Learning Objectives
- Use Gaussian elimination to solve a systems of equations represented as an augmented matrix
- Interpret the solution to a system of equations represented as an augmented matrix
We have seen how to write a system of equations with an augmented matrix, and then how to use row operations and back-substitution to obtain row-echelon form. Now we will use Gaussian Elimination as a tool for solving a system written as an augmented matrix. In our first example, we will show you the process for using Gaussian Elimination on a system of two equations in two variables.
Example: Solving a 2 X 2 System by Gaussian Elimination
Solve the given system by Gaussian elimination.
[latex]\begin{array}{l}2x+3y=6\hfill \\ \text{ }x-y=\frac{1}{2}\hfill \end{array}[/latex]
Solution
First, we write this as an augmented matrix.
[latex]\left[\begin{array}{rr}\hfill 2& \hfill 3\\ \hfill 1& \hfill -1\end{array}\text{ }|\text{ }\begin{array}{r}\hfill 6\\ \hfill \frac{1}{2}\end{array}\right][/latex]
We want a 1 in row 1, column 1. This can be accomplished by interchanging row 1 and row 2.
[latex]{R}_{1}\leftrightarrow {R}_{2}\to \left[\begin{array}{rrr}\hfill 1& \hfill -1& \hfill \\ \hfill 2& \hfill 3& \hfill \end{array}|\begin{array}{rr}\hfill & \hfill \frac{1}{2}\\ \hfill & \hfill 6\end{array}\right][/latex]
We now have a 1 as the first entry in row 1, column 1. Now let’s obtain a 0 in row 2, column 1. This can be accomplished by multiplying row 1 by [latex]-2[/latex], and then adding the result to row 2.
[latex]-2{R}_{1}+{R}_{2}={R}_{2}\to \left[\begin{array}{rrr}\hfill 1& \hfill -1& \hfill \\ \hfill 0& \hfill 5& \hfill \end{array}|\begin{array}{rr}\hfill & \hfill \frac{1}{2}\\ \hfill & \hfill 5\end{array}\right][/latex]
We only have one more step, to multiply row 2 by [latex]\frac{1}{5}[/latex].
[latex]\frac{1}{5}{R}_{2}={R}_{2}\to \left[\begin{array}{rrr}\hfill 1& \hfill -1& \hfill \\ \hfill 0& \hfill 1& \hfill \end{array}|\begin{array}{cc}& \frac{1}{2}\\ & 1\end{array}\right][/latex]
Use back-substitution. The second row of the matrix represents [latex]y=1[/latex]. Back-substitute [latex]y=1[/latex] into the first equation.
[latex]\begin{array}{l}x-\left(1\right)=\frac{1}{2}\hfill \\ \text{ }x=\frac{3}{2}\hfill \end{array}[/latex]
The solution is the point [latex]\left(\frac{3}{2},1\right)[/latex].
Try It
Solve the given system by Gaussian elimination.
[latex]\begin{array}{l}4x+3y=11\hfill \\ \text{ }\text{}\text{}x - 3y=-1\hfill \end{array}[/latex]
Solution
[latex]\left(2,1\right)[/latex]
In our next example, we will solve a system of two equations in two variables that is dependent. Recall that a dependent system has an infinite number of solutions and the result of row operations on it’s augmented matrix will be an equation such as [latex]0=0[/latex]. We also review writing the general solution to a dependent system.
Example: Solving a Dependent System
Solve the system of equations.
[latex]\begin{array}{l}3x+4y=12\\ 6x+8y=24\end{array}[/latex]
Solution
Perform
row operations on the augmented matrix to try and achieve
row-echelon form.
[latex]A=\left[\begin{array}{llll}3\hfill & \hfill & 4\hfill & \hfill \\ 6\hfill & \hfill & 8\hfill & \hfill \end{array}|\begin{array}{ll}\hfill & 12\hfill \\ \hfill & 24\hfill \end{array}\right][/latex]
[latex]\begin{array}{l}\hfill \\ \begin{array}{l}-\frac{1}{2}{R}_{2}+{R}_{1}={R}_{1}\to \left[\begin{array}{llll}0\hfill & \hfill & 0\hfill & \hfill \\ 6\hfill & \hfill & 8\hfill & \hfill \end{array}|\begin{array}{ll}\hfill & 0\hfill \\ \hfill & 24\hfill \end{array}\right]\hfill \\ {R}_{1}\leftrightarrow {R}_{2}\to \left[\begin{array}{llll}6\hfill & \hfill & 8\hfill & \hfill \\ 0\hfill & \hfill & 0\hfill & \hfill \end{array}|\begin{array}{ll}\hfill & 24\hfill \\ \hfill & 0\hfill \end{array}\right]\hfill \end{array}\hfill \end{array}[/latex]
The matrix ends up with all zeros in the last row: [latex]0y=0[/latex]. Thus, there are an infinite number of solutions and the system is classified as dependent. To find the generic solution, return to one of the original equations and solve for [latex]y[/latex].
