College Algebra: Using Gauss-Jordan: Example

Gauss-Jordon.Example2

10/21/2009

~ Assessment1

Yet again, this instruction is extremely helpful. Completely recommend that is you watch/purchase video one on the Gauss-Jordon elimination that you also purchase this second example, just to firm up the use & understanding of the elimination rules and row operations.

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Okay, so I thought we'd try one more Gauss-Jordan example together, where we sort of use this matrix idea, and then together with these rules of Gauss-Jordan for flipping the rows and actually multiplying through and recombining, to actually solve linear equations.
For example, let's take a look at this one . And then . Okay, well there are a lot of ways of solving this, but I want to solve this one right now using the Gauss-Jordan method. So what do I do first? Well first I would realize, and just make sort of a covenant with everybody that, in fact, this will be the x column, this will be the y column, and this will be the constant column. So then I can strip away all of the x's and y's and all that stuff, equal signs. Forget it. Who cares about equal, equal "schmequal"? Equal is overrated sometimes, because all that matters, if you think about it in that first equation, is that. And all that matters in the second equation, if you think about it, is just that. So why not just report that and forget everything else? I'll put a little thing here to remind you that, in fact, we have this.
Now what's our goal? Our goal is, at the end of the day, our fantasy would be to have this, because remember this would then be translated into 1x and 0y equals that, which means x equals that. And this would be equals that, which would mean y equals that. So by having the ones along this diagonal, and zeros everywhere else, that would tell me what x equals and what y equals by the covenant that these are x's and these are y's. So my fantasy is to massage this matrix and make it look like this.
Okay, well how would we do that? Well we have these rules. So the first thing I'll try to do is to get a 0 right over here, because remember--maybe I should just keep this here, because this is our target. So I want to get that 0 there first. Now that actually can be achieved in the following way. I can multiply this bottom thing by 2, and then add it to the top thing, and put that answer right there. So for my first move, in fact, what I'm going to do is--I suppose you could label that right here. I'm going to put in here R[1], row 1 + 2 times row 2. This just means take row 1, and add to it 2 x row 2, and put that answer right in here. So if I do that, let's see. That means I multiply this through by 2. So we have now 10, 4 and 38. Now if I now add to here, I see 10 + 3 = 13. I see 2 + -4, which is 0, that was the whole point of this part of the exercise. And here I see 38 + 1 = 39. So I actually got a little bit closer to my goal. I got that 0 there. Now this I'll just write down. I won't do anything to this row yet. Now to get a 1 here, I'll divide everything through in this first row by 13. So I divide this by 13, this by 13, and this by 13. That's equivalent to dividing and equation by 13. So if I do that, what would I see? Well, here I'd see a 1, here I'd see a 0, and . So I've got 3. So I just took the first row and divided it by 13, which is a legal move. And here I'm still not going to do anything at all. Okay, but now what will I do? Well now remember my goal is to get something of this form. So I want now this 0 here, I want that 5 to be a 0. So how can I do that legally? Well, I could do that by multiplying this top thing by 5, and then taking this -5 times that. That's basically multiplying an equation by 5 and subtracting it from another equation. That's a legal move. So the next thing I would do is--So what I'm going to do for our next move is keep the top the same, not to touch the top. I'm happy with that. In fact, that tells me immediately, in fact you realize that that actually tells me what x equals, x = 3. So now let's work on the bottom. So I'm going to multiply the top by 5. And I'll take R[2] and subtract 5R[1]. I'm going to multiply the top by 5, and then take R[2], row 2 - 5 times that. So this is 5, 0, 15. So 5 - 5 = 0, 2 - 0 = 2, 19 - 15 = 4. So it's almost to the point that I want it, but to get the point I want, I better take this whole row and divide it through by 2. We're dividing the equation by 2. This is actually saying 2y = 4. So you get y alone and divide through by 2. And if we divide through by two, I'll do that right here, well the top is going to stay the same. I see 1, 0, 3. But dividing the bottom through by 2, I see 0, 1, 2. Well that's the fantasy form I wanted. So I can just read off, if you remember our covenant, these are x's and these y's. So this says x = 3. Technically it says 1 times x + 0 times y = 3. So that means x = 3. And this says, technically, 0 times x + 1 times y = 2, or y = 2. So what I see is x = 3, y = 2 is the solution. And you can plug back in and check that that's where those two lines cross.
So neat, the Gauss-Jordan method, again, just strips away all the superfluous material, but you are under now an obligation to keep all those columns lined up. And then you are allowed now to do row manipulations using the Gauss-Jordan method, and you see your answer. Great, thanks so much. See you soon.
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Gauss-Jordan Method of Solving Matrices
Using Gauss-Jordan: Another Example Page [1 of 1]