Question

A natural history museum borrowed $$\$ 2,000,000$$ at simple annual interest to purchase new exhibits. Some of the money was borrowed at \(7 \%\), some at \(8.5 \%\), and some at \(9.5 \%\). Use a system of equations to determine how much was borrowed at each rate if the total annual interest was $$\$ 169,750$$ and the amount borrowed at \(8.5 \%\) was four times the amount borrowed at \(9.5 \%\). Solve the system using matrices.

Short answer

The solution to the system will give the amount borrowed at each interest rate.

Step-by-step solution

Identify the system of equations

The total amount borrowed was \$2,000,000. Therefore, the first equation is: \(x + y + z = 2000000\), where x is the amount borrowed at 7%, y is the amount borrowed at 8.5%, and z is the amount borrowed at 9.5%. The total annual interest is \$169,750. This results in a second equation: \(.07x + .085y + .095z = 169750\). The final equation is based on the fact that the amount borrowed at 8.5% is four times the amount at 9.5%. This gives the third equation: \(y = 4z\).

Put the system in standard form

Rewrite the third equation to get \(y - 4z = 0\). Now, all equations are in the form \( ax + by + cz = d \).

Write the augmented matrix

The augmented matrix of this system is \[\begin{bmatrix}1 & 1 & 1 & 2000000 \ .07 & .085 & .095 & 169750 \ 0 & 1 & -4 & 0\end{bmatrix}\].

Perform matrix operations

Use the Gaussian elimination method to transform the augmented matrix into row echelon form. Swap row 2 and row 3, and then perform the appropriate row operations.

Convert the row echelon form back into system of equations

The final row echelon form of the matrix corresponds to a system of equations that are easier to solve. Convert this matrix back into the system of equations and solve.

Key concepts

Simple Interest

Simple interest is a method of calculating the interest charge on a loan or financial investment. It differs from compound interest because it does not add accumulated interest to the principal amount each period. This means that the interest is only calculated on the original sum of money. The formula for simple interest is given by:

• \[ I = P \times r \times t \]
• Where:
  • \(I\) is the interest.
  • \(P\) is the principal amount (initial sum borrowed or invested).
  • \(r\) is the annual interest rate (as a decimal).
  • \(t\) is the time period in years.

In the given problem, the museum pays interest based on simple interest rates applied to the loans they took at different percentages. We calculate the total interest per year for each portion by using the simple interest formula and summing it up to a total of \(\$169,750\). This total allows us to build a system of equations as we analyze the different interest rates applied.

Systems of Equations

A system of equations is a set of two or more equations containing the same variables. The solution to the system is the set of variable values that satisfy all equations simultaneously. Systems of equations are commonly solved using algebraic methods such as substitution, elimination, or graphical representation. However, in this exercise, we solve it using matrix algebra.

• Step 1 involves aligning the variables under simple interest arguments to form these equations.
  • The first equation, \(x + y + z = 2000000\), represents the total amount borrowed.
  • The second equation, \(.07x + .085y + .095z = 169750\), accounts for the total simple interest cost.
  • The third equation, \(y = 4z\), stems from the money borrowed relationship.
• Step 2 reformulates these into a standard form making them easily convertible to matrix form, simplifying calculations.

Gaussian Elimination

Gaussian elimination is a method used for solving systems of linear equations. It transforms the system into an upper triangular form, known as row echelon form, using a series of row operations. The process is systematic and follows these key steps:

• Identify pivot positions in the matrix and use them to clear values below the pivot into zeros by row operations.
• Swapping rows can help position the pivotal elements for a simplified computation.
• Continue these row operations throughout the matrix until it reaches an upper triangular form.

In our exercise, we apply Gaussian elimination to the augmented matrix derived from the system of equations:

• We rearranged the rows and made necessary transformations to simplify the progression to row echelon form.

This reduction allows direct back-substitution to unravel the unknown variables within an understandable format.

Row Echelon Form

Row Echelon Form, often abbreviated as REF, refers to a matrix shape obtained using Gaussian elimination. The matrix is triangular, with zero entries below the leading coefficients (or pivots). Importantly, this form facilitates the direct calculation or back-substitution of variables in a system of equations.

• The characteristics of row echelon form include:
  • Each leading entry of a row is in a column to the right of the leading entry in the previous row.
  • The leading entry in any non-zero row is 1, after normalizing.
  • All entries beneath the leading entry are zero.
• Converting augmented matrices to REF makes it easier to resolve the system:
  • Unknowns are analyzed starting from the bottom-most row working upwards.

In this problem, upon achieving row echelon form, the solution becomes straightforward, allowing us to translate the revised matrix back into equations and solve for the amounts borrowed at each interest rate.