Chapter 3.5, Problem 1GP

a.

To determine

To organize the given data in a matrix.

Answer to Problem 1GP

  $\left(\begin{array}{cc}7.1& 6.6\\ \begin{array}{l}6.8\\ 3.2\\ 1.1\end{array}& \begin{array}{l}5.9\\ 2.2\\ 1.4\end{array}\end{array}\right)$

Explanation of Solution

Given:

Data for U.S. Census

  

Concept used:

A matrix is a rectangular array of scalars usually represented as:

  $A=\left(\begin{array}{ccc}{a}_{11}& \dots & a_{1n}\\ ⋮& \ddots & ⋮\\ a_{m1}& \cdots & a_{mn}\end{array}\right)$

Calculation:

To write the given data in a matrix form, add a column of male entries and a column of female entries.

That is, the matrix would be

  $\left(\begin{array}{cc}7.1& 6.6\\ \begin{array}{l}6.8\\ 3.2\\ 1.1\end{array}& \begin{array}{l}5.9\\ 2.2\\ 1.4\end{array}\end{array}\right)$

The matrix obtained is a $4\times 2$ matrix.

Conclusion:

Therefore, matrix form of the given data is $\left(\begin{array}{cc}7.1& 6.6\\ \begin{array}{l}6.8\\ 3.2\\ 1.1\end{array}& \begin{array}{l}5.9\\ 2.2\\ 1.4\end{array}\end{array}\right)$ .

b.

To determine

To add the elements in the columns.

Answer to Problem 1GP

  $18.2$ million, $16.1$ million

Explanation of Solution

Given:

Data for U.S. Census

  

Matrix form:

  $\left(\begin{array}{cc}7.1& 6.6\\ \begin{array}{l}6.8\\ 3.2\\ 1.1\end{array}& \begin{array}{l}5.9\\ 2.2\\ 1.4\end{array}\end{array}\right)$

Concept used:

A matrix is a rectangular array of scalars usually represented as:

  $A=\left(\begin{array}{ccc}{a}_{11}& \dots & a_{1n}\\ ⋮& \ddots & ⋮\\ a_{m1}& \cdots & a_{mn}\end{array}\right)$

Calculation:

Adding the columns of the matrix so obtained in part (a.).

  $\left(\begin{array}{cc}7.1& 6.6\\ \begin{array}{l}6.8\\ 3.2\\ 1.1\end{array}& \begin{array}{l}5.9\\ 2.2\\ 1.4\end{array}\end{array}\right)$

For male column,

  $7.1+6.8+3.2+1.1=18.2$

For female column,

  $6.6+5.9+2.2+1.4=16.1$

Conclusion:

Therefore, total number of males are $18.2$ million and females are $16.1$ million.

c.

To determine

To add the elements in the rows.

Answer to Problem 1GP

  $13.7$ million, $12.7$ million, $5.4$ million and $2.5$ million.

Explanation of Solution

Given:

Data for U.S. Census

  

Matrix form:

  $\left(\begin{array}{cc}7.1& 6.6\\ \begin{array}{l}6.8\\ 3.2\\ 1.1\end{array}& \begin{array}{l}5.9\\ 2.2\\ 1.4\end{array}\end{array}\right)$

Concept used:

A matrix is a rectangular array of scalars usually represented as:

  $A=\left(\begin{array}{ccc}{a}_{11}& \dots & a_{1n}\\ ⋮& \ddots & ⋮\\ a_{m1}& \cdots & a_{mn}\end{array}\right)$

Calculation:

Adding the columns of the matrix so obtained in part (a.).

  $\left(\begin{array}{cc}7.1& 6.6\\ \begin{array}{l}6.8\\ 3.2\\ 1.1\end{array}& \begin{array}{l}5.9\\ 2.2\\ 1.4\end{array}\end{array}\right)$

For first row, population in the age group 0-19:

  $7.1+6.6=13.7$ million

For second row, population in the age group 20-39:

  $6.8+5.9=12.7$

For third row, population in the age group 40-59:

  $3.2+2.2=5.4$

For fourth row, population in the age group 60+:

  $1.1+1.4=2.5$

Conclusion:

Therefore, total population in the respective age groups is $13.7$ million, $12.7$ million, $5.4$ million and $2.5$ million.

d.

To determine

To find the average of rows and columns.

Answer to Problem 1GP

  $\frac{34.3}{2}$ , $\frac{34.3}{4}$

Explanation of Solution

Given:

Data for U.S. Census

  

Matrix form:

  $\left(\begin{array}{cc}7.1& 6.6\\ \begin{array}{l}6.8\\ 3.2\\ 1.1\end{array}& \begin{array}{l}5.9\\ 2.2\\ 1.4\end{array}\end{array}\right)$

Concept used:

A matrix is a rectangular array of scalars usually represented as:

  $A=\left(\begin{array}{ccc}{a}_{11}& \dots & a_{1n}\\ ⋮& \ddots & ⋮\\ a_{m1}& \cdots & a_{mn}\end{array}\right)$

Calculation:

For the given data,

  $\left(\begin{array}{cc}7.1& 6.6\\ \begin{array}{l}6.8\\ 3.2\\ 1.1\end{array}& \begin{array}{l}5.9\\ 2.2\\ 1.4\end{array}\end{array}\right)$

Average of columns $=\frac{18.2+16.1}{2}$

  $=\frac{34.3}{2}$

Average of rows $=\frac{13.7+12.7+5.4+2.5}{4}$

  $=\frac{34.3}{2}$

Conclusion:

Therefore, the average of columns is $\frac{34.3}{2}$ and average of rows is $\frac{34.3}{4}$ .