Question

(1) The table below shows the calories, fat, and carbohydrates per ounce for three brands of cereal.

Calories Fat Carbohydrates

Cereal Brand A 10 0 11

Cereal Brand B 100 0.2 22.5

Cereal Brand C 130 5.6 19

Total Required 400 6.2 92

(i) Using the information in the table above, write a system of three equations.

(ii) Write a system in matrix form Ax = b.

(iii) Using the Cramer's Rule or Inverse Method, find the amount of each brand of cereal that will give the level of nutrition required. 

Answer 1

(i)

$$\def\arraystretch{1.5} \begin{array}{c} 10x+100y+130z=400 \\ 0x+0.2y+5.6z=6.2 \\ 11x+22.5y+19z=92 \end{array}$$

(ii)

$$A=\begin{pmatrix} 10& 100& 130 \\ 0& 0.2& 5.6 \\ 11 & 22.5 & 19 \end{pmatrix}, X=\begin{pmatrix} x\\ y \\ z \end{pmatrix}, B=\begin{pmatrix} 400\\ 6.2 \\ 92 \end{pmatrix},$$

$$AX=B$$

$$\begin{pmatrix} 10& 100& 130 \\ 0& 0.2& 5.6 \\ 11 & 22.5 & 19 \end{pmatrix}\begin{pmatrix} x\\ y \\ z \end{pmatrix}=\begin{pmatrix} 400\\ 6.2 \\ 92 \end{pmatrix}$$

(iii)

$$\Delta=\det A=\begin{vmatrix} 10& 100& 130 \\ 0& 0.2& 5.6 \\ 11 & 22.5 & 19 \end{vmatrix}$$

$$=10\begin{vmatrix} 0.2 & 5.6 \\ 22.5 & 19 \end{vmatrix}-0\begin{vmatrix} 100 & 130 \\ 22.5 & 19 \end{vmatrix}+11\begin{vmatrix} 100 &130 \\ 0.2 & 5.6 \end{vmatrix}$$

$$=-1222-0+5874=4652\not=0$$

$$\Delta_1=\begin{vmatrix} 400& 100& 130 \\ 6.2 & 0.2& 5.6 \\ 92 & 22.5 & 19 \end{vmatrix}$$

$$=400\begin{vmatrix} 0.2 & 5.6 \\ 22.5 & 19 \end{vmatrix}-6.2\begin{vmatrix} 100 & 130 \\ 22.5 & 19 \end{vmatrix}+92\begin{vmatrix} 100 &130 \\ 0.2 & 5.6 \end{vmatrix}$$

$$=-48880+6355+49128=6603$$

$$\Delta_2=\begin{vmatrix} 10& 400& 130 \\ 0& 6.2& 5.6 \\ 11 & 92 & 19 \end{vmatrix}$$

$$=10\begin{vmatrix} 6.2 & 5.6 \\ 92 & 19 \end{vmatrix}-0\begin{vmatrix} 400 & 130 \\ 92 & 19 \end{vmatrix}+11\begin{vmatrix} 400 &130 \\ 6.2 & 5.6 \end{vmatrix}$$

$$=-3974-0+15774=11800$$

$$\Delta_3=\begin{vmatrix} 10& 100& 400 \\ 0& 0.2& 6.2 \\ 11 & 22.5 & 92 \end{vmatrix}$$

$$=10\begin{vmatrix} 0.2 & 6.2 \\ 22.5 & 92 \end{vmatrix}-0\begin{vmatrix} 100 & 400 \\ 22.5 & 92 \end{vmatrix}+11\begin{vmatrix} 100 &400 \\ 0.2 & 6.2 \end{vmatrix}$$

$$=-1211-0+5940=4729$$

$$x=\dfrac{\Delta_1}{\Delta}=\dfrac{6603}{4652}=1.41939$$

$$y=\dfrac{\Delta_2}{\Delta}=\dfrac{11800}{4652}=2.53654$$

$$z=\dfrac{\Delta_3}{\Delta}=\dfrac{4729}{4652}=1.01655$$