1. **Stating the problem:** We have four clinics each requiring a combination of four drugs (A, B, C, D) to meet their total treatment units. The goal is to find the allocation values of each drug that satisfy the treatment needs. 2. **Formulating the system of equations:** From the table: - Mahalapye: $2A + 1B + 3C + 1D = 25$ - Francistown: $1A + 3B + 2C + 2D = 26$ - Maun: $3A + 2B + 1C + 1D = 24$ - Gantsi: $1A + 2B + 1C + 3D = 23$ 3. **Matrix form:** Let $\mathbf{x} = \begin{bmatrix} A \\ B \\ C \\ D \end{bmatrix}$ and $\mathbf{b} = \begin{bmatrix} 25 \\ 26 \\ 24 \\ 23 \end{bmatrix}$. The coefficient matrix $\mathbf{A}$ is: $$ \mathbf{A} = \begin{bmatrix} 2 & 1 & 3 & 1 \\ 1 & 3 & 2 & 2 \\ 3 & 2 & 1 & 1 \\ 1 & 2 & 1 & 3 \end{bmatrix} $$ So the system is: $$ \mathbf{A} \mathbf{x} = \mathbf{b} $$ 4. **Using the inverse matrix $\mathbf{A}^{-1}$ to solve for $\mathbf{x}$:** Given: $$ \mathbf{A}^{-1} = \begin{bmatrix} \frac{1}{16} & -\frac{5}{16} & \frac{3}{8} & \frac{1}{16} \\ -\frac{5}{16} & \frac{9}{16} & \frac{1}{8} & -\frac{5}{16} \\ \frac{3}{8} & \frac{1}{8} & -\frac{1}{4} & -\frac{1}{8} \\ \frac{1}{16} & -\frac{5}{16} & -\frac{1}{8} & \frac{9}{16} \end{bmatrix} $$ Calculate: $$ \mathbf{x} = \mathbf{A}^{-1} \mathbf{b} $$ 5. **Performing the multiplication:** $A = \frac{1}{16} \times 25 + (-\frac{5}{16}) \times 26 + \frac{3}{8} \times 24 + \frac{1}{16} \times 23$ $= \frac{25}{16} - \frac{130}{16} + \frac{72}{8} + \frac{23}{16} = \frac{25 - 130 + 144 + 23}{16} = \frac{62}{16} = 3.875$ $B = (-\frac{5}{16}) \times 25 + \frac{9}{16} \times 26 + \frac{1}{8} \times 24 + (-\frac{5}{16}) \times 23$ $= -\frac{125}{16} + \frac{234}{16} + 3 - \frac{115}{16} = \frac{-125 + 234 - 115}{16} + 3 = \frac{-6}{16} + 3 = -0.375 + 3 = 2.625$ $C = \frac{3}{8} \times 25 + \frac{1}{8} \times 26 + (-\frac{1}{4}) \times 24 + (-\frac{1}{8}) \times 23$ $= \frac{75}{8} + \frac{26}{8} - 6 - \frac{23}{8} = \frac{75 + 26 - 23}{8} - 6 = \frac{78}{8} - 6 = 9.75 - 6 = 3.75$ $D = \frac{1}{16} \times 25 + (-\frac{5}{16}) \times 26 + (-\frac{1}{8}) \times 24 + \frac{9}{16} \times 23$ $= \frac{25}{16} - \frac{130}{16} - 3 + \frac{207}{16} = \frac{25 - 130 + 207}{16} - 3 = \frac{102}{16} - 3 = 6.375 - 3 = 3.375$ 6. **Final answer:** $$ \boxed{\begin{bmatrix} A \\ B \\ C \\ D \end{bmatrix} = \begin{bmatrix} 3.875 \\ 2.625 \\ 3.75 \\ 3.375 \end{bmatrix}} $$ This means the allocation values for the drugs are approximately: - Drug A: 3.875 units - Drug B: 2.625 units - Drug C: 3.75 units - Drug D: 3.375 units These satisfy the treatment needs of all four clinics.