Question

A city zoo borrowed $$\$ 2,000,000$$ 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 equations to determine how much was borrowed at each rate if the total annual interest was $$\$ 186,000$$ and the amount borrowed at \(8 \%\) was twice the amount borrowed at \(12 \%\). Solve the system using matrices.

Short answer

The zoo borrowed \$ 800,000 at \( 8\% \), \$ 700,000 at \( 9\% \), and \$ 500,000 at \( 12\% \).

Step-by-step solution

Form Equations

Create three equations that represent the financial conditions. Let's say \( x \), \( y \) and \( z \) represent the amount of money borrowed at \( 8\% \), \( 9\% \) and \( 12\% \) respectively. So, 1) \( x + y + z = 2000000 \), 2) \( 0.08x +0.09y +0.12z = 186000 \), 3) \( x = 2z \).

Convert System of Equations into Matrix Form

The equation system can be represented as a 2D matrix: \[ \begin{bmatrix} 1 & 1 & 1 \ 0.08 & 0.09 & 0.12 \ 1 & 0 & -2 \ end{bmatrix} \begin{bmatrix} x \ y \ z \ end{bmatrix} = \begin{bmatrix} 2000000 \ 186000 \ 0 \ end{bmatrix} \]

Solve Matrices

Utilize matrix operations or a tool that can solve linear equation systems like the Gauss-Jordan elimination method. Based on the Gauss-Jordan elimination, the amounts borrowed at each interest rate would be \( x = \$ 800,000 \), \( y = \$ 700,000 \) and \( z = \$ 500,000 \).

Interpret the Results

These results mean that the zoo borrowed \$ 800,000 at \( 8\% \), \$ 700,000 at \( 9\% \), and \$ 500,000 at \( 12\% \).

Key concepts

System of Equations

A system of equations is a collection of two or more equations with a common set of variables. In everyday problems, like the one in the city zoo example, it helps solve for unknowns in cases where multiple conditions are applied. The zoo needed to figure out the amount of money borrowed at different interest rates so that the total borrowed amount and the total annual interest matched the conditions given in the problem.
Here's how to set up a system of equations:

• Identify the variables, like how we used \( x \), \( y \), and \( z \) here for amounts borrowed at different interest rates.
• Formulate equations using the conditions given, such as total borrowed amount and interest calculations.
• Use these equations to find the values of these variables that satisfy all equations simultaneously.

In our example:

• The sum of all borrowed amounts equaled the total amount, giving equation 1: \( x + y + z = 2000000 \).
• The sum of the interest of each loan needed to equal the total interest, leading to equation 2: \( 0.08x + 0.09y + 0.12z = 186000 \).
• The amount borrowed at 8% was double the amount borrowed at 12%, forming equation 3: \( x = 2z \).

Matrix Operations

Matrix operations are mathematical calculations that involve matrices, which are rectangular arrays of numbers. They are used to solve systems of equations efficiently. In this exercise, we translated the system of linear equations into a matrix form, which is often more manageable for computations.
The three main components needed:

• The coefficients of the variables form the **coefficient matrix**.
• The variables form the **variable matrix**.
• The constants on the right side of the equations form the **constant matrix**.

For example, the system of equations from the zoo problem was converted into the following matrix form:\[\begin{bmatrix} 1 & 1 & 1 \ 0.08 & 0.09 & 0.12 \ 1 & 0 & -2 \end{bmatrix}\begin{bmatrix} x \ y \ z \end{bmatrix} =\begin{bmatrix} 2000000 \ 186000 \ 0 \end{bmatrix}\]Once in matrix form, we can perform operations such as addition, subtraction, and, most importantly, Gauss-Jordan elimination to solve for variable values. It transforms the matrix to make solutions clearer.

Gauss-Jordan Elimination

Gauss-Jordan elimination is a systematic method for solving linear systems of equations by performing operations on their corresponding matrix. The aim is to convert the matrix into a row-echelon form and then further into reduced row-echelon form, where solutions to the system become obvious.
Key steps include:

• Use row operations to simplify and transform the matrix, aiming for zeros below and above the diagonal (leading) elements in the matrix.
• Normalize the leading coefficients to 1 by dividing the row by the coefficient value.
• Continue simplifying until each row has a leading 1 and all other entries in its column are zero.

In our example:

• We started with matrix form from the previous section.
• Through a series of row swaps, scaling, and adding or subtracting rows, the matrix is transformed until the solution is straightforward to read, shown in the resulting form where \( x = \\( 800,000 \), \( y = \\) 700,000 \), and \( z = \$ 500,000 \).

This method is especially powerful for larger systems or when using a computational tool, providing a clear path to the solution.