Chapter 12.2, Problem 28PPSE

Solution

a.

To determine: The matrix to show the number of each type of flower in each arrangement if a florist makes three special floral arrangements. One uses three lilies, the second uses three lilies and four carnations. The third one uses four daisies and three carnations. Lilies cost $\$2.15$ each, carnations cost $\$0.90$ each and daisies cost $\$1.30$ each.

The required matrix is $\left[\begin{array}{ccc}3& 0& 0\\ 3& 4& 0\\ 0& 3& 4\end{array}\right]$ .

Given information:

A florist makes three special floral arrangements. One uses three lilies, the second uses three lilies and four carnations. The third one uses four daisies and three carnations. Lilies cost $\$2.15$ each, carnations cost $\$0.90$ each and daisies cost $\$1.30$ each.

Explanation :

It is given that the first floral arrangement has three lilies, the second one has three lilies and four carnations and third one has four daisies and three carnations.

This arrangement can be represented by a $3\times 3$ matrix to represent each type of flower in each arrangement.

  $\begin{array}{c}\\ \text{Arrangement 1}\\ \text{Arrangement 2}\\ \text{Arrangement 3}\end{array}\left[\begin{array}{ccc}\text{lilies}& \text{carnations}& \text{daisies}\\ 3& 0& 0\\ 3& 4& 0\\ 0& 3& 4\end{array}\right]$

Thus, the required matrix is $\left[\begin{array}{ccc}3& 0& 0\\ 3& 4& 0\\ 0& 3& 4\end{array}\right]$ .

b.

To determine: The matrix to show the cost of each type of flower in each arrangement if a florist makes three special floral arrangements. One uses three lilies, the second uses three lilies and four carnations. The third one uses four daisies and three carnations. Lilies cost $\$2.15$ each, carnations cost $\$0.90$ each and daisies cost $\$1.30$ each.

The required matrix is $\left[\begin{array}{c}2.15\\ 0.90\\ 1.30\end{array}\right]$ .

Given information:

A florist makes three special floral arrangements. One uses three lilies, the second uses three lilies and four carnations. The third one uses four daisies and three carnations. Lilies cost $\$2.15$ each, carnations cost $\$0.90$ each and daisies cost $\$1.30$ each.

Explanation :

It is given that the lilies cost $\$2.15$ each, carnations cost $\$0.90$ each and daisies cost $\$1.30$ each.

This arrangement can be represented by a $3\times 1$ matrix to represent cost of each type of flower in each arrangement.

  $\begin{array}{c}\text{lilies}\\ \text{carnations}\\ \text{daisies}\end{array}\left[\begin{array}{c}2.15\\ 0.90\\ 1.30\end{array}\right]$

Thus, the required matrix is $\left[\begin{array}{c}2.15\\ 0.90\\ 1.30\end{array}\right]$ .

c.

To calculate: The matrix to show the cost of each floral arrangement if a florist makes three special floral arrangements. One uses three lilies, the second uses three lilies and four carnations. The third one uses four daisies and three carnations. Lilies cost $\$2.15$ each, carnations cost $\$0.90$ each and daisies cost $\$1.30$ each.

The required matrix is $\left[\begin{array}{c}6.45\\ 10.05\\ 7.90\end{array}\right]$ .

Given information:

A florist makes three special floral arrangements. One uses three lilies, the second uses three lilies and four carnations. The third one uses four daisies and three carnations. Lilies cost $\$2.15$ each, carnations cost $\$0.90$ each and daisies cost $\$1.30$ each.

Formula Used:

Use the matrix multiplication.

Calculation :

The number of each type of flower in each arrangement can be shown as $\left[\begin{array}{ccc}3& 0& 0\\ 3& 4& 0\\ 0& 3& 4\end{array}\right]$ .

The cost of each type of flower in each arrangement can be shown as $\left[\begin{array}{c}2.15\\ 0.90\\ 1.30\end{array}\right]$ .

The cost of each floral arrangement is calculated as $\left[\begin{array}{ccc}3& 0& 0\\ 3& 4& 0\\ 0& 3& 4\end{array}\right]\left[\begin{array}{c}2.15\\ 0.90\\ 1.30\end{array}\right]$ .

  $\begin{array}{c}\left[\begin{array}{ccc}3& 0& 0\\ 3& 4& 0\\ 0& 3& 4\end{array}\right]\left[\begin{array}{c}2.15\\ 0.90\\ 1.30\end{array}\right]=\left[\begin{array}{c}3\left(2.15\right)+0\left(0.90\right)+001.30\\ 3\left(2.15\right)+4\left(0.90\right)+0\left(1.30\right)\\ 0\left(2.15\right)+3\left(0.90\right)+4\left(1.30\right)\end{array}\right]\\ =\left[\begin{array}{c}6.45+0+0\\ 6.45+3.60+0\\ 0+2.70+5.20\end{array}\right]\\ =\left[\begin{array}{c}6.45\\ 10.05\\ 7.90\end{array}\right]\end{array}$

Thus, the required matrix is $\left[\begin{array}{c}6.45\\ 10.05\\ 7.90\end{array}\right]$ .