Chapter 4.2, Problem 14PPS

(a)

To determine

The matrix C to represent the current prices.

Answer to Problem 14PPS

The matrix for the current prices is $C=\left[\begin{array}{ccc}0.95& 1& 1.05\\ 0.75& 0.80& 0.85\\ 0.75& 0.80& 0.85\\ 1.00& 1.10& 1.20\end{array}\right]$ .

Explanation of Solution

Given:

The given table is shown in Table 1

Table 1

Drink	Small	Medium	Large
Soda	0.95	1	1.05
Iced Tea	0.75	0.80	0.85
Lemonade	0.75	0.80	0.85
Coffee	1	1.10	1.20

Calculation:

To write the matrix that represent the current prices consider the table shown in table 1 as,

  $C=\left[\begin{array}{ccc}0.95& 1& 1.05\\ 0.75& 0.80& 0.85\\ 0.75& 0.80& 0.85\\ 1.00& 1.10& 1.20\end{array}\right]$

(b)

To determine

To Find: The scalar that can determine a matrix $N$ to represent the new prices.

Answer to Problem 14PPS

The required scalar is $1.1$ .

Explanation of Solution

Consider the matrix is,

  $C=\left[\begin{array}{ccc}0.95& 1& 1.05\\ 0.75& 0.80& 0.85\\ 0.75& 0.80& 0.85\\ 1.00& 1.10& 1.20\end{array}\right]$

Consider that the prices are increased by 10% so the new prices are,

  $\begin{array}{c}1+10\%=1+\frac{10}{100}\\ =1.1\end{array}$

(c)

To determine

To Find: The matrix N.

Answer to Problem 14PPS

The required matrix $N=1.1\left[\begin{array}{ccc}1.045& 1.1& 1.155\\ 0.825& 0.880& 0.935\\ 0.825& 0.880& 0.935\\ 1.100& 1.210& 1.320\end{array}\right]$ ,

Explanation of Solution

Consider the matrix is,

  $C=\left[\begin{array}{ccc}0.95& 1& 1.05\\ 0.75& 0.80& 0.85\\ 0.75& 0.80& 0.85\\ 1.00& 1.10& 1.20\end{array}\right]$

Consider the scalar 1.1 then,

  $\begin{array}{c}C=1.1\left[\begin{array}{ccc}0.95& 1& 1.05\\ 0.75& 0.80& 0.85\\ 0.75& 0.80& 0.85\\ 1.00& 1.10& 1.20\end{array}\right]\\ =\left[\begin{array}{ccc}1.045& 1.1& 1.155\\ 0.825& 0.880& 0.935\\ 0.825& 0.880& 0.935\\ 1.100& 1.210& 1.320\end{array}\right]\end{array}$

(d)

To determine

To Find: The difference $N-C$ and what does this represent.

Answer to Problem 14PPS

The matrix is $\left[\begin{array}{ccc}0.095& 0.100& 0.105\\ 0.075& 0.080& 0.085\\ 0.075& 0.080& 0.085\\ 0.100& 0.11& 0.120\end{array}\right]$ and it shows the increase in price.

Explanation of Solution

Consider the matrix is,

  $C=\left[\begin{array}{ccc}0.95& 1& 1.05\\ 0.75& 0.80& 0.85\\ 0.75& 0.80& 0.85\\ 1.00& 1.10& 1.20\end{array}\right]$

Consider the other matrix,

  $N=\left[\begin{array}{ccc}1.045& 1.1& 1.155\\ 0.825& 0.880& 0.935\\ 0.825& 0.880& 0.935\\ 1.100& 1.210& 1.320\end{array}\right]$

Then, $N-C$ is,

  $\left[\begin{array}{ccc}1.045& 1.1& 1.155\\ 0.825& 0.880& 0.935\\ 0.825& 0.880& 0.935\\ 1.100& 1.210& 1.320\end{array}\right]-\left[\begin{array}{ccc}0.95& 1& 1.05\\ 0.75& 0.80& 0.85\\ 0.75& 0.80& 0.85\\ 1.00& 1.10& 1.20\end{array}\right]=\left[\begin{array}{ccc}0.095& 0.100& 0.105\\ 0.075& 0.080& 0.085\\ 0.075& 0.080& 0.085\\ 0.100& 0.11& 0.120\end{array}\right]$

The matrix shows the increase in price.