[latex]\begin{array}{l}3x+4y=12\hfill \\ \text{ }4y=12 - 3x\hfill \\ \text{ }y=3-\frac{3}{4}x\hfill \end{array}[/latex]
So the solution to this system is [latex]\left(x,3-\frac{3}{4}x\right)[/latex].
Now, we will take row-echelon form a step farther to solve a 3 by 3 system of linear equations. The general idea is to eliminate all but one variable using row operations and then back-substitute to solve for the other variables.
Example: Solving a System of Linear Equations Using Matrices
Solve the system of linear equations using matrices.
[latex]\begin{array}{c}\begin{array}{l}\hfill \\ \hfill \\ x-y+z=8\hfill \end{array}\\ 2x+3y-z=-2\\ 3x - 2y - 9z=9\end{array}[/latex]
Solution
First, we write the augmented matrix.
[latex]\left[\begin{array}{rrr}\hfill 1& \hfill -1& \hfill 1\\ \hfill 2& \hfill 3& \hfill -1\\ \hfill 3& \hfill -2& \hfill -9\end{array}\text{ }|\text{ }\begin{array}{r}\hfill 8\\ \hfill -2\\ \hfill 9\end{array}\right][/latex]
Next, we perform row operations to obtain row-echelon form.
[latex]\begin{array}{rrrrr}\hfill -2{R}_{1}+{R}_{2}={R}_{2}\to \left[\begin{array}{rrrrrr}\hfill 1& \hfill & \hfill -1& \hfill & \hfill 1& \hfill \\ \hfill 0& \hfill & \hfill 5& \hfill & \hfill -3& \hfill \\ \hfill 3& \hfill & \hfill -2& \hfill & \hfill -9& \hfill \end{array}|\begin{array}{rr}\hfill & \hfill 8\\ \hfill & \hfill -18\\ \hfill & \hfill 9\end{array}\right]& \hfill & \hfill & \hfill & \hfill -3{R}_{1}+{R}_{3}={R}_{3}\to \left[\begin{array}{rrrrrr}\hfill 1& \hfill & \hfill -1& \hfill & \hfill 1& \hfill \\ \hfill 0& \hfill & \hfill 5& \hfill & \hfill -3& \hfill \\ \hfill 0& \hfill & \hfill 1& \hfill & \hfill -12& \hfill \end{array}|\begin{array}{rr}\hfill & \hfill 8\\ \hfill & \hfill -18\\ \hfill & \hfill -15\end{array}\right]\end{array}[/latex]
The easiest way to obtain a 1 in row 2 of column 1 is to interchange [latex]{R}_{2}[/latex] and [latex]{R}_{3}[/latex].
[latex]\text{Interchange}{R}_{2}\text{and}{R}_{3}\to \left[\begin{array}{rrrrrrr}\hfill 1& \hfill & \hfill -1& \hfill & \hfill 1& \hfill & \hfill 8\\ \hfill 0& \hfill & \hfill 1& \hfill & \hfill -12& \hfill & \hfill -15\\ \hfill 0& \hfill & \hfill 5& \hfill & \hfill -3& \hfill & \hfill -18\end{array}\right][/latex]
Then
[latex]\begin{array}{l}\\ \begin{array}{rrrrr}\hfill -5{R}_{2}+{R}_{3}={R}_{3}\to \left[\begin{array}{rrrrrr}\hfill 1& \hfill & \hfill -1& \hfill & \hfill 1& \hfill \\ \hfill 0& \hfill & \hfill 1& \hfill & \hfill -12& \hfill \\ \hfill 0& \hfill & \hfill 0& \hfill & \hfill 57& \hfill \end{array}|\begin{array}{rr}\hfill & \hfill 8\\ \hfill & \hfill -15\\ \hfill & \hfill 57\end{array}\right]& \hfill & \hfill & \hfill & \hfill -\frac{1}{57}{R}_{3}={R}_{3}\to \left[\begin{array}{rrrrrr}\hfill 1& \hfill & \hfill -1& \hfill & \hfill 1& \hfill \\ \hfill 0& \hfill & \hfill 1& \hfill & \hfill -12& \hfill \\ \hfill 0& \hfill & \hfill 0& \hfill & \hfill 1& \hfill \end{array}|\begin{array}{rr}\hfill & \hfill 8\\ \hfill & \hfill -15\\ \hfill & \hfill 1\end{array}\right]\end{array}\end{array}[/latex]
The last matrix represents the equivalent system.
