▪ Solving a System with Gauss-Jordan Elimination

Learning Outcomes

Use Gauss-Jordan elimination to solve a system 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 use Gaussian elimination as a tool for solving a system written as an augmented matrix. Now we will see how to use Gauss-Jordan elimination, which is very similar to Gaussian elimination. We will review the same examples we used in the previous section.

Example: Solving a 2 X 2 System by Gauss-Jordan Elimination

Solve the given system by Gauss-Jordan elimination.

\begin{array}{l} 2 x + 3 y = 6 \\ x - y = \frac{1}{2} \end{array}

$\begin{array}{l}2x+3y=6\hfill \\ \text{ }x-y=\frac{1}{2}\hfill \end{array}$

Show Solution

First, we write this as an augmented matrix.

[\begin{array}{ccc} 2 & 3 & 6 \\ 1 & - 1 & \frac{1}{2} \end{array}]

$\left[\begin{array}{cc|c}\hfill 2& \hfill 3& \hfill 6\\ \hfill 1& \hfill -1& \hfill \frac{1}{2}\\ \end{array}\right]$

We want a 1 in row 1, column 1. This can be accomplished by interchanging row 1 and row 2.

R_{1} \leftrightarrow R_{2} \to [\begin{array}{ccc} 1 & - 1 & \frac{1}{2} \\ 2 & 3 & 6 \end{array}]

${R}_{1}\leftrightarrow {R}_{2}\to \left[\begin{array}{cc|c}\hfill 1& \hfill -1& \hfill \frac{1}{2}\\ \hfill 2& \hfill 3& \hfill 6\\ \end{array}\right]$

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 $-2$ and then adding the result to row 2.

- 2 R_{1} + R_{2} = R_{2} \to [\begin{array}{ccc} 1 & - 1 & \frac{1}{2} \\ 0 & 5 & 5 \end{array}]

$-2{R}_{1}+{R}_{2}={R}_{2}\to \left[\begin{array}{cc|c}\hfill 1& \hfill -1& \hfill \frac{1}{2}\\ \hfill 0& \hfill 5& \hfill 5\\ \end{array}\right]$

Next, we need a 1 in row 2, column 2. This can be accomplished by multiplying row 2 by $\frac{1}{5}$ .

$\frac{1}{5}{R}_{2}={R}_{2}\to \left[\begin{array}{cc|c}\hfill 1& \hfill -1& \hfill \frac{1}{2}\\ \hfill 0& \hfill 1& \hfill 1\\ \end{array}\right]$

Lastly, we need a 0 in row 1, column 2, and we can accomplish it by simply adding row 2 to row 1.

${R}_{1}+{R}_{2}={R}_{1}\to \left[\begin{array}{cc|c}\hfill 1& \hfill 0& \hfill \frac{3}{2}\\ \hfill 0& \hfill 1& \hfill 1\\ \end{array}\right]$

Or the reduced row-echelon form can be found by using “rref” on a graphing calculator:

$\left[\begin{array}{cc|c}\hfill 1& \hfill 0& \hfill \frac{3}{2}\\ \hfill 0& \hfill 1& \hfill 1 \end{array}\right]$

The first row of the matrix represents

x = \frac{3}{2}

$x=\frac{3}{2}$ and the second row of the matrix represents

y = 1

$y=1$ . No back-substitute is needed in Gauss-Jordan elimination.

The solution is the point $\left(\frac{3}{2},1\right)$ as we found in the previous section.

Example: Solving a Dependent System

Solve the system of equations.

\begin{array}{l} 3 x + 4 y = 12 \\ 6 x + 8 y = 24 \end{array}

$\begin{array}{l}3x+4y=12\\ 6x+8y=24\end{array}$

Show Solution

First, we write this as an augmented matrix.

