Chapter 8.1, Problem 102E

To determine

The investment on Breeding Facility of a zoo.

Answer to Problem 102E

  $\begin{array}{l}\boxed{\$1200,000}\\ \boxed{\$200,000}\\ \boxed{\$600,000}\end{array}$

Explanation of Solution

Given information:

A city zoo borrowed $\$\text{2},000,000$ at simple annual interest to construct a breeding facility. Some of the money was borrowed at $8\%$ some at $9\%$ , and some at $12\%$ . Use a system of linear equations to determine how much was borrowed at each rate given that the total annual interest was $\$\text{186},000$ and the amount borrowed at $8\%$ was twice the amount borrowed at $12\%$ . Solve the system of linear equations using matrices.

Calculation: let $x,y,z$ amount is borrowed at interest rate of $8\%,9\%$ and $12\%$

Interest rate be considered as variable,

Total amount is $\$\text{2},000,000$

Total interest is $\$\text{186},000$

  $x+y+z=\text{2},000,000$

  $0.08x+0.09y+0.12z=186000$

It is given that amount borrowed at $8\%$ was twice the amount borrowed at $12\%$ .

So $x=2z$

The system of equations is,

  $\{\begin{array}{l}x+y+z=\text{2},000,000\\ 0.08x+0.09y+0.12z=186000\\ x-2z=0\end{array}$

The augmented matrix of above system of equations,

  $\left[\begin{array}{ccccc}1& 1& 1& ⋮& 2000,000\\ 0.08& 0.09& 0.12& ⋮& 186,000\\ 1& 0& -2& ⋮& 0\end{array}\right]$

Now apply Gauss-Jordan elimination method.

Interchange row one and three,

  $R_1\iff R_3$ $\left[\begin{array}{ccccc}1& 0& -2& ⋮& 0\\ 0.08& 0.09& 0.12& ⋮& 186,000\\ 1& 1& 1& ⋮& 2000,000\end{array}\right]$

  $\begin{array}{l}-0.08R_1+R_2\to R_1\\ R_3-R_1\to R_3\end{array}$ $\left[\begin{array}{ccccc}1& 0& -2& ⋮& 0\\ 0& 0.09& 0.28& ⋮& 186,000\\ 0& 1& 3& ⋮& 2000,000\end{array}\right]$

  $\begin{array}{l}{R}_2\iff R_3\\ R_3-0.09R_2\to R_3\end{array}$ $\left[\begin{array}{ccccc}1& 0& -2& ⋮& 0\\ 0& 1& 3& ⋮& 2000,000\\ 0& 0& 0.01& ⋮& 6000\end{array}\right]$

  $1/0.01R_3\to R_3$ $\left[\begin{array}{ccccc}1& 0& -2& ⋮& 0\\ 0& 1& 3& ⋮& 2000,000\\ 0& 0& 1& ⋮& 600,000\end{array}\right]$

  $\begin{array}{l}{R}_1+2R_3\to R_1\\ R_2-3R_3\to R_2\end{array}$ $\left[\begin{array}{ccccc}1& 0& 0& ⋮& 1200,000\\ 0& 1& 0& ⋮& 200,000\\ 0& 0& 1& ⋮& 600,000\end{array}\right]$

Hence,

  $\begin{array}{l}x=1200,000\\ y=200,000\\ z=600,000\end{array}$

The amount invested at interest rate $8\%$ is $\boxed{\$1200,000}$

The amount invested at interest rate $9\%$ is $\boxed{\$200,000}$

The amount invested at interest rate $12\%$ is $\boxed{\$600,000}$