[latex]\begin{array}{l}\text{ }x-y+z=8\hfill \\ \text{ }y - 12z=-15\hfill \\ \text{ }z=1\hfill \end{array}[/latex]
Using back-substitution, we obtain the solution as [latex]\left(4,-3,1\right)[/latex].
Recall that there are three possible outcomes for solutions to linear systems. In the previous example, the solution [latex]\left(4,-3,1\right)[/latex] represents a point in three dimensional space. This point represents the intersection of three planes. In the next example, we solve a system using row operations and find that it represents a dependent system. A dependent system in 3 dimensions can be represented by two planes that are identical, much like in 2 dimensions where a dependent system represents two lines that are identical.
Example: Solving a 3 x 3 Dependent System
Solve the following system of linear equations using Gaussian Elimination.
[latex]\begin{array}{r}\hfill -x - 2y+z=-1\\ \hfill 2x+3y=2\\ \hfill y - 2z=0\end{array}[/latex]
Solution
Write the augmented matrix.
[latex]\left[\begin{array}{rrr}\hfill -1& \hfill -2& \hfill 1\\ \hfill 2& \hfill 3& \hfill 0\\ \hfill 0& \hfill 1& \hfill -2\end{array}\text{ }|\text{ }\begin{array}{r}\hfill -1\\ \hfill 2\\ \hfill 0\end{array}\right][/latex]
First, multiply row 1 by [latex]-1[/latex] to get a 1 in row 1, column 1. Then, perform row operations to obtain row-echelon form.
[latex]-{R}_{1}\to \left[\begin{array}{rrrrrrr}\hfill 1& \hfill & \hfill 2& \hfill & \hfill -1& \hfill & \hfill 1\\ \hfill 2& \hfill & \hfill 3& \hfill & \hfill 0& \hfill & \hfill 2\\ \hfill 0& \hfill & \hfill 1& \hfill & \hfill -2& \hfill & \hfill 0\end{array}\right][/latex]
[latex]{R}_{2}\leftrightarrow {R}_{3}\to \left[\begin{array}{rrrrr}\hfill 1& \hfill & \hfill 2& \hfill & \hfill -1\\ \hfill 0& \hfill & \hfill 1& \hfill & \hfill -2\\ \hfill 2& \hfill & \hfill 3& \hfill & \hfill 0\end{array}\text{ }|\begin{array}{rr}\hfill & \hfill 1\\ \hfill & \hfill 0\\ \hfill & \hfill 2\end{array}\right][/latex]
[latex]-2{R}_{1}+{R}_{3}={R}_{3}\to \left[\begin{array}{rrrrrr}\hfill 1& \hfill & \hfill 2& \hfill & \hfill -1& \hfill \\ \hfill 0& \hfill & \hfill 1& \hfill & \hfill -2& \hfill \\ \hfill 0& \hfill & \hfill -1& \hfill & \hfill 2& \hfill \end{array}|\begin{array}{rr}\hfill & \hfill 1\\ \hfill & \hfill 0\\ \hfill & \hfill 0\end{array}\right][/latex]
[latex]{R}_{2}+{R}_{3}={R}_{3}\to \left[\begin{array}{rrrrrr}\hfill 1& \hfill & \hfill 2& \hfill & \hfill -1& \hfill \\ \hfill 0& \hfill & \hfill 1& \hfill & \hfill -2& \hfill \\ \hfill 0& \hfill & \hfill 0& \hfill & \hfill 0& \hfill \end{array}|\begin{array}{rr}\hfill & \hfill 2\\ \hfill & \hfill 1\\ \hfill & \hfill 0\end{array}\right][/latex]
The last matrix represents the following system.
[latex]\begin{array}{l}\text{ }x+2y-z=1\hfill \\ \text{ }y - 2z=0\hfill \\ \text{ }0=0\hfill \end{array}[/latex]
We see by the identity [latex]0=0[/latex] that this is a dependent system with an infinite number of solutions. We then find the generic solution. By solving the second equation for [latex]y[/latex] and substituting it into the first equation we can solve for [latex]z[/latex] in terms of [latex]x[/latex].
