A city zoo borrowed $2,000,000 at simple annual interest to construct a breeding facility.Some of the money was borrowed at 9%, some at 11%, and some at 14%.Use a system of equations to determine how much was borrowed at 9% if the total annual interest was $216,000 and the amount borrowed at 9% was twice the amount borrowed at 14%.Solve the system using matrices.
A) $400,000 was borrowed at 9%
B) $800,000 was borrowed at 9%
C) $300,000 was borrowed at 9%
D) $1,600,000 was borrowed at 9%
E) $1,700,000 was borrowed at 9%

Question

Quizplus · Accepted Answer

Let x, y, and z represent the amounts borrowed at 9%, 11%, and 14% respectively. The given conditions can be translated into the following system of equations:1. $x + y + z = 2,000,000$ (The total amount borrowed)2. $0.09x + 0.11y + 0.14z = 216,000$ (The total interest)3. $x = 2z$ (The amount at 9% is twice the amount at 14%)Using matrices to solve this system, we first express it in matrix form as AX = B, where A is the coefficient matrix, X is the variable matrix, and B is the constant matrix.$$A = \begin{bmatrix} 1 & 1 & 1 \ 0.09 & 0.11 & 0.14 \ 1 & 0 & -2 \end{bmatrix}, X = \begin{bmatrix} x \ y \ z \end{bmatrix}, B = \begin{bmatrix} 2000000 \ 216000 \ 0 \end{bmatrix}$$Solving this system (either by using matrix inversion, row reduction, or any other suitable method), we find that $x = 800,000$, which means $800,000 was borrowed at 9%.

A City Zoo Borrowed $2,000,000 at Simple Annual Interest to Construct