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Algebra II Textbook26 chapters | 256 lessons

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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** 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:

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.

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.

To accomplish this, we can multiply our third row by 3/2 and add it to the first row. Multiplying the third row by 3/2, we get 0, 0, 3/2, and 6. Adding this to the first row, we get a new first row of 1, 1/2, 0, and 15/2. Now we can multiply our second row by -1/2 and add it to this new first row to get rid of our 1/2. Multiplying the second row by -1/2, we get 0, -1/2, 0, and -7/6. Adding to the first row, we get 1, 0, 0, and 19/3.

Our matrix is now in reduced row echelon form. Our bottom diagonal is all 0s. Our leading numbers are all 1s. And every column with a leading 1 has a zero everywhere else.

Now we can easily finish up our problem by solving for our variables. We have *x* = 19/3, *y* = 7/3, and *z* = 4. That was easy, wasn't it? There wasn't that much work left after we changed our matrix into reduced row echelon form.

What have we learned? We've learned that **linear systems** are a collection of linear equations. The method of Gauss-Jordan elimination is one way to solve linear systems. **Gauss-Jordan elimination** involves creating an augmented matrix of both sides of our equations, changing this matrix into reduced row echelon form, then finishing up the problem to find our solution.

**Reduced row echelon form** 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. After we've changed our matrix into reduced row echelon form, it is a simple process to finish up and find the solution.

Upon completing this lesson, you should be able to:

- Define linear systems and reduced row echelon form
- Explain how to use Gauss-Jordan elimination to solve linear systems

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Algebra II Textbook26 chapters | 256 lessons

- What is a Matrix? 5:39
- How to Write an Augmented Matrix for a Linear System 4:21
- How to Perform Matrix Row Operations 5:08
- Matrix Notation, Equal Matrices & Math Operations with Matrices 6:52
- How to Solve Inverse Matrices 6:29
- How to Solve Linear Systems Using Gaussian Elimination 6:10
- How to Solve Linear Systems Using Gauss-Jordan Elimination 5:00
- Multiplicative Inverses of Matrices and Matrix Equations 4:31
- How to Take a Determinant of a Matrix 7:02
- Solving Systems of Linear Equations in Two Variables Using Determinants 4:54
- Solving Systems of Linear Equations in Three Variables Using Determinants 7:41
- Using Cramer's Rule with Inconsistent and Dependent Systems 4:05
- How to Evaluate Higher-Order Determinants in Algebra 7:59
- Go to Algebra II: Matrices and Determinants

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