Watch videos with subtitles in your language, upload your videos, create your own subtitles! Click here to learn more on "how to Dotsub"

MOOC_LinearAlgebra_Lesson03

0 (0 Likes / 0 Dislikes)
Hello, and welcome to Lesson 3 of this Introduction to Linear Algebra with Wolfram U. The topic for this lesson is row reduction and echelon forms. Let's begin with a brief overview of the lesson. Here's a matrix, and it looks like it's got pretty random entries; some are large, some are small. The goal of this lesson is to transform such a matrix to a simpler echelon form. That's like a much more ordered set of entries. So you see over here, this looks much more orderly than the earlier matrix. For example, you've got the 1s along over here, and the last column looks like it's got some very interesting and useful information. The question is: how do you obtain an echelon form? Well, an echelon form can be obtained by using the process of row reduction. Fortunately, we have a very nice function called RowReduce in the Wolfram Language, which will give you the reduced echelon form of a matrix, like the one you see over here. The point is that once you have this reduced echelon form, you can solve linear systems. Let's begin with a formal definition of a reduced echelon form. A matrix is said to be in a reduced echelon form if it has a few properties. The first property is that all nonzero rows are above any rows with all zeros. You can think of the rows with all zeros as being kind of redundant. You don't want them to be on top at all. Next, the leading entry in each nonzero row must be 1. So, you always start with a 1 in each nonzero row. Next, each leading entry of a row is to the right of the leading entry of the row above it. That's where the word echelon comes from— it's an orderly arrangement, like groups of soldiers in an army. Finally, you want each leading 1 to be the only nonzero entry in its column. Here's an example of a matrix that is in a reduced echelon form. You can see over here the zeros are all at the end and then every row begins with a 1. If you look at any column with a 1 in it, then that's the only nonzero entry. The fact is that every matrix has a unique reduced echelon form, which can be found using RowReduce. Before going into RowReduce, let's just look at an example of how you might actually find the reduced echelon form using elementary row operations, just by manipulating the rows and getting the answer. Here's the matrix. You want to reduce it to echelon form. What you do is you take the second row, R2. Then you do R2 minus twice R1. That becomes 0 over here. After that, you divide by -3 to get that to be a 1. The -3 becomes a 1. Then you still want to make that 5 into a 0, so you do R1 + (-5)R2, so you get a 0 over there. Finally, you do R3. Goes to R3 + R1 - R2, and you get back this nice echelon form.  You can check the answer with RowReduce and you see that you get back just the same result. Let's try and use RowReduce directly. Here's a matrix. The matrix has three rows and four columns. Apply RowReduce to it, and sure enough, you get back a reduced echelon form. You see that this echelon form is so much nicer in structure than the matrix that you began with. A slightly larger example now: here's a random 8-by-9 matrix. I used SeedRandom just to make sure I get roughly the same answer every time. Here's the random matrix with 8 rows and 9 columns. I apply RowReduce to it and what you get back is a very nice simple form. Once we know how to use RowReduce, we can now talk about solving linear systems.  The question is: how do you solve a linear system using row reduction? Well, the steps are you first of all write in the augmented matrix with both the left- and right-hand sides of the system. Apply RowReduce to it, and then, looking at the output from RowReduce, you can write down the solution set. In fact, there are three possibilities. The first possibility is that you actually have a unique solution for your system, just one solution, or you might have infinitely many solutions, and both of those are said to give you a consistent system. The third case is where you have no solutions, and that's referred to as having an inconsistent system. Just to summarize and make sure we understand terminology, you can either have a unique solution or you can have infinitely many solutions or you can have no solution at all. One example of each. Here is a system of equations, a linear system, three equations, three unknowns: <i>x</i>, <i>y</i> and <i>z</i>. You first of all set up the augmented matrix for the system. Make sure it looks right. You have {3, 5, 11, 46} in the first row— that looks correct— {2, -3, 4, 8} second row, and {1, 6, -7, -38} in the third row. All that looks just right. When you apply RowReduce to it, look at the matrix form, and now it's clear that this first one over here is just a placeholder for <i>x</i>, so <i>x</i> is 1; that's a <i>y</i>, so <i>y</i> is 2; and that's a <i>z</i>, so <i>z</i> is 3. So <i>x</i> = 1, <i>y</i> = 2 and <i>z</i> = 3 is a very nice, simple, unique solution for this system of equations. On to a slightly harder example, where we actually have infinitely many solutions. Here's a system. Write in the augmented matrix again. Make sure it looks right with MatrixForm. For example, the first equation is 5<i>x</i> + 2<i>y</i> + 11<i>z</i> = 4. So you've got {5, 2, 11, 4}, etc. Apply RowReduce to it. Look at the matrix form and you see that the last row is just a bunch of zeros; that says you actually have infinitely many solutions. The question is: how do you write them down? This first row over here says that <i>x</i> = -25<i>z</i> + 10, because the <i>z</i> goes across the right-hand side, and <i>y</i> = 57<i>z</i> - 23. For each value of <i>z</i>, you actually get a value of <i>x</i> and a value of <i>y</i>. For example, if <i>z</i> is 1, then if you plug back over there, <i>x</i> is -15, and you plug back over there and <i>y</i> is 34. In fact, you have infinitely many solutions, each one corresponding to a different value of <i>z</i>. The question is: how did this example come about? What I did was I simply took the first two equations over here and I added them, I summed them to get the third here. So 5 + 7 = 12, 2 + 3 = 5, 11 + 4 = 15 and 4 + 1 = 5. The third equation is in fact redundant over here; it's got no new information, and hence, you actually have infinitely many solutions to that system of two equations. That's one way you might get infinitely many solutions. The question is: can we then go ahead and construct an example where you have an inconsistent system, like, can I make this slightly different to have no solutions at all? Well I could just make the 5 something else and that's exactly what I'll do on the next slide. I take the first two equations but I replace the 5 by a 6, and that's going to give us an inconsistent system. The sign of inconsistency is you have a row that looks like zeros all the way up there where you have a 1. That says 0 = 1, which is never possible. You have a system. Write the matrix for it, the augmented matrix. Look at the matrix form and make sure it's all OK. It does look OK. Apply RowReduce and you see over here that the last row is 0 = 1. That's no good; there is no solution for this system, but of course, we knew that— we just constructed it the way we wanted to, to make sure that there is no solution. So the 6 over here made sure that there is no solution.  So, this is a system which has no solution. That brings me to the end of this lesson. The main point is that every matrix can be reduced to an echelon form by using elementary row operations, which are nothing but what you do if you actually solve the equation by hand. Namely, you could, let's say, interchange two equations, or you might add one multiple of an equation to another, etc. Those are called elementary row operations. Of course, within the Wolfram Language, we have a nice function called RowReduce, which will let you do much more and get to the reduced echelon form in no time at all. Now, the point about this row reduction is that it can be used to determine whether there are any solutions at all for a system and how many solutions are there. So, RowReduce and row reduction are a great way to solve linear systems. Of course, row reduction is a bit more technical than using LinearSolve, but it's a very powerful way of understanding linear systems. That's the end of this lesson, and next lesson we'll talk about vectors in <i>n</i>-dimensional space.  Do review this very important lesson on row reduction. I hope you've enjoyed it and learned a lot from it. Thank you very much. I'll stop over here.

Video Details

Duration: 10 minutes and 34 seconds
Country:
Language: English
License: Dotsub - Standard License
Genre: None
Views: 6
Posted by: wolfram on Sep 30, 2020

MOOC_LinearAlgebra_Lesson03

Caption and Translate

    Sign In/Register for Dotsub to translate this video.