Q. 84 - Precalculus Enhanced with Graphing Utilities

Question

School Loan Interest Jamal and Stephanie each have school loans issued from the same two banks. The amounts borrowed and the monthly interest rates are given next (interest is compounded monthly):

(a) Write a matrix A for the amounts borrowed by each student and a matrix B for the monthly interest rates.

(b) Compute AB and interpret the results.

(c) Let $C=\left[\begin{array}{c}1\\ 1\end{array}\right]$. Compute $A(C+B)$and interpret the results.

Step-by-Step Solution

Answer

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

(b)  $AB=\left[\begin{array}{c}62\\ 50.3\end{array}\right]$

(c) $A(C+B)=\left[\begin{array}{c}7062\\ 6350.3\end{array}\right]$

 

Step 1. Given information

Jamal and stephanie have loan at leader 1 =$\$4000 and \$2500$ with rate of interest $0.011\%(1.1\%)$

Jamal and stephanie have loan at leader 2 =$\$3000 and \$3800$ with rate of interest  $0.006\%(0.6\%)$

Step 2. (a) Write a matrix A for the amounts borrowed by each student and a matrix B for the monthly interest rates.

The amount borrowed by jamal and stephanie can be written in the matrix form

$A=\left[\begin{array}{cc}4000& 3000\\ 2500& 3800\end{array}\right]$

Again the monthly interest rates can also be written in the matrix form

$B=\left[\begin{array}{c}0.011\\ 0.006\end{array}\right]$

Step 3. (b) Compute AB and interpret the results.

We calculate the matrix AB by multiplaying the two matrices

$$\begin{gathered} AB=\left[\begin{array}{cc}4000& 3000\\ 2500& 3800\end{array}\right]\left[\begin{array}{c}0.011\\ 0.006\end{array}\right] \\ =\left[\begin{array}{c}4000(0.011)+3000(0.006)\\ 2500(0.011)+3800(0.006)\end{array}\right] \\ AB=\left[\begin{array}{c}62\\ 50.3\end{array}\right] \end{gathered}$$

Thus,after multiplaying the two matrices,we get the amount of intrest for each student that is,$\$62$for jamal and $\$50.30$ for stephanie

Step 4. Let C = 1 1 .Compute A ( C + B ) and interpret the result

We calculate the matrix by multiplaying $A(C+B)=AC+AB$

First find the value of AC by multiplaying the matrices

  $$\begin{gathered} AC=\left[\begin{array}{cc}4000& 3000\\ 2500& 3800\end{array}\right]\left[\begin{array}{c}1\\ 1\end{array}\right] \\ =\left[\begin{array}{c}4000\left(1\right)[phone]\mathrm{}\mathrm{}\\ \mathrm{}\left(1\right)+3800\left(1\right)\end{array}\right] \\ AC=\left[\begin{array}{c}7000\\ 6300\end{array}\right] \end{gathered}$$

Now, we have the value of $AB=\left[\begin{array}{c}62\\ 50.3\end{array}\right] \left[from\right(b\left)\right]$

Now add $AC and AB$to get the value of $A(C+B)$

$$\begin{gathered} AC+AB=\left[\begin{array}{c}7000\\ 6300\end{array}\right]+\left[\begin{array}{c}62\\ 50.3\end{array}\right] \\ =\left[\begin{array}{c}7000+62\\ 6300+50.3\end{array}\right] \\ =\left[\begin{array}{c}7062\\ 6350.3\end{array}\right] \end{gathered}$$