Okay, here we are on the next slide, and we're going to determine the determinants in two equations and we're going to use the cross multiply method. All right, so let's clear this slide and we're going to talk about that. So here I have two equations, three I plus I two equal 14 and two I one plus three equals seven. And one of the another way that we can find our variables, in this case, I one and I two is by doing some, what we call a matrix or cross multiplying. So if you look Add my equation here. When we want to find dy one, and we want to find I two, we have something called the system determinant which is represented by D here.

Alright, so we have to find that first. So basically all we do is we take this matrix so notice I've got three over here, then one, then two, and then three again. All right. So we take our coefficients, and we make a matrix. That's cool. It's clear the scratch slide.

Alright, so now the first thing we want do is we want to multiply this way as I show you in this way. So, three times three is nine. Two times one is two. So I want to subtract nine minus two and get seven. This is seven is my system, determinant okay? Which is the D again in this equation alright have cleared the slide.

So now, if we look at this equation, I equals d one over D and I two equals di two over D All right. So now, I if I want to find I won I've got To find the eye. So again, we do the matrix, but this time, since I'm looking for this unknown, what I do is I substitute the answers right here in the matrix. All right, so now I use the this and I just multiply 14 times three. And then I go up this way, but I subtract seven times one. So if I go through the math, 14 times three is 42.

Seven times one is seven. I'm subtracting 42 minus seven, and my answer is 35. So now, I found my Di which is 35. And all I do is plug in so I won equals d over d. Vi over seven because we found seven up here. Okay, we found our di one is 35 we substitute up their 35 by seven equals five and there's my answer. Ah.

Now I want to find di two. And again I got to cross multiply, but notice I keep Korean two as my coefficients here. But since I'm looking for di two, I substitute here with the answers in the two equations, 14 and seven. And now all again, all I do is cross multiply three times seven minus Two times 14, so this way and I go up this way. So three times seven is 21. Two times 14 is 2821 minus 28 is a minus seven.

So now I want to find I two. So all I do is substitute, well, we know D is seven, and we just solve for di two, that's a minus seven. So I use this so I got minus seven divided by seven is a minus one. So di, sorry, I two is a minus one. So now if we want to check to make sure we're right, we just plug in our rock answers for our variables and we look Get the first equation three is one plus i two equals 14. Well, what do we know?

We know I want is five, and I two is a minus one. So three times five is 15. plus a minus one is 14. So that checks. All right, let's do the next one. Let's do this one, too. I one plus three equals seven.

Again, I want equals five. And I want I'm sorry, I two equals A minus one. So two times five is 10. Plus a mind plus three times a minus one, which is a minus three. That also equals seven. So that checks also, and guess what we can do that wicked nice website that I could have gave you in the previous discussion.

You could go in there, plug them in and see if you get the same answers. In fact, we're going to do that right now. Okay, here we are. So I go to this URL right there and select here. My matrix, it stays at two by three. So I'm going to set the matrix.

Alright, so now let's, let's minimize, not minimize, let's shrink this screen here. There we go. And try to make it so we can see it. Okay, and what and so I'm just going to put in three 213 14 and seven and hit saw. And then we go just what we got. Let me see if I can five and one right there.

Five and a minus one. Bingo. Bingo. Bingo. Bingo. So we did it right.

Okay, let's go on. Alright. Why don't you try doing this problem right here. And we want to find the system determinant dx d y, the value of x the value of y. Take a few minutes to do it, and try to do it out. I mean, you could go to that website again.

Gave you previously with the answers and plug it in. In fact, if you want, do it by hand and plug it in, and then if you want to continue, I actually do the solutions for you on the next slide, too. Alrighty. All right. So take, hit the pause button, do the exercises. We'll see you on the next slide.

All right, looking at this, give you this question here. I mean that those two equations there, the first thing you got to do is find the system determine it. So we go here and hear one times one minus one times one is the answer. So we get up, we get two. All right, then we find dx. And again, we multiply this way and then we multiply that way.

And we add two times one is two, minus one. minus A times A minus one is a plus eight. So my answer is 10. So I want to find x. So it's dx over D. So 10 divided by two equals five d y. Again, we cross multiply, one times eight is eight, minus one times two is a minus two.

So d y becomes six. And then we just plug in six divided by two is three, so y equals three. And then we can do our checks. To make sure that they work they solve the equation. And then again, I'm not going to do it this time, but you can go to that website. And you can plug in these numbers as I showed you before, and you should get you should get x and y at y three In five, so it should work.

Alright, we're going to stop here and go on. Okay, let's stop here and do some problems to reinforce what we just went over. And notice I said you can use any method. So basically, we can use the matrix method, which we went over the last slide or two, or the one in the very beginning where I multiplied the equation to get rid of one of the variables, I solve for that unknown. I found that unknown, and then I plugged into one of the equations to find the second variable. So those are the two we're talking about.

And then when all is said and done, on the next slide, I'm going to give you the answer. But then just for the heck of it, plug it into that website that I get You plug these in and see what you get. You should, everything should match. Everything should check. Okay, so let's stop here, do the problems. And then when you continue, you will see the answers.

All right, here are the answers to the equations. Like I said, use any method, but no matter what method you use, I mean, you got to get the same answers. All right? I mean, just because you use one method, and then another method, you shouldn't get two different answers. So again, use the method I mentioned to you about in the previous slide, you can plug it into the on that website that I gave you. And that's pretty much yet.

So we're going to conclude here on this lecture. We've got one more section, or one more lecture in this section, I should say and Again, we were, we're giving you something on simultaneous linear equations. But we can go much, much deeper. And that's not what this course is about this course is to give you an overall understanding of mathematics. So with that said, I'm going to wish everyone a wonderful time. And we're going to end this lecture and we'll see on the next one