Chapter 4, Problem 15STP

(a)

To determine

To find: The system of equations to model the situation.

Answer to Problem 15STP

The system of equations to model the situation is given below.

Explanation of Solution

Given:

The cost of each jet printer is $\$200$ .

The cost of each laser printer is $\$500$ .

  $x$ represents the number of ink jet printers.

  $y$ represents the number of laser printers.

The system of equation that model the given situation is given as:

  $\begin{array}{c}x+y=40\text{number of boxes for the printers}\\ \text{200}x+500y=11,600\text{total price for the printers}\end{array}$

So the system of equations are shown above.

(b)

To determine

To find: The matrix equation form that can use to solve the above system of equation.

Answer to Problem 15STP

The matrix equation is given below.

Explanation of Solution

The matrix of the system’s coefficient $A$ is given as:

  $\left[\begin{array}{cc}1& 1\\ 200& 500\end{array}\right]$

The system can be written in a matrix equation form is given as:

  $\begin{array}{l}AX=b\\ \left[\begin{array}{cc}1& 1\\ 200& 500\end{array}\right]\cdot \left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}40\\ 11600\end{array}\right]\end{array}$

Therefore the matrix equation is given above.

(c)

To determine

To find: The inverse of the coefficient matrix and hence solve the matrix equation.

Answer to Problem 15STP

The answer is given below.

Explanation of Solution

The inverse of coefficient $A$ is calculated as:

  $\begin{array}{c}{A}^{-1}=\frac{1}{1\cdot 500-1\cdot 300}\left[\begin{array}{cc}500& -1\\ -200& 1\end{array}\right]\\ =\frac{1}{300}\left[\begin{array}{cc}500& -1\\ -200& 1\end{array}\right]\\ =\left[\begin{array}{cc}\frac{500}{300}& \frac{-1}{300}\\ \frac{-200}{300}& \frac{1}{300}\end{array}\right]\end{array}$

The system can be rewritten as:

  $\begin{array}{c}{A}^{-1}AX=A^{-1}b\\ X=A^{-1}b\end{array}$

Further simplified as:

  $\begin{array}{c}\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{cc}\frac{500}{300}& -\frac{1}{300}\\ \frac{-200}{300}& \frac{1}{300}\end{array}\right]\cdot \left[\begin{array}{c}40\\ 11,600\end{array}\right]\\ =\left[\begin{array}{c}\frac{500}{300}\cdot 40-\frac{1}{300}\cdot 11,600\\ -\frac{200}{300}\cdot 40+\frac{1}{300}\cdot 11,600\end{array}\right]\\ =\left[\begin{array}{c}28\\ 12\end{array}\right]\end{array}$

Further simplified as:

  $\begin{array}{c}x=28\\ y=12\end{array}$