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MOOC_LinearAlgebra_Lesson03

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Hello, and welcome to lesson three 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. So here’s a matrix and it looks like it's got pretty random entry, 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 got the ones 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. And the point is, that once you have the echelon form you can solve linear systems. So, 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 non-zero rows are above any rules with all zeros. So you can think of the rows with all zeros as being kind of redundant. You do not want them to be on top at all. Next, the leading entry in each non-zero row must be one. So, you always start with a one in each non-zero 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 like an orderly arrangement, like groups of soldiers in an army. Finally, you want each leading one to be the only non-zero 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 rule begins with the one. If you look at any column with a one in it, then that's the only non-zero entry. The fact is that every matrix has a unique reduced echelon form, which can be found using RowReduce. Before going into row reduce, let’s just look at an example of how you might actually find the reduced echelon form using elementary row operation, 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 zero over here. After that you divide by -3 to get that to be a 1. The -3 becomes a 1. Then he still want to make that 5 into 0, so you do rule 1, plus -5, times R2. So you get a zero over there. Finally, you do row three. 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 row reduce 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 over here that this echelon form is so much nicer and structured than the matrix that you began with. A slightly larger example now, so here's a random eight by nine matrix. I use a SeedRandom just to make sure I get roughly the same answer every time. Here's the random matrix with eight rows and nine columns. I apply RowReduce to it and what you get back is a very nice, simple form. Once we know how to use row reduce, we can now talk about solving linear systems. So the question is, how do you solve a linear system using row reduction? Well, the steps are, you first of all, write on 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 set 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. 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: x, y, and z. You first of all setup the augmented matrix 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 x. So x is one. That's a y, so y is two, and that’s a z, so z is three. So x=1, y=2, z=3 is a very nice, simple, unique solution for this system of equations. On to a slightly harder example, we actually have infinitely many solutions. Here's a system, right on the augmented matrix again. Make sure it looks right with matrix form. For example, the first equation is 5x+2y+11=4. So you have 5, 2, 11, 4, etcetera. 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 x=-25z+10, because the z goes across the right-hand side. y=57z-3. For each value of z you actually get a value of x and a value of y. For example, if z is one, then if you plug back over there, x is -15, and you plug back over there and y is 34. You have infinitely many solutions, each one corresponding to a different value of C. 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 30 here. So 5+7=12, 2+3=5, 11+4=15, and 4+1=5. The third equation is 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 five something else and that's exactly what I do on the next slide. Take the first two equations but I replace the five by a six, 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 then where you have a one. That says 0=1, which is never possible. Your system, right, the matrix for it, the augmented matrix, look at the matrix form and make sure it's all okay. It does look okay. 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 the way we wanted to, to make sure that there is no solutions. So the six 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 solved the equation by hand. You could, let’s say, interchange two equations or you might add one multiple of an equation to another, etcetera. So those are called elementary row operations. Of course, within the Wolfram Language, we have a nice function called row reduce, which will let you do much more and get to the reduced echelon form in no time at all. Now, the point about this load 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 row reduce and row reduction are a great way to solve linear systems, of course, row reduction is a bit more technical than using linear solve but it’s a very powerful way of understanding linear system. That's the end of this lesson and the next lesson we'll talk about vectors in n-dimensional space. Do review this very important lesson on row reduction and I hope you've enjoyed it and learned a lot from it. Thank you very much, I stop over here.

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Duration: 10 minutes and 34 seconds
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Posted by: wolfram on Sep 30, 2020

MOOC_LinearAlgebra_Lesson03

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