 Welcome to our lecture series linear algebra done openly. This is a video series based upon the typical topics covered in a linear algebra course following the open-source textbook of the same name linear algebra done openly. This book has been written and curated by myself, Dr. Andrew Missildine of Southern Utah University. And the story behind this book in this series is actually quite a fun one here. Over the years, in many semesters, I've been teaching linear algebra at Southern Utah University. At SUU, the class is called Math 2270. Over the years, I've been developing lecture notes for the class that have eventually evolved into the textbook connected to this lecture series here. But one thing that kind of makes this open-source textbook very different than other either commercial textbooks or open textbooks you might be able to find, not just on linear algebra but on any topic really, is that this textbook is actually partially developed by the students themselves. And while the original source for this course, of course, was based upon my own lecture notes, over the years, students have contributed week by week, little bit by little bit, new examples, new homework questions, new explanations, new graphics, and many other types of contributions they've made to this textbook that's evolved into something greater than I would have ever been able to produce on my own. And so these videos, this is the first video of the series that'll be a companion to the the textbook itself. If you're interested in reading the textbook, you can find a link to the textbook in the description of this video below and you'll find that link in all of the videos inside of this series. And also, if you're at all interested in contributing this textbook at all, feel free to reach out to me and I would be glad to accept anyone's contributions, not just my own students. And so this is the first video of section 1.1 in our textbook entitled Linear Systems. At the very beginning of linear algebra, we're going to talk about the idea of a system of linear equations. So what does one mean by a linear equation? Well, we'll actually talk a lot more about that in the next section, section 1.2. But for the current situation, by a linear equation, we mean an equation of the following form. We have, I have the equation a1x1 plus a2x2 plus a3x3 plus a4x4 all the way up to anxn and this equals b. Where n here is just going to be some positive integer. So we have some number of variables and the x's here are going to be our variables x1, x2, x3 up to xn. We potentially could have any number of variables in a linear equation. Now for the most part, when we actually do these calculations by hand, we'll have two variables, three variables, four variables. We really won't be ever be doing like a thousand variables in a single equation, only because we're students and we want to at some point, you know, maybe do something other than our linear algebra homework. But be aware in practice, it's very possible that we could have systems with thousands, if not millions of equations and maybe variables. But again, I get a little ahead of myself here. These x's will represent variables. That is, these are placeholders for numbers to be placed in their stead. They're just substitutes for what the numbers potentially could be. As the name variable suggests, they are able to vary. On the other hand, the numbers a1, a2, a3 up to an, these are commonly referred to as the coefficients of the linear equation. These are going to be fixed numbers. For the purpose of this section by a fixed number here, we mean a real number. Then on the right hand side, there'll also be some other number b. Not every linear equation is written in this form, but an equation is going to be a linear equation if it can be written in this form, if it's equivalent to this form. By a system of linear equations, we have a set or a collection of equations which have common variables. We'll see some specific examples on the next slides. When one has a system of equations, you usually start off with some curly brace. You have the first equation, second equation, third equation that you'll write there. When we look for a solution to the system of equations, what this means is we have an assignment. So x1 equals something, like 1, x2 equals something like a 3. x3 equals something like a 0, etc. We have an assignment for each of the variables for which we're going to plug those assignments into the first equation, into the second equation, into the third equation, and we do this for all of the equations that are present in the system. Now to be a solution of the equation, the assignment of the variables has to make the left hand side equal to the right hand side. We often say that it solves the equation or the equation is satisfied by that assignment. So a solution to a system will be a solution to each and every equation in the system. Again, we'll see an example of this in just a moment. The solution set to a system of equations will be the set of all solutions to the equations. And as we study systems of linear equations, or sometimes called linear systems for short, as we study linear systems, we're often going to want to know what the solution set is, the set of all solutions to said linear system of equations. And another bit of vocabulary, we say that two systems of equations are equivalent if they have the same solution set. So the equations could be different, but if the same set of numbers solves one and the other, we would say the two systems are equivalent. So consider the following example, we have a system of two equations. There's two equations and two unknowns. On the previous slide, we often, you saw that the variables were called x1, x2, x3. This is sort of a generic way. When we have a small subset of variables, typically we'll use, not subscripts, we'll use alphabetic letters like x and y. So if you have two variables, we'll typically call that x and y. If we have three variables, we'll often call that x, y and z. If there's a fourth variable, maybe w. But once you get more than four variables, you're typically going to start using that subscript notation like x1, x2, x3, x4, x5. But we might even use that subscript notation even when we have a small set of equations of variables, I should say. So this right here is a system with two variables, x and y, and has two equations. And so commonly this is what we refer to as a two by two linear system, a two by two system here. And one convention I should mention for you is that we're always going to list first the number of equations, and then we list second the number of variables. Now in this situation, the number of variables and equations is the same. So we mixed it up, we'd be right for the wrong reason. But when we talk about an n by an n by n system, that would be say we have m rows and n variables, and we'll talk some more about that later on. So is we're supposed to test for this system of equations is the is the assignment three one a solution to this linear system. But what that means is you're just going to plug three in for x and one in for y into the first equation. And when you do that, you get three, if you replace the x with a three, five times one, we replace y with a one. And if you simplify the right hand side, you get three plus five, which is equal to eight, you'll notice that this is the right hand side. So sure enough, x equals three, y equals one, is an assignment that satisfies the first equation. Three one is a solution to the first equation. But to be a solution to the system, we have to also check that it satisfies the second equation. If