▪ 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.

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

Solve the given system by Gauss-Jordan elimination.

[latex]\begin{array}{l}2x+3y=6\hfill \\ \text{ }x-y=\frac{1}{2}\hfill \end{array}[/latex]

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Example: Solving a Dependent System

Solve the system of equations.

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

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Example: Solving a System of Linear Equations using Matrices

Solve the following system of linear equations using Gaussian Elimination.

[latex]\begin{array}{r}\hfill x-y+z=8\\ \hfill 2x+3y-z=-2\\ \hfill 3x-2y-9z=9\end{array}[/latex]

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First, we write the augmented matrix.

[latex]\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][/latex]

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

[latex]\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}[/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}{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][/latex]

Then

[latex]\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}[/latex]

Finally,

[latex]\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}[/latex]

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

[latex]\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][/latex]

The last matrix represents the equivalent system.

[latex]\begin{array}{l}\text{ }x=4\hfill \\ \text{ }y=-3\hfill \\ \text{ }z=1\hfill \end{array}[/latex]

So, we obtain the solution as [latex]\left(4,-3,1\right)[/latex].

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]

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Write the augmented matrix.

[latex]\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][/latex]

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

[latex]-{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][/latex]

[latex]{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][/latex]

[latex]-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][/latex]

[latex]{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][/latex]

[latex]-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][/latex]

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

[latex]\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][/latex]

The last matrix represents the following system.

[latex]\begin{array}{l}\text{ }x+3z=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 first equation for [latex]x[/latex] and the second equation for [latex]y[/latex] in terms of [latex]z[/latex].

[latex]\begin{array}{l}\text{ }x=1-3z\hfill \\ \text{ }y=2z\hfill \end{array}[/latex]

So, the generic solution is [latex]\left(1-3z, 2z, z\right)[/latex].

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 [latex](x,y)[/latex] 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 [latex](my+b, y)[/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](m_{1}z+b_{1}, m_{2}z+b_{2}, z)[/latex].

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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 [latex]\left[A\right],\left[B\right],\left[C\right]\text{,} \dots[/latex].
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.

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

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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?

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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?

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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}{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 either perform row operation or use “rref” to achieve reduced 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]

[latex]-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][/latex]

[latex]-\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][/latex]

[latex]-{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][/latex]

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

[latex]\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][/latex]

The last matrix represents the following system:

[latex]\begin{array}{r}\hfill x=3,000\\ \hfill y=1,000\\ \hfill z=6,000\end{array}[/latex]

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