# How to Solve Linear Systems Using Gauss-Jordan Elimination Video

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• 0:01 Linear Systems
• 0:44 Gauss-Jordan Elimination
• 1:21 Reduced Row Echelon Form
• 3:49 Finishing Up
• 4:09 Lesson Summary
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Lesson Transcript
Instructor: Yuanxin (Amy) Yang Alcocer

Amy has a master's degree in secondary education and has taught math at a public charter high school.

You will come across simple linear systems and more complex ones as you progress in math. Watch this video lesson to learn how you can use Gauss-Jordan elimination to help you solve these linear systems.

## Linear Systems

Linear systems are a collection of linear equations. We can have linear systems in two variables, linear systems in three variables, and linear systems in even more variables. The key to all these systems is that there are no exponents to any of the variables. In other words, all the variables are to the first degree. As you progress in math, you will come across linear systems of all kinds, both simple and more complex. Being able to solve them easily will serve you well. So, keep watching and you will learn how to use Gauss-Jordan elimination to help you solve your linear systems.

To help us demonstrate this technique, let's work on this linear system in three variables:

## Gauss-Jordan Elimination

So, the process of Gauss-Jordan elimination involves creating an augmented matrix of both sides of our equations, changing this matrix into reduced row echelon form (I will explain this later), then finishing up the problem to find our solution.

Let's begin. We first create our augmented matrix. Our first row is 2, 1, -3, and 3. Our second row is -2, 2, 3, and 4. Our final third row is 0, -3, 2, and 1.

Now that we've created our augmented matrix, it is now time to change it up.

## Reduced Row Echelon Form

We want to change it into its reduced row echelon form. What is this form? It is when our matrix has zeros on the lower diagonal and the first nonzero number in each row is 1. Also, if a column has a leading 1, then all the other numbers also need to be 0.

To begin, the lower diagonal in our matrix includes the numbers -2 and -3. We need to change these to 0. Let's do that first.

To change the -2 to a 0, we can add the first and second equations together. Doing that we get a new second row of 0, 3, 0, and 7. Now we can add the new second row to the third row to get a new third row of 0, 0, 2, and 8. We now have this matrix:

Now, let's change all our leading numbers to 1s. We divide the first row by 2 to get 1, 1/2, -3/2, and 3/2. We divide the second row by 3 to get 0, 1, 0, and 7/3. We divide our third row by 2 to get 0, 0, 1, and 4. Our matrix now looks like this:

We look at our leading 1s and we see that in the second column we need to change the 1/2 to 0, and in the third column we need to change the -3/2 to 0 as well. We remember that in reduced row echelon form any column with a leading 1 needs to have 0 for all the other numbers.

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