 Welcome back to our lecture series Math 2170, Applied Linear Algebra for Students at Southern Utah University. As usual, I'll be your professor today, Dr. Andrew Missildine. In this lecture, we're going to talk about section 8.3 from the textbook. We're going to introduce the idea of systems of linear equations. We've really been talking about things that are surrounding this idea of linear equations, but now it's official that we should talk about them. And so, what have we talked about recently in chapter 8 here? Well, we've learned about linear transformations. We've also learned about affine transformations, but focusing on linear transformations. We've mostly focused on the following problem. We have this linear transformation from, call it T, it goes from Rn to Rm. And the idea is, if I put some vector x into the machine, it transforms it and produces a T of x. That is, you're given some type of input and then you produce some type of output from it. Like so, that's what our transformation does. The question that we really want to focus on now is going the other way around because generally speaking, if you have a function, I give you an input, we can calculate the output. That's fairly simple stuff. But what if we want to go the other way around? What if we know the output, right? What input, if any, actually maps on to that point? That's the question we want to investigate right now. So suppose we have some vector b, and so this will belong to the co-domain, Rm. So b is some vector in the target space. And we're curious, is there some vector x, which, if applied to the transformation T, gives us that specified vector b? Now, if T is a linear transformation, then we know that it can be represented by matrix multiplication. So we're trying to solve this equation T of x equals b, but that's equivalent to trying to solve the matrix equation ax equals b, right? So is there some vector x, which if multiplied by a gives you b? So in some essence, we're trying to factor the vector b using a fixed matrix a. Now, suppose that our matrix a has the following column vectors. We have a1, a2, all the way up to an. And we should mention throughout all this that a, of course, is an m by n matrix in this consideration, right? So the column vectors of a are a1, a2, up to an. And let's also suppose the vector x has as its entries x1, x2, all the way down to xn, like so. Now, if we take, if we take, of course, this matrix with these column vectors and this vector with those entries, then multiplying the matrix by the vector is equivalent to this linear combination right here. Remember, the product of a matrix by a vector is the linear combination of the column vectors of said matrix, using as the coefficients the entry of the vector. So a times x is the same thing as x1 times a1 plus x2 times a2, and adding all this way up to xn times an. So this, this matrix equation is equivalent to this vector equation. And we have to solve this vector equation, all right? So just stepping along through these definitions here. So the next step here is actually to recognize what are these column vectors look like. So for ai, so this is the ith column in there, we'll say that the first entry, the first entry is going to be a1i, then we have a2i all the way down to ami, like so. And so we'll use a double index to talk about the the entries in these column vectors. So if we look at this vector combination up here, this linear combination. So if you look at just the first entry, right on the left hand side, you're going to get x1 times a11, then you're going to get x2 times a12. And this goes all the way down to xn, a1n. Okay, then the second entry looks something like x, excuse me, x1, a21. And then we get an x2, a22, all the way down to xn, a2n. And then this pattern is going to continue and continue and continue and continue until the very end, you have an x1, am1, you're then going to add to that x2, am2. And then you proceed all the way down to xn, amn. So if you were to combine all of those vectors on the left hand side into one single vector, you get this thing. Then on the right hand side, we still have the vector b, but let's expand it as well. Let's say the first entry of b is b1, then b2, by our usual convention, of course, bn, right there, excuse me, bm. There's m entries there. And so we have this vector equal to that vector. Well, the only way that two vectors are equal is that their corresponding parts are equal to each other. So it's got to be that the first entry of this vector agrees with this vector right here, with that entry. And therefore that gives us an equation. It's in fact a linear equation, x1, a11 plus x2, a12, all the way up to xn, a1n, that equals b1. And we do that for each of the coordinates. So we look at the second entry on the left, the second entry on the right, they have to be equal if the vectors are equal. So that gives us a second linear equation. And this proceeds all the way down until we get to the bottom right there. And so this right here in the blue box now is what we call a linear system of equations. I should say a system of linear equations or just a linear system for short. It's a system, it's a list of equations and each of them are linear equations. Now there's multiple variables perhaps in play here, because the b vectors are fixed. That's the target. And the numbers a are also fixed, all right? And in which case it's the x's that we don't know. So we have these equations with x's. Now since the a's are known, these are actually coefficients. And so most likely we would actually put them in front of the variable, in which case we get something that looks like the following. A linear equation is an equation that looks like the following. a1 times x1 plus a2 times x2 plus all the way up to an times xn equals b. So that's just a single linear equation with n variables. A system of linear equations that means we have some list of linear equations with common variables. The numbers a are coefficients. We have this number b over here. And we're trying to solve for that. Now to find a solution to a system of linear equations, we're looking for an evaluation of the variable that simultaneously solves all of the equations in the system, right? So it's solution for the first equation, a solution for the second equation, solution for the third equation. That gives us a solution to the system of equations. The solution set is in the collection of all the solutions. That's all the points, all the vectors x that satisfy every equation in the system. And we say that two systems of equations are equivalent if they have the same solution set. You'll often see me draw a little twiddle sign between the two systems that say they're equivalent. That is, we've changed the system, but the solutions didn't change. And that's really what we care about, all right? So, I mean, it should be mentioned that there's a natural geometric interpretation of systems of linear equations because linear equations have geometric interpretations, like take the equation x plus 5y equals 8. Now normally in this course, we call the entries of the vector here x1, x2, all the way up to xn, right? But for this situation, to make the geometry more explicit, I'm going to use the traditional x, y, and z for coordinates in the planar in three-space. So, if you take the equation x plus 5y equals 8, you might prefer it in the more conventional slope-intercept form. So, y equals negative one-fifth plus eight-fifths, something like that. So, we have a, I forgot the x there, whoops, an az. So, we have an x-intercept, y-intercept, excuse me, of eight-fifths, something like this. You have this negative one-fifth slope, go down one over five, something like that. And so, that's illustrated here in this yellow line on the screen. So, we can graph this thing. And similarly, the equation 2x minus y equals 5, we can graph that thing and we get this equation, the equation of the line gives you this picture of the line. This is actually why we call them linear equations, that in two dimensions, the graph of a linear equation, that is two variables here, is a real McCoy line. And so, as we try to find the solution to the system, we have to investigate, what are the solutions of this line? Well, that's just, the solutions of that equation are just the points on this line, everything we draw on the yellow line, that's a solution to the equation. Similarly, with the 2x minus y equals 5, the solutions to that equation are the points on this line. So, the graph is the solution set. Now, a solution to the system of equations is going to be any pair, any point that solves both the first and the second equation, right? So, we're looking for points which are on the yellow line and on the green line. So, solving a system of equations, it's geometrically looking for the intersection. So, geometrically, we can see that the intersection between these two lines is going to be 3, 1. I want to verify this algebraically. If we put 3, 1 into the first equation, you end up with 3 for x, 1, 4, y. Let's try to simplify this thing. You're going to get 3 plus 5, which is in fact equal to 8, which is the right-hand side. We have that the right-hand side is equal to the left-hand side. And therefore, we have that 3, 1 as a solution to the first equation, all right? Well, we have to check the second one, too. What happens when we plug in 3, 1 into the second equation? Well, the left-hand side is going to look like 2 times 3 minus 1, all right? 2 times 3 is 6, minus 1 is a 5. That is the right-hand side. And so, we've then verified that 3, 1 is then a solution for the second equation as well. And that's what we saw geometrically right here, all right? That we have a point on both of the lines. It simultaneously solves the system of equations. It simultaneously solves both equations. So that's our solution. And as we can see geometrically, there's only the one solution. There's only one point that lives on both lines.