[\begin{array}{ccc} 3 & 4 & 12 \\ 6 & 8 & 24 \end{array}]

$\left[\begin{array}{cc|c}\hfill 3& \hfill 4& \hfill 12\\ \hfill 6& \hfill 8& \hfill 24\\ \end{array}\right]$

Perform row operations on the augmented matrix to try and achieve reduced row-echelon form.

- \frac{1}{2} R_{2} + R_{1} = R_{1} \to [\begin{array}{ccc} 0 & 0 & 0 \\ 6 & 8 & 24 \end{array}]

$-\frac{1}{2}{R}_{2}+{R}_{1}={R}_{1}\to \left[\begin{array}{cc|c}\hfill 0& \hfill 0& \hfill 0\\ \hfill 6& \hfill 8& \hfill 24\\ \end{array}\right]$

R_{1} \leftrightarrow R_{2} \to [\begin{array}{ccc} 6 & 8 & 24 \\ 0 & 0 & 0 \end{array}]

${R}_{1}\leftrightarrow {R}_{2}\to \left[\begin{array}{cc|c}\hfill 6& \hfill 8& \hfill 24\\ \hfill 0& \hfill 0& \hfill 0\\ \end{array}\right]$

Now multiply $\frac{1}{6}$ to row 1 to obtsin 1 in row 1, column 1.

$\frac{1}{6}{R}_{1}={R}_{1}\to \left[\begin{array}{cc|c}\hfill 1& \hfill \frac{4}{3}& \hfill 4\\ \hfill 0& \hfill 0& \hfill 0\\ \end{array}\right]$

Or the reduced row-echelon form can be found by using “rref” on a graphing calculator:

$\left[\begin{array}{cc|c}\hfill 1& \hfill \frac{4}{3}& \hfill 4\\ \hfill 0& \hfill 0& \hfill 0\end{array}\right]$

The matrix ends up with all zeros in the last row: $0y=0$ . 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 $y$ .

\begin{array}{l} x + \frac{4}{3} y = 4 \\ \frac{4}{3} y = 4 - x \\ y = 3 - \frac{3}{4} x \end{array}

$\begin{array}{l}x+\frac{4}{3}y=4\hfill \\ \frac{4}{3}y=4 - x\hfill \\ y=3-\frac{3}{4}x\hfill \end{array}$

So the solution to this system is $\left(x,3-\frac{3}{4}x\right)$ .

Example: Solving a System of Linear Equations using Matrices

Solve the following system of linear equations using Gaussian Elimination.

$\begin{array}{r}\hfill x-y+z=8\\ \hfill 2x+3y-z=-2\\ \hfill 3x-2y-9z=9\end{array}$

Show Solution

First, we write the augmented matrix.

$\left[\begin{array}{ccc|c}\hfill 1& \hfill -1& \hfill 1& \hfill 8\\ \hfill 2& \hfill 3& \hfill -1& \hfill -2\\ \hfill 3& \hfill -2& \hfill -9& \hfill 9\end{array}\right]$

Next, we perform row operations to obtain reduced row-echelon form.

$\begin{array}{rrrrr}\hfill -2{R}_{1}+{R}_{2}={R}_{2}\to \left[\begin{array}{ccc|c}\hfill 1& \hfill -1& \hfill 1& \hfill 8\\ \hfill 0& \hfill 5& \hfill -3& \hfill -18\\ \hfill 3& \hfill -2& \hfill -9& \hfill 9\end{array}\right]& \hfill & \hfill & \hfill & \hfill -3{R}_{1}+{R}_{3}={R}_{3}\to \left[\begin{array}{ccc|c}\hfill 1& \hfill -1& \hfill 1& \hfill 8\\ \hfill 0& \hfill 5& \hfill -3& \hfill -18\\ \hfill 0& \hfill 1& \hfill -12& \hfill -15\end{array}\right]\end{array}$

The easiest way to obtain a 1 in row 2 of column 1 is to interchange ${R}_{2}$ and ${R}_{3}$ .

