Example 11 from Algebra and Trigonometry, 11.6 Solving Systems with Gaussian Elimination

Applying 3 × 3 Matrices to Finance

Ava invests a total of $10,000 in three accounts, one paying 5% interest, another paying 8% interest, and the third paying 9% interest. The annual interest earned on the three investments last year was $770. The amount invested at 9% was twice the amount invested at 5%. How much was invested at each rate?

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Solution
We have a system of three equations in three variables. Let $x$ be the amount invested at 5% interest, let $y$ be the amount invested at 8% interest, and let $z$ be the amount invested at 9% interest. Thus,

$\begin{array}{l} x + y + z = 10, 000 \\ 0.05 x + 0.08 y + 0.09 z = 770 \\ 2 x - z = 0 \end{array}$
As a matrix, we have

$[\begin{array}{r} 1 & 1 & 1 \\ 0.05 & 0.08 & 0.09 \\ 2 & 0 & −1 \end{array} | \begin{array}{r} 10, 000 \\ 770 \\ 0 \end{array}]$
Now, we perform Gaussian elimination to achieve row-echelon form.

$\begin{array}{l} \begin{array}{l} −0.05 R_{1} + R_{2} = R_{2} \to [\begin{array}{r} 1 & 1 & 1 \\ 0 & 0.03 & 0.04 \\ 2 & 0 & −1 \end{array} | \begin{array}{r} 10,000 \\ 270 \\ 0 \end{array}] \end{array} \\ −2 R_{1} + R_{3} = R_{3} \to [\begin{array}{r} 1 & 1 & 1 \\ 0 & 0.03 & 0.04 \\ 0 & −2 & −3 \end{array} | \begin{array}{r} 10,000 \\ 270 \\ −20,000 \end{array}] \\ \frac{1}{0.03} R_{2} = R_{2} \to [\begin{array}{r} 0 & 1 & 1 \\ 0 & 1 & \frac{4}{3} \\ 0 & −2 & −3 \end{array} | \begin{array}{r} 10,000 \\ 9,000 \\ −20,000 \end{array}] \\ 2 R_{2} + R_{3} = R_{3} \to [\begin{array}{r} 1 & 1 & 1 \\ 0 & 1 & \frac{4}{3} \\ 0 & 0 & - \frac{1}{3} \end{array} | \begin{array}{r} 10,000 \\ 9,000 \\ −2,000 \end{array}] \end{array}$
The third row tells us $- \frac{1}{3} z = −2,000;$ thus $z = 6,000.$
The second row tells us $y + \frac{4}{3} z = 9,000.$ Substituting $z = 6,000,$ we get
$\begin{array}{r} y + \frac{4}{3} (6,000) = 9,000 \\ y + 8,000 = 9,000 \\ y = 1,000 \end{array}$
The first row tells us $x + y + z = 10, 000.$ Substituting $y = 1, 000$ and $z = 6, 000,$ we get
$\begin{array}{l} x + 1, 000 + 6, 000 = 10,000 \\ x = 3,000 \end{array}$
The answer is $3,000 invested at 5% interest, $1,000 invested at 8%, and $6,000 invested at 9% interest.

Your turn

The book's Try It right after this example: same idea, new numbers. Only the answer is given.

A small shoe company took out a loan of $1,500,000 to expand their inventory. Part of the money was borrowed at 7%, part was borrowed at 8%, and part was borrowed at 10%. The amount borrowed at 10% was four times the amount borrowed at 7%, and the annual interest on all three loans was $130,500. Use matrices to find the amount borrowed at each rate.

How did it go?

Your turn

The book's Try It right after this example: same idea, new numbers. Only the answer is given.

Solution

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Solution

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