## Precalculus (6th Edition) Blitzer

a. Step 1. Assume the amounts of Food-A, Food-B, and Food-C are $x,y,z$ respectively. We can set up a system of equations using the desired quantities of each vitamins: $\begin{cases} 3x+y+3z=14 \\ 7x+5y+8z=32 \\ x+3y+2z=9 \end{cases}$ Rearrange the equations: $\begin{cases} x+3y+2z=9 \\ 3x+y+3z=14 \\ 7x+5y+8z=32 \end{cases}$ Step 2. Write a matrix of the system and perform row the operations: $\begin{bmatrix} 1 & 3 & 2 & | & 9 \\ 3 & 1 & 3 & | & 14 \\ 7 & 5 & 8 & | & 32 \end{bmatrix} \begin{array} ..\\3R1-R2\to R2\\7R1-R3\to R3 \end{array}$ $\begin{bmatrix} 1 & 3 & 2 & | & 9 \\ 0 & 8 & 3 & | & 13 \\ 0 & 16 & 6 & | & 31 \end{bmatrix} \begin{array} ..\\..\\R3-2R2\to R3 \end{array}$ $\begin{bmatrix} 1 & 3 & 2 & | & 9 \\ 0 & 8 & 3 & | & 13 \\ 0 & 0 & 0 & | & 5 \end{bmatrix} \begin{array} ..\\..\\.. \end{array}$ Step 3. The last row gives $0=5$, which is not possible. Thus, we have an inconsistent system and there is no solution. In other words, no combination of these foods can provide exactly the quantities specified. b. If the riboflavin requirement is increased by 5 mg, we have: $\begin{cases} x+3y+2z=9 \\ 3x+y+3z=14 \\ 7x+5y+8z=32+5 \end{cases}$ and the matrix becomes: $\begin{bmatrix} 1 & 3 & 2 & | & 9 \\ 3 & 1 & 3 & | & 14 \\ 7 & 5 & 8 & | & 37 \end{bmatrix} \begin{array} ..\\3R1-R2\to R2\\7R1-R3\to R3 \end{array}$ $\begin{bmatrix} 1 & 3 & 2 & | & 9 \\ 0 & 8 & 3 & | & 13 \\ 0 & 16 & 6 & | & 26 \end{bmatrix} \begin{array} ..\\..\\R3-2R2\to R3 \end{array}$ $\begin{bmatrix} 1 & 3 & 2 & | & 9 \\ 0 & 8 & 3 & | & 13 \\ 0 & 0 & 0 & | & 0 \end{bmatrix} \begin{array} ..\\..\\.. \end{array}$ The last row gives $0=0$. Thus, we have a dependent system with an unlimited number of solutions of $x,y,z$ values.