Chapter 12, Problem 26CCSR

Solution

a.

To determine: The matrix equation for the given data.

The obtained matrix is $\left[\begin{array}{ccc}2& 3& 3\\ 3& 3& 3\\ 4& 1& 2\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}24\\ 27\\ 20\end{array}\right]$ .

Given Information:

The table is defined,

Food	Protein	Fat	Carbohydrates
A	$\text{2 g/oz}$	$3\text{ g/oz}$	$4\text{ g/oz}$
B	$3\text{ g/oz}$	$3\text{ g/oz}$	$1\text{ g/oz}$
C	$3\text{ g/oz}$	$3\text{ g/oz}$	$2\text{ g/oz}$

Calculation:

Consider the given table,

Food	Protein	Fat	Carbohydrates
A	$\text{2 g/oz}$	$3\text{ g/oz}$	$4\text{ g/oz}$
B	$3\text{ g/oz}$	$3\text{ g/oz}$	$1\text{ g/oz}$
C	$3\text{ g/oz}$	$3\text{ g/oz}$	$2\text{ g/oz}$

Suppose that x is the amount of food $\text{A}$ , y is the amount of food B , and z is the amount of food C .

Now, write the system of equations.

  $\begin{array}{c}2x+3y+3z=24\\ 3x+3y+3z=27\\ 4x+y+2z=20\end{array}$

Write the matrix form of the system.

  $\begin{array}{c}AX=B\\ \left[\begin{array}{ccc}2& 3& 3\\ 3& 3& 3\\ 4& 1& 2\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}24\\ 27\\ 20\end{array}\right]\end{array}$

Therefore, the obtained matrix equation $\left[\begin{array}{ccc}2& 3& 3\\ 3& 3& 3\\ 4& 1& 2\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}24\\ 27\\ 20\end{array}\right]$ .

b.

To calculate: The solution of the system of equations.

The obtained solution is $\left(x,y,z\right)=\left(3,4,2\right)$ .

Given Information:

Refer the matrix equation from the previous part (a).

  $\left[\begin{array}{ccc}2& 3& 3\\ 3& 3& 3\\ 4& 1& 2\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}24\\ 27\\ 20\end{array}\right]$

Calculation:

Consider the given information,

  $\left[\begin{array}{ccc}2& 3& 3\\ 3& 3& 3\\ 4& 1& 2\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}24\\ 27\\ 20\end{array}\right]$

Rewrite the equation as following.

  $\begin{array}{c}{A}^{-1}AX=A^{-1}B\\ \left(I\right)X=A^{-1}B\\ X=A^{-1}B\end{array}$

Use the row reduced form on augmented matrix to find the solution.

  $\begin{array}{l}\left[\begin{array}{cccc}2& 3& 3& 24\\ 3& 3& 3& 27\\ 4& 1& 2& 20\end{array}\right]\underrightarrow{R_1\to R_2-R_1}\left[\begin{array}{cccc}1& 0& 0& 3\\ 3& 3& 3& 27\\ 4& 1& 2& 20\end{array}\right]\underrightarrow{R_2\to \frac{1}{3}{R}_2}\left[\begin{array}{cccc}1& 0& 0& 3\\ 1& 1& 1& 9\\ 4& 1& 2& 20\end{array}\right]\\ \\ \underrightarrow{R_2\to R_2-R_1}\left[\begin{array}{cccc}1& 0& 0& 3\\ 0& 1& 1& 6\\ 4& 1& 2& 20\end{array}\right]\underrightarrow{R_3\to R_3-R_2}\left[\begin{array}{cccc}1& 0& 0& 3\\ 0& 1& 1& 6\\ 4& 0& 1& 14\end{array}\right]\\ \\ \underrightarrow{R_3\to 4R_1-R_3}\left[\begin{array}{cccc}1& 0& 0& 3\\ 0& 1& 1& 6\\ 0& 0& 1& 2\end{array}\right]\underrightarrow{R_2\to R_2-R_3}\left[\begin{array}{cccc}1& 0& 0& 3\\ 0& 1& 0& 4\\ 0& 0& 1& 2\end{array}\right]\end{array}$

Hence, the obtained result is $\left(x,y,z\right)=\left(3,4,2\right)$ .

c.

To calculate: The number of ounces of each food are required.

A is 3 ounces, the food for B is 4 ounces and the food for C is 2 ounces.

Given Information:

Refer the matrix equation from the previous part (a).

  $\left[\begin{array}{ccc}2& 3& 3\\ 3& 3& 3\\ 4& 1& 2\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}24\\ 27\\ 20\end{array}\right]$

Calculation:

Consider the given information,

  $\left[\begin{array}{ccc}2& 3& 3\\ 3& 3& 3\\ 4& 1& 2\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}24\\ 27\\ 20\end{array}\right]$

The obtained solution in the previous part for the above solution is $\left(x,y,z\right)=\left(3,4,2\right)$ . This implies that the food for A is 3 ounces, the food for B is 4 ounces and the food for C is 2 ounces.

Hence, the obtained result is “A is 3 ounces, the food for B is 4 ounces and the food for C is 2 ounces”.