67 units - Corn: 233.33 units - Rice: 126.67 units This verifies the solution using the matrix inverse method.