Chapter 8 - Section 8.1 - Matrix Solutions to Linear Systems - Exercise Set - Page 894: 47

4 ounces of food A, $\frac{1}{2}$ ounces of food B and 1 ounce of food C should be used.

Let $x$ be the number of ounces of food A, $y$ be the number of ounces of food B and $z$ be the number of ounces of food C. It is provided that three foods allow exactly 660 calories, 25 grams of protein, and 425 milligrams of vitamin C. Therefore, from the provided table, the system of equations is: $\begin{align} & 40x+200y+400z=660 \\ & 5x+2y+4z=25 \\ & 30x+10y+300z=425 \end{align}$ First write the augmented matrix for the given system of equations: Augmented matrix is: $\left[ \begin{matrix} 40 & 200 & 400 & 660 \\ 5 & 2 & 4 & 25 \\ 30 & 10 & 300 & 425 \\ \end{matrix} \right]$ Now, reduce the matrix to row echelon form by using row operation ${{R}_{1}}\to \frac{1}{40}{{R}_{1}}$, we get $\left[ \begin{matrix} 1 & 5 & 10 & \frac{33}{2} \\ 5 & 2 & 4 & 25 \\ 30 & 10 & 300 & 425 \\ \end{matrix} \right]$ ${{R}_{2}}\to {{R}_{2}}-5{{R}_{1}}$, we get $\left[ \begin{matrix} 1 & 5 & 10 & \frac{33}{2} \\ 0 & -23 & -46 & \frac{-115}{2} \\ 30 & 10 & 300 & 425 \\ \end{matrix} \right]$ ${{R}_{3}}\to {{R}_{3}}-30{{R}_{1}}$, we get $\left[ \begin{matrix} 1 & 5 & 10 & \frac{33}{2} \\ 0 & -23 & -46 & \frac{-115}{2} \\ 0 & -140 & 0 & -70 \\ \end{matrix} \right]$ ${{R}_{2}}\to \frac{-1}{23}{{R}_{2}}$, we get $\left[ \begin{matrix} 1 & 5 & 10 & \frac{33}{2} \\ 0 & 1 & 2 & \frac{5}{2} \\ 0 & -140 & 0 & -70 \\ \end{matrix} \right]$ ${{R}_{3}}\to {{R}_{3}}+140{{R}_{2}}$, we get $\left[ \begin{matrix} 1 & 5 & 10 & \frac{33}{2} \\ 0 & 1 & 2 & \frac{5}{2} \\ 0 & 0 & 280 & 280 \\ \end{matrix} \right]$ ${{R}_{3}}\to \frac{1}{280}{{R}_{3}}$, we get $\left[ \begin{matrix} 1 & 5 & 10 & \frac{33}{2} \\ 0 & 1 & 2 & \frac{5}{2} \\ 0 & 0 & 1 & 1 \\ \end{matrix} \right]$ ${{R}_{2}}\to {{R}_{2}}-2{{R}_{3}}$, we get $\left[ \begin{matrix} 1 & 5 & 10 & \frac{33}{2} \\ 0 & 1 & 0 & \frac{1}{2} \\ 0 & 0 & 1 & 1 \\ \end{matrix} \right]$ ${{R}_{1}}\to {{R}_{1}}-10{{R}_{3}}$, we get $\left[ \begin{matrix} 1 & 5 & 0 & \frac{13}{2} \\ 0 & 1 & 0 & \frac{1}{2} \\ 0 & 0 & 1 & 1 \\ \end{matrix} \right]$ ${{R}_{1}}\to {{R}_{1}}-5{{R}_{2}}$, we get $\left[ \begin{matrix} 1 & 0 & 0 & 4 \\ 0 & 1 & 0 & \frac{1}{2} \\ 0 & 0 & 1 & 1 \\ \end{matrix} \right]$ Thus, $x=4,y=\frac{1}{2},z=1$ Hence, $4\text{ ounces}$ of food A, $\frac{1}{2}\text{ ounces}$ of food B and $1\text{ ounces}$ of food C should be used.