Chapter 4.2, Problem 21PPS

(a)

To determine

To Find: The matrix for each of the library’s number of books.

Answer to Problem 21PPS

The required matrix are,

  $\begin{array}{l}A=\left[\begin{array}{c}10000\\ 5000\\ 5000\end{array}\right]\\ B=\left[\begin{array}{c}15000\\ 10000\\ 2500\end{array}\right]\\ C=\left[\begin{array}{c}4000\\ 700\\ 800\end{array}\right]\end{array}$

Explanation of Solution

Given:

The library A has 10000 novels, 5000 biographies and 5000 children’s book.

The library B has 15000 novels, 10000 biographies and 2500 children’s book.

The library C has 4000 novels, 700 biographies and 800 children’s book.

Calculation:

From the given data the required matrix for the library as,

  $\begin{array}{l}A=\left[\begin{array}{c}10000\\ 5000\\ 5000\end{array}\right]\\ B=\left[\begin{array}{c}15000\\ 10000\\ 2500\end{array}\right]\\ C=\left[\begin{array}{c}4000\\ 700\\ 800\end{array}\right]\end{array}$

(b)

To determine

To Find: The total number of each type of book in all the libraries.

Answer to Problem 21PPS

The matrix for total of each type of book in the library is $\left[\begin{array}{c}29000\\ 15700\\ 8300\end{array}\right]$ .

Explanation of Solution

Consider the matrices for the library is,

  $\begin{array}{l}A=\left[\begin{array}{c}10000\\ 5000\\ 5000\end{array}\right]\\ B=\left[\begin{array}{c}15000\\ 10000\\ 2500\end{array}\right]\\ C=\left[\begin{array}{c}4000\\ 700\\ 800\end{array}\right]\end{array}$

Then, the matrix for each type of book in all the library is,

  $\left[\begin{array}{c}10000\\ 5000\\ 5000\end{array}\right]+\left[\begin{array}{c}15000\\ 10000\\ 2500\end{array}\right]+\left[\begin{array}{c}4000\\ 700\\ 800\end{array}\right]=\left[\begin{array}{c}29000\\ 15700\\ 8300\end{array}\right]$

(c)

To determine

To Find: The number of books that library A has more than that of library C.

Answer to Problem 21PPS

The required matrix is $\left[\begin{array}{c}6000\\ 4300\\ 4200\end{array}\right]$ .

Explanation of Solution

Consider the matrices for the library is,

  $\begin{array}{l}A=\left[\begin{array}{c}10000\\ 5000\\ 5000\end{array}\right]\\ B=\left[\begin{array}{c}15000\\ 10000\\ 2500\end{array}\right]\\ C=\left[\begin{array}{c}4000\\ 700\\ 800\end{array}\right]\end{array}$

Then, the matrix for the each type of book that library A has more than that of library C is,

  $\left[\begin{array}{c}10000\\ 5000\\ 5000\end{array}\right]-\left[\begin{array}{c}4000\\ 700\\ 800\end{array}\right]=\left[\begin{array}{c}6000\\ 4300\\ 4200\end{array}\right]$

(d)

To determine

To Find: The matrix $A+B$ and its relevance.

Answer to Problem 21PPS

The matrix is $\left[\begin{array}{c}25000\\ 15000\\ 7500\end{array}\right]$ and the matrix gives the total of each type of book in library A and B.

Explanation of Solution

Consider the matrices for the library is,

  $\begin{array}{l}A=\left[\begin{array}{c}10000\\ 5000\\ 5000\end{array}\right]\\ B=\left[\begin{array}{c}15000\\ 10000\\ 2500\end{array}\right]\\ C=\left[\begin{array}{c}4000\\ 700\\ 800\end{array}\right]\end{array}$

Then, matrix $A+B$ is,

  $\left[\begin{array}{c}10000\\ 5000\\ 5000\end{array}\right]+\left[\begin{array}{c}15000\\ 10000\\ 2500\end{array}\right]=\left[\begin{array}{c}25000\\ 15000\\ 7500\end{array}\right]$

The matrix gives the total of each type of book in library A and B.