$\text{Interchange }{R}_{2}\text{ and }{R}_{3}\to \left[\begin{array}{ccc|c}\hfill 1& \hfill -1& \hfill 1& \hfill 8\\ \hfill 0& \hfill 1& \hfill -12& \hfill -15\\ \hfill 0& \hfill 5& \hfill -3& \hfill -18\end{array}\right]$

Then

$\begin{array}{l}\\ \begin{array}{rrrrr}\hfill -5{R}_{2}+{R}_{3}={R}_{3}\to \left[\begin{array}{ccc|c}\hfill 1& \hfill -1& \hfill 1& \hfill 8\\ \hfill 0& \hfill 1& \hfill -12& \hfill -15\\ \hfill 0& \hfill 0& \hfill 57& \hfill 57\end{array}\right]& \hfill & \hfill & \hfill & \hfill -\frac{1}{57}{R}_{3}={R}_{3}\to \left[\begin{array}{ccc|c}\hfill 1& \hfill -1& \hfill 1& \hfill 8\\ \hfill 0& \hfill 1& \hfill -12& \hfill -15\\ \hfill 0& \hfill 0& \hfill 1& \hfill 1\end{array}\right]\end{array}\end{array}$

Finally,

$\begin{array}{l}\\ \begin{array}{rrrrr}\hfill 12{R}_{3}+{R}_{2}={R}_{2}\to \left[\begin{array}{ccc|c}\hfill 1& \hfill -1& \hfill 1& \hfill 8\\ \hfill 0& \hfill 1& \hfill 0& \hfill -3\\ \hfill 0& \hfill 0& \hfill 1& \hfill 1\end{array}\right]& \hfill & \hfill & \hfill & \hfill {R}_{2}-{R}_{3}+{R}_{1}={R}_{1}\to \left[\begin{array}{ccc|c}\hfill 1& \hfill 0& \hfill 0& \hfill 4\\ \hfill 0& \hfill 1& \hfill 0& \hfill -3\\ \hfill 0& \hfill 0& \hfill 1& \hfill 1\end{array}\right]\end{array}\end{array}$

Or the reduced row-echelon form can be found by using “rref” on a graphing calculator:

$\left[\begin{array}{ccc|c}\hfill 1& \hfill 0& \hfill 0& \hfill 4\\ \hfill 0& \hfill 1& \hfill 0& \hfill -3\\ \hfill 0& \hfill 0& \hfill 1& \hfill 1\end{array}\right]$

The last matrix represents the equivalent system.

$\begin{array}{l}\text{ }x=4\hfill \\ \text{ }y=-3\hfill \\ \text{ }z=1\hfill \end{array}$

So, we obtain the solution as $\left(4,-3,1\right)$ .

Example: Solving a 3 x 3 Dependent System

Solve the following system of linear equations using Gaussian Elimination.

$\begin{array}{r}\hfill -x - 2y+z=-1\\ \hfill 2x+3y=2\\ \hfill y - 2z=0\end{array}$

Show Solution

Write the augmented matrix.

$\left[\begin{array}{ccc|c}\hfill -1& \hfill -2& \hfill 1& \hfill -1\\ \hfill 2& \hfill 3& \hfill 0& \hfill 2\\ \hfill 0& \hfill 1& \hfill -2& \hfill 0\end{array}\right]$

First, multiply row 1 by $-1$ to get a 1 in row 1, column 1. Then, perform row operations to obtain reduced row-echelon form.

