Chapter 7, Problem 73RE

Solution

a.

To determine: The matrix of order of $1\times 4$ , where $N$ representing the number of shares of each stock.

The obtained matrix is $N=\left[\begin{array}{cccc}200& 400& 600& 250\end{array}\right]$ .

Given Information:

The shareholding is defined as,

  $\begin{array}{l}A:200\text{ shares }- \$80/\text{shares}\\ B:400\text{ shares }- \$120/\text{shares}\\ C:600\text{ shares }- \$200/\text{shares}\\ D:250\text{ shares }- \$300/\text{shares}\end{array}$

Calculation:

Consider the given information,

  $\begin{array}{l}A:200\text{ shares }- \$80/\text{shares}\\ B:400\text{ shares }- \$120/\text{shares}\\ C:600\text{ shares }- \$200/\text{shares}\\ D:250\text{ shares }- \$300/\text{shares}\end{array}$

Write the matrix form the given information shareholding.

  $N=\left[\begin{array}{cccc}200& 400& 600& 250\end{array}\right]$

Therefore, the obtained matrix is $N=\left[\begin{array}{cccc}200& 400& 600& 250\end{array}\right]$ .

b.

To determine: The matrix of the order of $1\times 4$ , name $P$ .

The matrix is $P=\left[\begin{array}{cccc}80& 120& 200& 300\end{array}\right]$ .

Given Information:

The share holding the price per share is defined as,

  $\begin{array}{l}A:200\text{ shares }- \$80/\text{shares}\\ B:400\text{ shares }- \$120/\text{shares}\\ C:600\text{ shares }- \$200/\text{shares}\\ D:250\text{ shares }- \$300/\text{shares}\end{array}$

Calculation:

Consider the given information,

  $\begin{array}{l}A:200\text{ shares }- \$80/\text{shares}\\ B:400\text{ shares }- \$120/\text{shares}\\ C:600\text{ shares }- \$200/\text{shares}\\ D:250\text{ shares }- \$300/\text{shares}\end{array}$

Write the matrix of the order of $1\times 4$ per share price holding.

  $P=\left[\begin{array}{cccc}80& 120& 200& 300\end{array}\right]$

Therefore, the obtained matrix is $P=\left[\begin{array}{cccc}80& 120& 200& 300\end{array}\right]$ .

c.

To determine: The matrix product that gives the total cost of the stocks tat he customer bought.

The obtained total cost is $\$259000$ .

Given Information:

The share holding the price per share is defined as,

  $\begin{array}{l}A:200\text{ shares }- \$80/\text{shares}\\ B:400\text{ shares }- \$120/\text{shares}\\ C:600\text{ shares }- \$200/\text{shares}\\ D:250\text{ shares }- \$300/\text{shares}\end{array}$

Calculation:

Consider the given information,

Refer the matrix form the previous parts.

  $N=\left[\begin{array}{cccc}200& 400& 600& 250\end{array}\right]$ and $P=\left[\begin{array}{cccc}80& 120& 200& 300\end{array}\right]$

The total cost of the shares is $N\cdot P^t$ .

  $\begin{array}{c}N\cdot P^t=\left[\begin{array}{cccc}200& 400& 600& 250\end{array}\right]{\left[\begin{array}{cccc}80& 120& 200& 300\end{array}\right]}^T\\ =\left[\begin{array}{cccc}200& 400& 600& 250\end{array}\right]\left[\begin{array}{c}80\\ 120\\ 200\\ 300\end{array}\right]\\ =16000+48000+120000+75000\\ =\$259000\end{array}$

Therefore, the obtained total cost is $\$259000$ .