Okay, hey, back again in this next lecture. All righty. In the, we ended up on the last lecture, we talked about determinants for two equations. Now we're going to do three equations. And basically the technique is the same. But it gets a little bit more complicated.

Well, I shouldn't use the word complicated. You've got to be a little bit more careful about your signs and your matrix. All right, because we've got one more variable, you just got to make sure things are lined up nice when you're doing it by hand. So again, nothing's changed as far as we can see. We're going to find the system determinant. Number one, and then we're going to find da di d di c just like we did previously, only there was two.

Now we've added a third one and then finally we're going to find the unknowns. In this example, ay ay ay ay ay ay b. And I see. All right, so with all that said, let's go on to the next slide and see how we do this. All right, here we go. So we've got our equation up here.

Our equations I should say, and we've got to create and create our matrix. So 15 minus five. This is assumed to be zero. There we go. minus five plus 20 minus five. There you go.

Okay. assumed to be zero minus five and 15 a go. So there's our matrix. And then what we do is, since we've we've got a multiply, we'll take the first two on the left and repeat. So first two on the left. Bingo.

Bingo. minus five minus 520 20 minus five, minus five, and then 1515. Again, minus five, minus five, zero, and zero. All right, so we created a matrix. All right. Now what we're going to do, the next step is we're going to multiple To apply diagonally starting from the left, so we're going to start start at the top part, the top, left hand, upper left hand corner of the matrix and go down just like I'm showing you here.

And you notice all the colors, all those numbers are in red. All right? And we're going to multiply them. Alright, so I show you here 15 times 20 times 15. Okay, and then we're going to add them, but let's, let's, let's not go there just yet. All right, so now we'll do the next one.

And we start over here, and I put them in green for you. So it's minus five times minus five times zero, and then we'll do the third one. Okay, cleaned off the slide. Now we'll do the third one, and we've got zero times a minus five times a minus five, right there. So we've done these in brackets for ones that I've checked on top. So now what we're going to do is we're going to subtract, see the subtraction right there, we're going to subtract.

So now what we're going to do is we're going to multiply, but we're going to multiply the opposite way. All right, and let me explain to you what it is. So what I'm going to do is I'm going to change the color on my stylus. So we're going to go over Here, right there. And we're going to multiply this time we're going up. We've got we're starting at the lower left hand corner, and we're going up to the right, so we're going this way.

Zero times 20 times zero. Then we're going to do, we're going to move over and we're going to do the next one. Okay, clean this screen off, and we're doing this one right there again, and we go up this way. On the next diagonal row, I'll call them and I have a minus five, minus five and a 15 minus five, minus five and 15 right there. And then we do the last one. One, which is right here.

And we have 15 minus five, minus 515 minus five minus five. Okay, now we're going to multiply them. Alright. So let me stop here, clean the slide now explain that. Okay, so again, if you remember the rules that we talked about originally, we've got a bracket around each one of these, right is my bracket. Here's my bracket right there.

And right there, show you another one over here, right there and right there. So there's a bracket around each group of numbers. So we want to do that first. So now we break it down. So this one here, when do that multiplication, it comes out to be 4500. All right, when I do this multiplication right there comes out to be zero because zero times any number is zero.

And so doesn't this one right here, okay, that also is zero. All right. So now I go over here and let's clear the slide again. I go over here. Remember we want to subtract. So again, zero times any number right here is zero.

All right, five, minus five. Okay, I cleaned the slide off, and again, we're going up this way. Here. All right, again, we got the brackets and we're multiplying I. And now we've we've, we're going to go up this way here, minus five times a minus five times 15. And then this last one right here plus a minus 15 times a minus five times a minus five.

So now we're working. We've already done this. So 4500 plus zero, plus zero minus zero, right there, because I've got a zero here. And when I do the math here, I get 375. He And 375 here. Remember, when I have two negative numbers and I multiply them, it becomes positive.

So they're positive 375. So now I've got 4500 minus a positive 750. And when I do the math, my answer is 3750 or 3750. Okay, on the next step here, we're going to find da. Alright, notice I put determinant system up there, which we found in the last slide, which was 3750. Now we're going to find da and again, the technique is very much the same.

All right, we're going to start here. Okay, we built up matrix so we're using the same matrix. Okay, but here's the difference. We want to find da. Okay we use da. To solve for ay ay ay ay ay at the very, very end.

