Chapter 11.4, Problem 84AYU

  $\left(a\right)$

To determine

To calculate: The matrix A that represent the amounts borrowed by each student and matrix B that represent the monthly interest rates.

Answer to Problem 84AYU

The matrices are $A=\left[\begin{array}{cc}4000& 3000\\ 2500& 3800\end{array}\right]\text{ and }B=\left[\begin{array}{c}0.011\\ 0.006\end{array}\right]$ .

Explanation of Solution

Given information:

The amount borrowed by Jamal and Stephanie from two banks for school loan and monthly interest rate of banks are provided below,

  $\begin{array}{ccc}& \text{Lender 1}& \text{Lender 2}\\ \text{Jamal}& \$4000& \$3000\\ \text{Stephanie}& \$2500& \$3800\end{array}\begin{array}{cc}& \text{Monthly Interest Rate}\\ \text{Lender 1}& 0.011\left(1.1\%\right)\\ \text{Lender 2}& 0.006\left(0.6\%\right)\end{array}$

Calculation:

Consider the amount borrowed by Jamal and Stephanie from two banks for school loan and monthly interest rate of banks are provided below,

  $\begin{array}{ccc}& \text{Lender 1}& \text{Lender 2}\\ \text{Jamal}& \$4000& \$3000\\ \text{Stephanie}& \$2500& \$3800\end{array}\begin{array}{cc}& \text{Monthly Interest Rate}\\ \text{Lender 1}& 0.011\left(1.1\%\right)\\ \text{Lender 2}& 0.006\left(0.6\%\right)\end{array}$

Let matrix A that represent the amounts borrowed by each student and matrix B that represent the monthly interest rates.

So, row 1 of matrix A denote the amount borrowed by Jamal from two banks. And row 2 of matrix Adenote the amount borrowed by Stephanie from two banks. Matrix B will be a column vector with monthly interest rates provided by banks.

Therefore,

  $A=\left[\begin{array}{cc}4000& 3000\\ 2500& 3800\end{array}\right]\text{ and }B=\left[\begin{array}{c}0.011\\ 0.006\end{array}\right]$

  $\left(b\right)$

To determine

To calculate: The value of $AB$ .

Answer to Problem 84AYU

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

Explanation of Solution

Given information:

The amount borrowed by Jamal and Stephanie from two banks for school loan and monthly interest rate of banks are provided below,

  $\begin{array}{ccc}& \text{Lender 1}& \text{Lender 2}\\ \text{Jamal}& \$4000& \$3000\\ \text{Stephanie}& \$2500& \$3800\end{array}\begin{array}{cc}& \text{Monthly Interest Rate}\\ \text{Lender 1}& 0.011\left(1.1\%\right)\\ \text{Lender 2}& 0.006\left(0.6\%\right)\end{array}$

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 matrixB.

Calculation:

Consider the amount borrowed by Jamal and Stephanie from two banks for school loan and monthly interest rate of banks are provided below,

  $\begin{array}{ccc}& \text{Lender 1}& \text{Lender 2}\\ \text{Jamal}& \$4000& \$3000\\ \text{Stephanie}& \$2500& \$3800\end{array}\begin{array}{cc}& \text{Monthly Interest Rate}\\ \text{Lender 1}& 0.011\left(1.1\%\right)\\ \text{Lender 2}& 0.006\left(0.6\%\right)\end{array}$

Let matrix A that represent the amounts borrowed by each student and matrix B that represent the monthly interest rates.

So, row 1 of matrix A denote the amount borrowed by Jamal from two banks. And row 2 of matrix A denote the amount borrowed by Stephanie from two banks. Matrix B will be a column vector with monthly interest rates provided by banks.

Therefore,

  $A=\left[\begin{array}{cc}4000& 3000\\ 2500& 3800\end{array}\right]\text{ and }B=\left[\begin{array}{c}0.011\\ 0.006\end{array}\right]$

Now,

  $\begin{array}{c}AB=\left[\begin{array}{cc}4000& 3000\\ 2500& 3800\end{array}\right]\left[\begin{array}{c}0.011\\ 0.006\end{array}\right]\\ =\left[\begin{array}{c}44+18\\ 27.5+22.8\end{array}\right]\\ =\left[\begin{array}{c}62\\ 50.3\end{array}\right]\end{array}$

Monthly compound interest for Jamal is 62 and for Stephanie is $50.3$ .

  $\left(c\right)$

To determine

To calculate: The value of $A\left(C+B\right)$ .

Answer to Problem 84AYU

The value of $A\left(C+B\right)$ is $\left[\begin{array}{c}7062\\ 6350.3\end{array}\right]$ .

Explanation of Solution

Given information:

The amount borrowed by Jamal and Stephanie from two banks for school loan and monthly interest rate of banks are provided below,

  $\begin{array}{ccc}& \text{Lender 1}& \text{Lender 2}\\ \text{Jamal}& \$4000& \$3000\\ \text{Stephanie}& \$2500& \$3800\end{array}\begin{array}{cc}& \text{Monthly Interest Rate}\\ \text{Lender 1}& 0.011\left(1.1\%\right)\\ \text{Lender 2}& 0.006\left(0.6\%\right)\end{array}$

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 matrixB.

Calculation:

Consider the amount borrowed by Jamal and Stephanie from two banks for school loan and monthly interest rate of banks are provided below,

  $\begin{array}{ccc}& \text{Lender 1}& \text{Lender 2}\\ \text{Jamal}& \$4000& \$3000\\ \text{Stephanie}& \$2500& \$3800\end{array}\begin{array}{cc}& \text{Monthly Interest Rate}\\ \text{Lender 1}& 0.011\left(1.1\%\right)\\ \text{Lender 2}& 0.006\left(0.6\%\right)\end{array}$

Let matrix A that represent the amounts borrowed by each student and matrix B that represent the monthly interest rates.

So, row 1 of matrix A denote the amount borrowed by Jamal from two banks. And row 2 of matrix A denote the amount borrowed by Stephanie from two banks. Matrix B will be a column vector with monthly interest rates provided by banks.

Therefore,

  $A=\left[\begin{array}{cc}4000& 3000\\ 2500& 3800\end{array}\right]\text{ and }B=\left[\begin{array}{c}0.011\\ 0.006\end{array}\right]$

Now, it is provided that $C=\left[\begin{array}{c}1\\ 1\end{array}\right]$

  $\begin{array}{c}A\left(C+B\right)=\left[\begin{array}{cc}4000& 3000\\ 2500& 3800\end{array}\right]\left(\left[\begin{array}{c}1\\ 1\end{array}\right]+\left[\begin{array}{c}0.011\\ 0.006\end{array}\right]\right)\\ =\left[\begin{array}{cc}4000& 3000\\ 2500& 3800\end{array}\right]\left[\begin{array}{c}1.011\\ 1.006\end{array}\right]\\ =\left[\begin{array}{c}7062\\ 6350.3\end{array}\right]\end{array}$

Monthly compound interest for Jamal is $7062$ and for Stephanie is $6350.3$ .