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Row reduce the matrices in Exercises 3 and 4 to reduced echelon form. Circle the pivot positions in the final matrix and in the original matrix, and list the pivot columns.$\left[\begin{array}{cccc}{1} & {2} & {3} & {4} \\ {4} & {5} & {6} & {7} \\ {6} & {7} & {8} & {9}\end{array}\right]$

$\left[\begin{array}{cccc}{1} & {0} & {-1} & {-2} \\ {0} & {1} & {2} & {3} \\ {0} & {0} & {0} & {0}\end{array}\right]$ Pivots are in row 1 $\operatorname{col} 1$ and row $2,$ col 2

Algebra

Chapter 1

Linear Equations in Linear Algebra

Section 2

Row Reduction and Echelon Forms

Introduction to Matrices

Oregon State University

Harvey Mudd College

Baylor University

Lectures

01:32

In mathematics, the absolu…

01:11

13:54

Row reduce the matrices in…

04:02

Use row operations to tran…

04:31

use elementary row operati…

09:11

02:08

08:47

04:53

01:28

02:07

13:43

Reduce the given matrix to…

So we're given the following matrix. We have 1234 5678 44567 Actually, and then 6789 And we're wanting to get into reduced echelon form. So first thing we're gonna do is try to focus on this first column. We focus on the first column. We acknowledge that we have a one here, so that's good. Eso this is going to be row equivalent to one zero zero. That's how we want to do it. And how are we going to do that? Well, we're going to use replacement, so we see that we're going to replace our to, um, minus four times are one, and we're gonna place are three by subtracting six are one. Okay, because that's how we get that. And then this will be left with 234 down here. We will have, um, two times for eight, so we'll have a negative three. Two times, uh, six is 12. So I have a negative five here. Then we'll end up getting three times negative for which is negative. 12. It would be a negative six, and then we have three times negative six, which is negative. 18. So that'll make this a negative 10, then. Lastly, we have four times Negative. Four. That's negative. 16. So we'll have a negative nine here, and then we get negative. 24 plus nine. So that's going to give us a pause. A negative 15. Okay, so then we want to go down to the next row, and we see that this is the next thing that we're focusing on. That negative three. So to get rid of that, to make it a one, we're going to skill this. We're going to scale. Row three wrote to. Actually, we're going to scale it by a negative one third. So we do. Negative one third times are two. And then just for the sake of making things simpler, we're going to, um, scale row three by 1/5. Okay. It'll just make it simpler, but we don't have to do it, so we'll get 100 234 one, 23 And this will be one too. Three. All right, Now, um, we can we see what we do is we can just cancel the top in the bottom room or that the second in the third row. This will be through replacement. We will dio are three minus are too. And then we, um we can also replace the top row are one minus two times are too, since this is a to and this is just a one, So using this row replacement will end up getting 10 00 This will be a one. This will be a zero, and we'll see. We'll get zeros across the board. So now this is what we have. Um And then we would end up getting zero here. This would be and negative one, and this would be a negative six plus force. And negative too. So this is about as reduced A We can make it on doing so. We now wanna look at pivot positions that's going to be the first non zero number in a row. So we see that there are two pivot positions. Um, and these pivot positions didn't change, so this would be a position, and this would be a position. They are the same. And, um, the pivot columns would be, uh, column one and calm to that would be the pivot positions or the pivot columns.

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