[latex]\begin{array}{l}\text{ }x+2y-z=1\hfill \\ \text{ }y=2z\hfill \\ \hfill \\ x+2\left(2z\right)-z=1\hfill \\ \text{ }x+3z=1\hfill \\ \text{ }z=\frac{1-x}{3}\hfill \end{array}[/latex]
Now we substitute the expression for [latex]z[/latex] into the second equation to solve for [latex]y[/latex] in terms of [latex]x[/latex].
[latex]\begin{array}{l}\text{ }y - 2z=0\hfill \\ \text{ }z=\frac{1-x}{3}\hfill \\ \hfill \\ y - 2\left(\frac{1-x}{3}\right)=0\hfill \\ \text{ }y=\frac{2 - 2x}{3}\hfill \end{array}[/latex]
The generic solution is [latex]\left(x,\frac{2 - 2x}{3},\frac{1-x}{3}\right)[/latex].
The General Solution to a Dependent 3 X 3 System
Recall that when you solve a dependent system of linear equations in two variables using elimination or substitution, you can write the solution [latex](x,y)[/latex] in terms of x, because there are infinitely many (x,y) pairs that will satisfy a dependent system of equations, and they all fall on the line [latex](x, mx+b)[/latex]. Now that you are working in three dimensions, the solution will represent a plane, so you would write it in the general form [latex](x, m_{1}x+b_{1}, m_{2}x+b_{2})[/latex].
Try It
Solve the system using Gaussian Elimination.
[latex]\begin{array}{c}x+4y-z=4\\ 2x+5y+8z=15\\ x+3y - 3z=1\end{array}[/latex]
Solution
[latex]\left(1,1,1\right)[/latex]
Q & A
Can any system of linear equations be solved by Gaussian elimination?
Yes, a system of linear equations of any size can be solved by Gaussian elimination.
How To: Given a system of equations, solve with matrices using a calculator.
- Save the augmented matrix as a matrix variable [latex]\left[A\right],\left[B\right],\left[C\right]\text{,} \dots[/latex].
- Use the ref( function in the calculator, calling up each matrix variable as needed.
Example: Solving Systems of Equations Using a Calculator
Solve the system of equations.
[latex]\begin{array}{r}\hfill 5x+3y+9z=-1\\ \hfill -2x+3y-z=-2\\ \hfill -x - 4y+5z=1\end{array}[/latex]
Solution
Write the augmented matrix for the system of equations.
[latex]\left[\begin{array}{rrr}\hfill 5& \hfill 3& \hfill 9\\ \hfill -2& \hfill 3& \hfill -1\\ \hfill -1& \hfill -4& \hfill 5\end{array}\text{ }|\text{ }\begin{array}{r}\hfill 5\\ \hfill -2\\ \hfill -1\end{array}\right][/latex]
On the matrix page of the calculator, enter the augmented matrix above as the matrix variable [latex]\left[A\right][/latex].
[latex]\left[A\right]=\left[\begin{array}{rrrrrrr}\hfill 5& \hfill & \hfill 3& \hfill & \hfill 9& \hfill & \hfill -1\\ \hfill -2& \hfill & \hfill 3& \hfill & \hfill -1& \hfill & \hfill -2\\ \hfill -1& \hfill & \hfill -4& \hfill & \hfill 5& \hfill & \hfill 1\end{array}\right][/latex]
Use the ref( function in the calculator, calling up the matrix variable [latex]\left[A\right][/latex].
[latex]\text{ref}\left(\left[A\right]\right)[/latex]
Evaluate.
[latex]\begin{array}{l}\hfill \\ \left[\begin{array}{rrrr}\hfill 1& \hfill \frac{3}{5}& \hfill \frac{9}{5}& \hfill \frac{1}{5}\\ \hfill 0& \hfill 1& \hfill \frac{13}{21}& \hfill -\frac{4}{7}\\ \hfill 0& \hfill 0& \hfill 1& \hfill -\frac{24}{187}\end{array}\right]\to \begin{array}{l}x+\frac{3}{5}y+\frac{9}{5}z=-\frac{1}{5}\hfill \\ \text{ }y+\frac{13}{21}z=-\frac{4}{7}\hfill \\ \text{ }z=-\frac{24}{187}\hfill \end{array}\hfill \end{array}
[/latex]
Using back-substitution, the solution is [latex]\left(\frac{61}{187},-\frac{92}{187},-\frac{24}{187}\right)[/latex].
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.
