Solving System of Equations using Matrices: Apply It 1

Giselle Miller

Solving System of Equations using Matrices: Apply It 1

Write the augmented matrix for a system of equations
Write the system of equations from an augmented matrix
Perform row operations on a matrix
Use matrix operations and row reductions to find solutions to systems of linear equations

Applications of Systems of Equations

Now we will turn to the applications for which systems of equations are used. In the next example we determine how much money was invested at two different rates given the sum of the interest earned by both accounts.

Setting up a system of equations in the following examples uses the same ideas you have used before to write a system of linear equations to model a situation. The only difference is that now you’ll use matrices and Gaussian elimination to solve the system.

Carolyn invests a total of [latex]$12,000[/latex] in two municipal bonds, one paying [latex]10.5\%[/latex] interest and the other paying [latex]12\%[/latex] interest. The annual interest earned on the two investments last year was [latex]$1,335[/latex]. How much was invested at each rate?

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

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

[latex]\begin{array}{l}\text{ }x+y+z=10,000\hfill \\ 0.05x+0.08y+0.09z=770\hfill \\ \text{ }2x-z=0\hfill \end{array}[/latex]

As a matrix, we have

[latex]\left[\begin{array}{ccc|c}\hfill 1& \hfill 1& \hfill 1& \hfill 10,000\\ \hfill 0.05& \hfill 0.08& \hfill 0.09& \hfill 770\\ \hfill 2& \hfill 0& \hfill -1& \hfill 0\end{array}\right][/latex]

Now, we perform Gaussian elimination to achieve row-echelon form.

[latex]\begin{array}{l}\begin{array}{l}\hfill \\ -0.05{R}_{1}+{R}_{2}={R}_{2}\to \left[\begin{array}{ccc|c}\hfill 1& \hfill 1& \hfill 1& \hfill 10,000\\ \hfill 0& \hfill 0.03& \hfill 0.04& \hfill 270\\ \hfill 2& \hfill 0& \hfill -1& \hfill 0\end{array}\right]\hfill \end{array}\hfill \\ -2{R}_{1}+{R}_{3}={R}_{3}\to \left[\begin{array}{ccc|c}\hfill 1& \hfill 1& \hfill 1& \hfill 10,000\\ \hfill 0& \hfill 0.03& \hfill 0.04& \hfill 270\\ \hfill 0& -2& \hfill -3& \hfill -20,000 \end{array}\right]\hfill \\ \frac{1}{0.03}{R}_{2}={R}_{2}\to \left[\begin{array}{ccc|c}\hfill 1& \hfill 1& \hfill 1& \hfill 10,000\\ \hfill 0& \hfill 1& \hfill \frac{4}{3}& \hfill 9,000\\ \hfill 0& -2& \hfill -3& \hfill -20,000 \end{array}\right]\hfill \\ 2{R}_{2}+{R}_{3}={R}_{3}\to \left[\begin{array}{ccc|c}\hfill 1& \hfill 1& \hfill 1& \hfill 10,000\\ \hfill 0& \hfill 1& \hfill \frac{4}{3}& \hfill 9,000\\ \hfill 0& 0& \hfill -\frac{1}{3}& \hfill -2,000 \end{array}\right]\hfill \end{array}[/latex]

The third row tells us [latex]-\frac{1}{3}z=-2,000[/latex]; thus [latex]z=6,000[/latex].

The second row tells us [latex]y+\frac{4}{3}z=9,000[/latex].

Substituting [latex]z=6,000[/latex], we get

[latex]\begin{array}{r}\hfill y+\frac{4}{3}\left(6,000\right)=9,000\\ \hfill y+8,000=9,000\\ \hfill y=1,000\end{array}[/latex]

The first row tells us [latex]x+y+z=10,000[/latex]. Substituting [latex]y=1,000[/latex] and [latex]z=6,000[/latex], we get
[latex]\begin{array}{l}x+1,000+6,000=10,000\hfill \\ \text{ }x=3,000\text{ }\hfill \end{array}[/latex]

The answer is [latex]$3,000[/latex] invested at [latex]5\%[/latex] interest, [latex]$1,000[/latex] invested at [latex]8\%[/latex], and [latex]$6,000[/latex] invested at [latex]9\%[/latex] interest.

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