we plug in three comma one, in that situation, well, you'd get two times three for x minus one for y. You see two times three is six minus one is going to give us five, that likewise checks out for the for the second equation. And so we see that we do in fact have a solution to the system of linear equations there. Now, this turned out to be the only solution to the system of equations. If we tried a different solution, for example, sorry, different assignments, like if we took zero comma three, well, you look at the first equation, you get zero plus five times three, that's equal to zero plus 15, which is equal to 15, that's not equal to eight. And so that would violate the first equation right there. So that would be not a solution. But be aware that to be a solution, we have to make sure that we satisfy all of the equations. And so even if we had something like the following, if we took eight comma zero, you'll see that when you plug it into the first equation, you get eight plus five times zero, which is equal to eight plus zero, which is eight, that checks out. If you plug into the second equation, though, you get two times eight minus zero, you're going to get 16, 16, sorry, which is not equal to five. And so we don't have a solution at eight zero because it didn't work for all of the equations. We have to make sure we get all of the equations satisfied to be a solution. And a geometric explanation of what's going on here is that if we think of our two equations as graphs in the plane, right, we could graph these things. The first equation remember was x plus five y equals eight. If we were to put this in slope intercept form, you get five y is equal to negative x plus eight, I just minus x from both sides, divide both sides by five, you get y equals negative one fifth x plus eight fifths. And so what you see there is a line whose slope is negative one fifth. That is, every time we go down one, we'll go to the right by five. And it's one intercept is eight fifths, eight fifths is a little bit less than 10 fifths, which would be two. And so you see this line right here in yellow. Its y intercept would be a little bit less than two, and it's going to go down one over five. Like so. The other line, the other equation remember was two x minus y equals five. If you put that in slope intercept form, you'll see negative y equals negative two x plus five, and times both sides by negative one y equals two x minus five. This one has a y intercept of negative five, which would be down here, a slope of two. So you go up two over one, connect the dots, you would get this green line that you see here on the screen. And these two lines intersect at a unique point, which is this point right here at three comma one. So geometrically what we're seeing here when we solve the system of equations, we're looking for a point which is on the first yellow line, which would be any of these points right here. But we also need to find a point which satisfies the second equation, which geometrically would be a point on that line right there. So we're looking for those for a point there. So geometrically as we try to solve this system of linear equations, we're looking for a point which is simultaneously on the first line and on the second line. And when we look at the graphs here, we can see that there's only one point that'll live on both of the lines. And that was this point three comma one. One could find that geometrically if we drew this thing well enough. But it turns out an algebraic technique of solving will be better. And we'll see that a little bit more in the future. Let's take a look at an example with three variables. This is now an example of a three by three system, three equations, three unknowns. We have the variables x, y, and z right here. It can be checked very quickly that the point two comma three comma four is in fact a solution. And we can see this by checking the arithmetic here if we plug in two for x three for y four for z into the first equation we'll get two plus two times three plus three times four. So we get two plus six plus 12. Like so two and six is an eight plus 12 is 20 that then verifies the right hand side. If we check the second equation, you're going to get negative two replace the x with the two, replace the y with a three. And you'll notice in the second equation there is no variable z right now this is still a three by three system there's three variables but it might be that some of the equations actually are absent of one of the variables that's okay. You won't have to plug anything in there you can think of it as there's a coefficient of zero in front of your z value there. We can just ignore it. So negative two times two is a negative four plus three is a negative one that checks out for the second equation. And then lastly for the third equation we get negative three times two for x negative six times y which is three now and five times z which will be a four. So we get negative three times two which is negative six negative six times three is a negative 18 and then lastly you're going to get five times four which is 20. If you take 20 and 18 that gives us a positive two negative six plus two is equal to negative four and that checks out right there. And so we can see that our our assignment two three four it's satisfied all three equations and this is a solution to the system of equations. And one if one were to try to find another solution you're not going to find one turns out this is the unique solution to the system of equations much like the previous example we saw that there was only one solution. Now if we looked at one third example of this let's take another three by three system here if we were to try we could we could actually find a couple of we can actually find multiple solutions this one. So for example if we take five negative one zero you plug that into the first equation you're going to get two times five minus y times negative negative negative y which y itself is negative one do be cautious there you want to make sure you get the double negative you're subtracting negative one and plus zero for z you'll get two times five which is 10 plus one that's equal to 11. So that worked out for the second equation you're going to get five plus three times negative one uh plus zero 10 times zero zero uh so you end up with five minus three which equals two that checks out and then lastly we get negative three times five plus two times negative one minus zero you'll get negative 15 minus two which adds up to negative 17. So we see that this is in fact a solution but we actually could try other values as well this one has multiple solutions a second solution you're going to see here is six to one if we put this into the first equation we get two times six minus two plus one that equals 12 minus two plus one 10 plus one that gives us 11 that works for the second equation we get six plus three times two minus 10 times one so you're going to get six plus six minus 10 that will be 12 minus 10 which is equal to two so that checks out and then for the last one you're going to get negative three times six plus two times two minus three times one and you see that turns out to be negative 18 plus four minus three you're going to get negative 18 plus one which is negative 17 again and so this one six to one is in fact also a solution to the system of equations so it is possible for a system to have multiple solutions in fact in this in this example one could actually prove that there are infinitely many solutions to this system of equations and we'll see in a later video exactly how one might do something like that