Welcome back, everyone. So in another video, we learned how to solve a system of equations by using something called the row echelon form. We wrote a matrix in a specific form where we had ones along the diagonal, and then we had zeros everywhere else. What I'm going to show you in this video is another way of solving a system of equations by writing a matrix in something called reduced row echelon form. Now it might seem like this is an entirely new process, but I'm actually going to show you it is really similar to this. We're going to use all the same steps and row operations and all of that stuff. And then we'll just actually work out this example here. So let's just go ahead and get started. Really, the main difference between these types of forms here is the matrix that you're trying to work towards. With row echelon form, we had ones along the diagonal, and we had zeros under the diagonal like we have over here. And for a reduced row echelon form, what you're trying to get is ones along the diagonal. So that's exactly the same, except you're trying to get zeros under and above the diagonal. So that's really the only difference here. And remember, these numbers can be anything. There's no restrictions on them. The key advantage to getting something in this format over here is that when we turn it back into a system of equations, you're done. You actually don't have to do any more work versus with the row echelon form. Once we converted something to equations, we had to use substitution to get your x and y values and things like that. With this form, you actually have no more work to do. So what I'm going to do in this video is we're actually just here to sort of look at the example that we did in the last video. We're going to sort of pick up from there. So what I mean by that is that in this example, we use this matrix with 1, 7, 14, negative 1, 2, and 4, and we use our different row operations like adding and multiplying. And then when we got to this point over here, we had row echelon form because we had ones and zeros. And then we turned it back into a system of equations. And what we got here was that the answer was y equals 2 and x equals 0. So what we're going to do here is we're going to sort of pick up from this step, and we're just going to go one step further because now what we have to do is get something in reduced row echelon form. So I have ones along the diagonal, zeros under. But now I'm going to actually have to figure out how to get this position to be a 0 because that would make it reduced row echelon form. Alright? So that's really all there is to it. We're actually just going to sort of pick up from this step, and we're going to have to just do one more sort of round of swapping, multiplying, and adding and things like that. And what we should expect is we'll get the same exact answer. Alright? Well, let's get started here. So remember, I've got this matrix. I've got 1, 7, 14, 01, and 2. So I've got these numbers already that are good, and then I just have to get this number over here to be a 0. How do I do that? Well, let's take a look at our row operations. Can I swap? Swapping isn't going to help here because that's going to mess up all the ones and zeros you already have. Can I multiply this equation by something to get this to be 0? You can't because the only thing is you would have to multiply this whole thing by 0, but you can't do that. So remember, the only way to get zeros in a position is you have to add. So we're going to have to add something to this equation over here to cancel out that 7. So we're going to have to add something to row 7 over here to cancel this out. What do I have to do? Well, if you take a look here, I've got 7. The only way to cancel it out is if I have negative 7. So remember, I can add a multiple of row 2 in order to cancel out this 7 over here. And what you have to do is you have to multiply this whole row 2 by negative 7. Multiply negative 7 times this number, it'll then cancel out this positive 7. This is going to be negative 7 times row 2. Alright? Now let's go ahead and work this out here. What does that turn out to be? Well, what happens is row 2 is completely unaffected, so this would be 1, 01, and 2. What does row 1 become? Well, one plus negative 7 times 0 is still just 1. So that's when that's so that's the useful thing of having these zeros over here is that when you multiply them by numbers, it actually doesn't do anything. So you still have just 1 here. What about this? Well, this is going to be 7 +-seven times 1, and that's actually going to cancel out to 0. And then what about this 14? This is going to be 14 plus negative 7 times 2, which ends up being negative 14. So So what you're actually going to see here is that this completely just cancels out to 0. This won't always happen, by the way, so this won't always just be 0. Sometimes you'll get other numbers, but in this particular case, that's what it worked out to be. So now what you'll see here is that we have this matrix in reduced row echelon form. I got ones along the diagonal and zeros everywhere else. So this is reduced row echelon form. And when you convert it back into a system of equations, we're going to see what happens here. With the x coefficients, I have 1 x. And then with the y coefficients, I have 0 y. So it's almost like it just doesn't exist. And this equals 0. So in other words, x equals 0. And then for the bottom row, the second row, this is 0 x, so that goes away, plus 1 y equals 2. In other words, that just ends up being y equals 2. These are exactly the numbers that we got when we did it using Gaussian elimination or row echelon form, but we just did sort of a slightly different way. So we got the same exact answers, and that's good. So, really, that's how to use this sort of reduced row echelon form. So, basically, to summarize, the key difference between these two different types of methods is how much work you have to do between the matrices and equations. With row echelon form, you have to do less work with the matrices, but then you'll have to do more work when you turn it back into equations. Whereas the reduced row echelon form, you're going to have to do more work with the matrices upfront. So you're going to have to do more row operations, but then you're going to have to do less work when it comes down to the equations. And it'll actually just sort of spit out your numbers right for you. Alright? So one last thing to point out here, this is sometimes called Gaussian elimination, this method here of row echelon form, whereas this one that we're working with is sometimes called Gauss-Jordan elimination for the mathematicians that sort of devised them. But that's really all there is to it. Thanks for watching, folks, and I'll see you in the next video.
Table of contents
- 0. Fundamental Concepts of Algebra3h 29m
- 1. Equations and Inequalities3h 27m
- 2. Graphs1h 43m
- 3. Functions & Graphs2h 17m
- 4. Polynomial Functions1h 54m
- 5. Rational Functions1h 23m
- 6. Exponential and Logarithmic Functions2h 28m
- 7. Measuring Angles39m
- 8. Trigonometric Functions on Right Triangles2h 5m
- 9. Unit Circle1h 19m
- 10. Graphing Trigonometric Functions1h 19m
- 11. Inverse Trigonometric Functions and Basic Trig Equations1h 41m
- 12. Trigonometric Identities 2h 34m
- 13. Non-Right Triangles1h 38m
- 14. Vectors2h 25m
- 15. Polar Equations2h 5m
- 16. Parametric Equations1h 6m
- 17. Graphing Complex Numbers1h 7m
- 18. Systems of Equations and Matrices1h 6m
- 19. Conic Sections2h 36m
- 20. Sequences, Series & Induction1h 15m
- 21. Combinatorics and Probability1h 45m
- 22. Limits & Continuity1h 49m
- 23. Intro to Derivatives & Area Under the Curve2h 9m
18. Systems of Equations and Matrices
Introduction to Matrices
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