Chapter 11.4, Problem 83AYU

  $\left(a\right)$

To determine

To calculate: The matrix A that represent the credit hours taken by each student and matrix B that represent the cost per credit hour.

Answer to Problem 83AYU

The matrices are $A=\left[\begin{array}{cc}6& 9\\ 3& 12\end{array}\right]\text{ and }B=\left[\begin{array}{c}80.00\\ 277.80\end{array}\right]$ .

Explanation of Solution

Given information:

The number of credit hours taken and the cost per credit hour for Nikki and Joe who take classes at a community college, LCCC and a local university, SIUE is provided below,

  $\begin{array}{ccc}& \text{LCCC}& \text{SIUE}\\ \text{Nikki}& 6& 9\\ \text{Joe}& 3& 12\end{array}\begin{array}{cc}& \text{Cost per credit hour}\\ \text{LCCC}& \$80.00\\ \text{SIUE}& \$277.80\end{array}$

Calculation:

Consider the number of credit hours taken and the cost per credit hour for Nikki and Joe who take classes at a community college, LCCC and a local university, SIUE is provided below,

  $\begin{array}{ccc}& \text{LCCC}& \text{SIUE}\\ \text{Nikki}& 6& 9\\ \text{Joe}& 3& 12\end{array}\begin{array}{cc}& \text{Cost per credit hour}\\ \text{LCCC}& \$80.00\\ \text{SIUE}& \$277.80\end{array}$

Let matrix A that represent the credit hours taken by each student and matrix B that represent the cost per credit hour.

So, row 1 of matrix A denote the credit hours taken by Nikki at two universities. And row 2 of matrix A denote the credit hours taken by Joe at two universities. Matrix B will be a column vector with cost per credit hour.

Therefore,

  $A=\left[\begin{array}{cc}6& 9\\ 3& 12\end{array}\right]\text{ and }B=\left[\begin{array}{c}80.00\\ 277.80\end{array}\right]$

  $\left(b\right)$

To determine

To calculate: The value of $AB$ .

Answer to Problem 83AYU

The value of $AB$ is $\left[\begin{array}{c}62\\ 50.3\end{array}\right]$ .

Explanation of Solution

Given information:

The number of credit hours taken and the cost per credit hour for Nikki and Joe who take classes at a community college, LCCC and a local university, SIUE is provided below,

  $\begin{array}{ccc}& \text{LCCC}& \text{SIUE}\\ \text{Nikki}& 6& 9\\ \text{Joe}& 3& 12\end{array}\begin{array}{cc}& \text{Cost per credit hour}\\ \text{LCCC}& \$80.00\\ \text{SIUE}& \$277.80\end{array}$

Formula used:

To find the product AB of two matrices A and B, the number of column in matrix A must be equal to number of rows in matrix B.

Calculation:

Consider the number of credit hours taken and the cost per credit hour for Nikki and Joe who take classes at a community college, LCCC and a local university, SIUE is provided below,

  $\begin{array}{ccc}& \text{LCCC}& \text{SIUE}\\ \text{Nikki}& 6& 9\\ \text{Joe}& 3& 12\end{array}\begin{array}{cc}& \text{Cost per credit hour}\\ \text{LCCC}& \$80.00\\ \text{SIUE}& \$277.80\end{array}$

Let matrix A that represent the credit hours taken by each student and matrix B that represent the cost per credit hour.

So, row 1 of matrix A denote the credit hours taken by Nikki at two universities. And row 2 of matrix A denote the credit hours taken by Joe at two universities. Matrix B will be a column vector with cost per credit hour.

Therefore,

  $A=\left[\begin{array}{cc}6& 9\\ 3& 12\end{array}\right]\text{ and }B=\left[\begin{array}{c}80.00\\ 277.80\end{array}\right]$

Now,

  $\begin{array}{c}AB=\left[\begin{array}{cc}6& 9\\ 3& 12\end{array}\right]\left[\begin{array}{c}80.00\\ 277.80\end{array}\right]\\ =\left[\begin{array}{c}480+2500.2\\ 240+3333.6\end{array}\right]\\ =\left[\begin{array}{c}2980.2\\ 3573.6\end{array}\right]\end{array}$

Tuition fee for Nikki is $\$2980.2$ and for Stephanie is $\$3573.6$ .