$-{R}_{1}\to \left[\begin{array}{ccc|c}\hfill 1& \hfill 2& \hfill -1& \hfill 1\\ \hfill 2& \hfill 3& \hfill 0& \hfill 2\\ \hfill 0& \hfill 1& \hfill -2& \hfill 0\end{array}\right]$

${R}_{2}\leftrightarrow {R}_{3}\to \left[\begin{array}{ccc|c}\hfill 1& \hfill 2& \hfill -1& \hfill 1\\ \hfill 0& \hfill 1& \hfill -2& \hfill 0\\ \hfill 2& \hfill 3& \hfill 0& \hfill 2\end{array}\right]$

$-2{R}_{1}+{R}_{3}={R}_{3}\to\left[\begin{array}{ccc|c}\hfill 1& \hfill 2& \hfill -1& \hfill 1\\ \hfill 0& \hfill 1& \hfill -2& \hfill 0\\ \hfill 0& \hfill -1& \hfill 2& \hfill 0\end{array}\right]$

${R}_{2}+{R}_{3}={R}_{3}\to\left[\begin{array}{ccc|c}\hfill 1& \hfill 2& \hfill -1& \hfill 1\\ \hfill 0& \hfill 1& \hfill -2& \hfill 0\\ \hfill 0& \hfill 0& \hfill 0& \hfill 0\end{array}\right]$

$-2{R}_{2}+{R}_{1}={R}_{1}\to\left[\begin{array}{ccc|c}\hfill 1& \hfill 0& \hfill 3& \hfill 1\\ \hfill 0& \hfill 1& \hfill -2& \hfill 0\\ \hfill 0& \hfill 0& \hfill 0& \hfill 0\end{array}\right]$

Or the reduced row-echelon form can be found by using “rref” on a graphing calculator:

$\left[\begin{array}{ccc|c}\hfill 1& \hfill 0& \hfill 3& \hfill 1\\ \hfill 0& \hfill 1& \hfill -2& \hfill 0\\ \hfill 0& \hfill 0& \hfill 0& \hfill 0\end{array}\right]$

The last matrix represents the following system.

$\begin{array}{l}\text{ }x+3z=1\hfill \\ \text{ }y - 2z=0\hfill \\ \text{ }0=0\hfill \end{array}$

We see by the identity $0=0$ that this is a dependent system with an infinite number of solutions. We then find the generic solution. By solving the first equation for $x$ and the second equation for $y$ in terms of $z$ .

$\begin{array}{l}\text{ }x=1-3z\hfill \\ \text{ }y=2z\hfill \end{array}$

So, the generic solution is $\left(1-3z, 2z, z\right)$ .

The General Solution to a Dependent 3 X 3 System in Gauss-Jordan Elimination

Recall that when you solve a dependent system of linear equations in two variables using elimination or substitution, you can write the solution $(x,y)$ in terms of y, because there are infinitely many (x,y) pairs that will satisfy a dependent system of equations, and they all fall on the line $(my+b, y)$ . Now that you are working in three dimensions, the solution will represent a plane, so you would write it in the general form $(m_{1}z+b_{1}, m_{2}z+b_{2}, z)$ .

Try It

Q & A

Can any system of linear equations be solved by Gauss-Jordan elimination?

Yes, a system of linear equations of any size can be solved by Gauss-Jordan elimination.

How To: Given a system of equations, solve with matrices using a calculator

Save the augmented matrix as a matrix variable $\left[A\right],\left[B\right],\left[C\right]\text{,} \dots$ .
Use the rref( function in the calculator, calling up each matrix variable as needed.

Example: Solving Systems of Equations Using a Calculator

Solve the system of equations.

$\begin{array}{r}\hfill 5x+3y+9z=-1\\ \hfill -2x+3y-z=-2\\ \hfill -x - 4y+5z=1\end{array}$

Show Solution

Write the augmented matrix for the system of equations.

$\left[\begin{array}{ccc|c}\hfill 5& \hfill 3& \hfill 9& \hfill -1\\ \hfill -2& \hfill 3& \hfill -1& \hfill -2\\ \hfill -1& \hfill -4& \hfill 5& \hfill 1\end{array}\right]$

On the matrix page of the calculator, enter the augmented matrix above as the matrix variable $\left[A\right]$ .

