1. **Problem Statement:** We have a bacteria culture with three types of bacteria (A, B, C) each requiring certain amounts of carbon, phosphate, and nitrogen daily. The total daily supply of these nutrients is given. We want to find the number of bacteria units of each type that can be supported. 2. **Set up the system of equations:** Let $x$, $y$, and $z$ be the number of bacteria units of types A, B, and C respectively. From the nutrient requirements: - Carbon: $2x + 4y + 3z = 100000$ - Phosphate: $3x + y + 5z = 135000$ - Nitrogen: $6x + 2y + 8z = 230000$ 3. **Write the system in matrix form:** $$\begin{bmatrix} 2 & 4 & 3 \\ 3 & 1 & 5 \\ 6 & 2 & 8 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 100000 \\ 135000 \\ 230000 \end{bmatrix}$$ 4. **Apply row reduced echelon form (RREF):** We perform Gaussian elimination to find the RREF of the augmented matrix: $$\left[ \begin{array}{ccc|c} 2 & 4 & 3 & 100000 \\ 3 & 1 & 5 & 135000 \\ 6 & 2 & 8 & 230000 \end{array} \right]$$ 5. **Step-by-step elimination:** - Divide row 1 by 2: $$\left[ \begin{array}{ccc|c} 1 & 2 & 1.5 & 50000 \\ 3 & 1 & 5 & 135000 \\ 6 & 2 & 8 & 230000 \end{array} \right]$$ - Replace row 2 by row 2 minus 3*row 1: $$3 - 3*1 = 0, \quad 1 - 3*2 = -5, \quad 5 - 3*1.5 = 0.5, \quad 135000 - 3*50000 = 135000 - 150000 = -15000$$ - Replace row 3 by row 3 minus 6*row 1: $$6 - 6*1 = 0, \quad 2 - 6*2 = 2 - 12 = -10, \quad 8 - 6*1.5 = 8 - 9 = -1, \quad 230000 - 6*50000 = 230000 - 300000 = -70000$$ Matrix now: $$\left[ \begin{array}{ccc|c} 1 & 2 & 1.5 & 50000 \\ 0 & -5 & 0.5 & -15000 \\ 0 & -10 & -1 & -70000 \end{array} \right]$$ - Divide row 2 by -5: $$\left[ \begin{array}{ccc|c} 1 & 2 & 1.5 & 50000 \\ 0 & 1 & -0.1 & 3000 \\ 0 & -10 & -1 & -70000 \end{array} \right]$$ - Replace row 3 by row 3 plus 10*row 2: $$0, -10 + 10*1 = 0, -1 + 10*(-0.1) = -1 -1 = -2, -70000 + 10*3000 = -70000 + 30000 = -40000$$ Matrix now: $$\left[ \begin{array}{ccc|c} 1 & 2 & 1.5 & 50000 \\ 0 & 1 & -0.1 & 3000 \\ 0 & 0 & -2 & -40000 \end{array} \right]$$ - Divide row 3 by -2: $$\left[ \begin{array}{ccc|c} 1 & 2 & 1.5 & 50000 \\ 0 & 1 & -0.1 & 3000 \\ 0 & 0 & 1 & 20000 \end{array} \right]$$ - Replace row 2 by row 2 plus 0.1*row 3: $$0, 1, 0, 3000 + 0.1*20000 = 3000 + 2000 = 5000$$ - Replace row 1 by row 1 minus 1.5*row 3: $$1, 2, 0, 50000 - 1.5*20000 = 50000 - 30000 = 20000$$ - Replace row 1 by row 1 minus 2*row 2: $$1, 0, 0, 20000 - 2*5000 = 20000 - 10000 = 10000$$ 6. **Final RREF matrix:** $$\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 10000 \\ 0 & 1 & 0 & 5000 \\ 0 & 0 & 1 & 20000 \end{array} \right]$$ 7. **Solution:** $$x = 10000, \quad y = 5000, \quad z = 20000$$ This means the culture can support 10,000 units of bacteria type A, 5,000 units of type B, and 20,000 units of type C daily given the nutrient supplies.