So now what I do is I take my solutions on the three simultaneous equation, and I substitute them for what's in the i a column. So for instance, 16 04, so I got 16 04. Now I go on to this column on my three linear equations, and you'll see that I got 520 minus five and then minus five and 15 right here. Okay, just to repeat what I said previously, okay, I have as if I, there's nothing there then that's represented as a zero. And then again, I take these two columns and repeat them as I show here. Okay.

Let me clear the slide off. All right, so I'm going down this way. Again, just reviewing and making sure that you understand. Since I'm looking for ay ay ay ay ay ay, ay, substitute these numbers in this column here. Okay. Now all I've done is I've reproduced this, if this is the same column, and I've just brought this over here, just for demonstration purposes, and you'll see Why, if you, if you keep, if you have good management techniques, when you solve these, if you do them by hand, you don't need to do that.

But I did this for showing you should represent, you know, displaying on how to do this. So that's why I've done that. So now what we're going to do, as I show you here in red, okay, is we're going to multiply the numbers down 16 times 20 times 15. Then I go to the next one minus five minus 15 times four. And then the last group of numbers 00 times five, which is here. All right.

Now what I'm going to do is is I'm, again, I'm just showing you over here, I'm going to subtract all right I'm going to subtract, and I'm going this way. So four times 20 times zero plus minus five times minus five times 16 times 15, zero times five right there, and that's going to equal da. So let's go to the next one and simplify it. Okay, so now I'm starting over here because remember, we're going to subtract these two. So from here back, I'm gonna add these up. So I've got 48 100 and zero.

All right, and then over here, I hate when I multiply this. I have have 04 and zero. Okay. So now all I do is just basic math 49 minus 400 and da equals 4500. That's it. Okay?

It's not difficult. The hard part about something like this is keeping things in order. And be careful about your signs if they're negative numbers in there with the brackets. Okay? All right, let's go find the IB. Okay, we're gonna find the IB now.

And everything's the same except for one thing. Okay, what are we looking for? We're looking for IB, aren't we? Alright, so now I substitute the answers that are in my three simultaneous equations for the coefficients for IB. All right, and since we over here also, but we spoke about this in the last slide. All right, so now, again, I do my matrix 15 minus five as soon to be zero.

Since I'm looking for IV, I substitute this in here. Zero appears zero minus 515, zero minus 515. Here they are. And then again, arm hygiene. Just take the two columns and just bring them over there. Okay, so that's it.

And I've done that on this one. Again, I spoke about that before, I don't mean to repeating myself, but that this is done here just for demonstration purposes for this lecture. So let's clear the slide. All right, and again, the technique is the same, I'm going to multiply down 15 times zero times 1516 times a minus five times zero and then zero times five times four. Right there. Now we're going to subtract.

All right, and we're going to use the other matrix and we're going to go up the other way. Here we go. All right. That's it. That is it. All right, so now, we're going to do math.

So we're going to simplify this. And since we have a zero, in each one of these groups of numbers, they're all going to be zero. So I'm going to get zero plus zero plus zero, because as you know, zero times any number is zero. Now I'm going to subtract the other group So I have zero times zero right here, zero times zero times 00. When I do this math here, I get a minus 300. And then I do this and I get a plus 1200.

All right. So what happens? Minus that becomes plus. And so therefore, 300 plus 1200 is 1500. So the IB is 1500 right there. Okay.

I, let's clear the slide and do the last one. Okay, we're trying to find di C here. So everything's is saying Except What do I do? I'm looking for I see. So now I substitute this right in this column here. All right, so now again, I multiply going from the upper left hand corner to the right.

So I got 15 times 20 dot four right there, times four rather. Then I got five times zero times zero. And then the last 116 minus five minus five, right here. Now I'm going to subtract and I'm going to skip track. Okay, but instead of going from the upper left to the right, I go from the lower left to the upper right. Right there.

Right there. And right there, okay. Now I combine them, which I'm showing you here. 1200 plus zero plus four is 1600. And then I'm going to subtract zero plus zero plus 100. So, excuse me.

1500 Equals D, di C. And here are my answers so far. All right. All right, he went on quick on this one because it's repetitive. Again, the only thing different is when we shift this group of numbers from the different variables that we're going to be solving for. Okay. All right.

Let's stop here, and go To the next part of this where we find ay ay ay ay ay ay ay. ay b and IC