$\left[A\right]=\left[\begin{array}{rrrr}\hfill 5& \hfill 3& \hfill 9& \hfill -1\\ \hfill -2& \hfill 3& \hfill -1& \hfill -2\\ \hfill -1& \hfill -4& \hfill 5& \hfill 1\end{array}\right]$

Use the rref( function in the calculator, calling up the matrix variable $\left[A\right]$ .

$\text{rref}\left(A\right)$

Evaluate.

$\text{rref}\left(\left[A\right]\right)=\left[\begin{array}{ccc|c}\hfill 1& \hfill 0& \hfill 0& \hfill \frac{61}{187}\\ \hfill 0& \hfill 1& \hfill 0& \hfill -\frac{92}{187}\\ \hfill 0& \hfill 0& \hfill 1& \hfill -\frac{24}{187}\end{array}\right]$

So, the solution is $\left(\frac{61}{187},-\frac{92}{187},-\frac{24}{187}\right)$ .

Applications of Systems of Equations

Now we will turn to the applications for which systems of equations are used. Again, we will use same examples but use Gauss-Jordan elimination instead of Gaussian elimination.

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?

Show Solution

We have a system of two equations in two variables. Let $x=$ the amount invested at 10.5% interest, and $y=$ the amount invested at 12% interest.

$\begin{array}{l}\text{ }x+y=12,000\hfill \\ 0.105x+0.12y=1,335\hfill \end{array}$

As a matrix, we have

$\left[\begin{array}{cc|c}\hfill 1& \hfill 1& \hfill 12,000\\ \hfill 0.105& \hfill 0.12& \hfill 1,335\\ \end{array}\right]$

and its reduced row-echelon form is

$\left[\begin{array}{cc|c}\hfill 1& \hfill 0& \hfill 7,000\\ \hfill 0& \hfill 1& \hfill 5,000\\ \end{array}\right]$

Then,

$\begin{array}{r}\hfill x=7,000\\ \hfill y=5,000 \end{array}$

So, $7,000 was invested at 10.5% interest and $5,000 was invested at 12% interest.

Try It

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?

Show 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,

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

As a matrix, we have

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

Now, we either perform row operation or use “rref” to achieve reduced row-echelon form.

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

$-3{R}_{3}={R}_{3}\to\left[\begin{array}{ccc|c}\hfill 1& \hfill 1& \hfill 1& \hfill 10,000\\ \hfill 0& \hfill 1& \hfill \frac{4}{3}& \hfill 9,000\\ \hfill 0& \hfill 0& \hfill 1& \hfill 6,000\end{array}\right]$

$-\frac{3}{4}{R}_{3}+{R}_{2}={R}_{2}\to \left[\begin{array}{ccc|c}\hfill 1& \hfill 1& \hfill 1& \hfill 10,000\\ \hfill 0& \hfill 1& \hfill 0& \hfill 1,000\\ \hfill 0& \hfill 0& \hfill 1& \hfill 6,000\end{array}\right]$

$-{R}_{2}-{R}_{3}+{R}_1={R}_{1}\to\left[\begin{array}{ccc|c}\hfill 1& \hfill 0& \hfill 0& \hfill 3,000\\ \hfill 0& \hfill 1& \hfill 0& \hfill 1,000\\ \hfill 0& \hfill 0& \hfill 1& \hfill 6,000\end{array}\right]$

Or the reduced row-echelon form can be found by using “rref” on a graphing calculator:

$\left[\begin{array}{ccc|c}\hfill 1& \hfill 0& \hfill 0& \hfill 3,000\\ \hfill 0& \hfill 1& \hfill 0& \hfill 1,000\\ \hfill 0& \hfill 0& \hfill 1& \hfill 6,000\end{array}\right]$

The last matrix represents the following system:

$\begin{array}{r}\hfill x=3,000\\ \hfill y=1,000\\ \hfill z=6,000\end{array}$

So, the answer is $3,000 invested at 5% interest, $1,000 invested at 8%, and $6,000 invested at 9% interest.

Module 6: Systems of Linear Equations