Example: Applying 2 × 2 Matrices to Finance
Carolyn invests a total of $12,000 in two municipal bonds, one paying 10.5% interest and the other paying 12% interest. The annual interest earned on the two investments last year was $1,335. How much was invested at each rate?
Solution
We have a system of two equations in two variables. Let [latex]x=[/latex] the amount invested at 10.5% interest, and [latex]y=[/latex] the amount invested at 12% interest.
[latex]\begin{array}{l}\text{ }x+y=12,000\hfill \\ 0.105x+0.12y=1,335\hfill \end{array}[/latex]
As a matrix, we have
[latex]\left[\begin{array}{rr}\hfill 1& \hfill 1\\ \hfill 0.105& \hfill 0.12\end{array}\text{ }|\text{ }\begin{array}{r}\hfill 12,000\\ \hfill 1,335\end{array}\right][/latex]
Multiply row 1 by [latex]-0.105[/latex] and add the result to row 2.
[latex]\left[\begin{array}{rr}\hfill 1& \hfill 1\\ \hfill 0& \hfill 0.015\end{array}\text{ }|\text{ }\begin{array}{r}\hfill 12,000\\ \hfill 75\end{array}\right][/latex]
Then,
[latex]\begin{array}{l}0.015y=75\hfill \\ \text{ }y=5,000\hfill \end{array}[/latex]
So [latex]12,000 - 5,000=7,000[/latex].
Thus, $5,000 was invested at 12% interest and $7,000 at 10.5% interest.
Example: 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?
Solution
We have a system of three equations in three variables. Let [latex]x[/latex] be the amount invested at 5% interest, let [latex]y[/latex] be the amount invested at 8% interest, and let [latex]z[/latex] be the amount invested at 9% 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}{rrr}\hfill 1& \hfill 1& \hfill 1\\ \hfill 0.05& \hfill 0.08& \hfill 0.09\\ \hfill 2& \hfill 0& \hfill -1\end{array}\text{ }|\text{ }\begin{array}{r}\hfill 10,000\\ \hfill 770\\ \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}{rrrrrr}\hfill 1& \hfill & \hfill 1& \hfill & \hfill 1& \hfill \\ \hfill 0& \hfill & \hfill 0.03& \hfill & \hfill 0.04& \hfill \\ \hfill 2& \hfill & \hfill 0& \hfill & \hfill -1& \hfill \end{array}|\begin{array}{rr}\hfill & \hfill 10,000\\ \hfill & \hfill 270\\ \hfill & \hfill 0\end{array}\right]\hfill \end{array}\hfill \\ -2{R}_{1}+{R}_{3}={R}_{3}\to \left[\begin{array}{rrrrrr}\hfill 1& \hfill & \hfill 1& \hfill & \hfill 1& \hfill \\ \hfill 0& \hfill & \hfill 0.03& \hfill & \hfill 0.04& \hfill \\ \hfill 0& \hfill & \hfill -2& \hfill & \hfill -3& \hfill \end{array}|\begin{array}{rr}\hfill & \hfill 10,000\\ \hfill & \hfill 270\\ \hfill & \hfill -20,000\end{array}\right]\hfill \\ \frac{1}{0.03}{R}_{2}={R}_{2}\to \left[\begin{array}{rrrrrr}\hfill 0& \hfill & \hfill 1& \hfill & \hfill 1& \hfill \\ \hfill 0& \hfill & \hfill 1& \hfill & \hfill \frac{4}{3}& \hfill \\ \hfill 0& \hfill & \hfill -2& \hfill & \hfill -3& \hfill \end{array}|\begin{array}{rr}\hfill & \hfill 10,000\\ \hfill & \hfill 9,000\\ \hfill & \hfill -20,000\end{array}\right]\hfill \\ 2{R}_{2}+{R}_{3}={R}_{3}\to \left[\begin{array}{rrrrrr}\hfill 1& \hfill & \hfill 1& \hfill & \hfill 1& \hfill \\ \hfill 0& \hfill & \hfill 1& \hfill & \hfill \frac{4}{3}& \hfill \\ \hfill 0& \hfill & \hfill 0& \hfill & \hfill -\frac{1}{3}& \hfill \end{array}|\begin{array}{rr}\hfill & \hfill 10,000\\ \hfill & \hfill 9,000\\ \hfill & \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 $3,000 invested at 5% interest, $1,000 invested at 8%, and $6,000 invested at 9% interest.
Try It
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.
Solution
$150,000 at 7%, $750,000 at 8%, $600,000 at 10%
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