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Video transcript

We've done a lot of work on multiplying, adding, subtracting and inverting matrices. So now let's delve a little into what a matrix is actually good for. And remember, all a matrix is is, a way of representing data. And all of those rules we learned, you can kind of view them as human-created rules. There's no fundamental thing in nature that says matrices have to be multiplied the way we learned. But I think you'll see as we progress into applications, that the way that matrix operations have been defined are actually quite useful. So let's go back to our Algebra 1 or Algebra 2. I forget when you tend to learn it. But let's go back to linear equations. So what were linear equations? Systems of linear equations. Well you had two lines, and you essentially had to figure out where the two lines intersected. So you might have had something like-- let me think of something-- 3x plus 2y. Is equal to 7. And then you might have, minus 6x plus 6y is equal to-- I need to do this in my head just to make sure that I get numbers that work out well-- equal to 6. I think this will work out well. And what was this problem essentially? Well this is a line, and this is a line. So you had to figure out where they intersected. And if you were to draw those two lines-- Actually let's draw them. Just because this is all about getting intuition, and seeing how it maps over into the matrix world. And the word 'matrix world' has a whole new meaning after 1999. Let's see, so if that's my coordinate axes, what is this? I always have to put everything into y equals mx plus b form for me to-- So this equation is what? It's y is equal to 3/2 x plus 7/2. So 7/2 is what? It's like 3 1/2 or something? So if that's 7/2, that's going to have a slope of 3 1/2. So it's a little bit steeper than a slope of 1. So it's going to look something like that. That's that line. And then this line is going to look like what? I'll do a different color. It's going to look like-- It's the same thing is as-- Oh, you know what? I did that wrong. Because that line, I just realized, is equal to minus 3x plus 7/2. Because when you take this to the other side, it becomes minus 3x divided by 2, so it's actually going to be downward sloping. So it will look something like this. It's going to be a little bit steeper than something that has a slope of negative 1, so I'm just approximating. So that line will look something like that. And this line, this will be y-- I'm just rewriting this-- y is equal to x plus 1, if I'm right. Yeah. Because this go to the other side. Divide everything by 6. y is equal to x plus 1, so its y intercept will be-- We said this was 3 and 1/2, so maybe if this is 1. And it has a slope of 1. Then it'll just look like something like this. And so when you solve a system of equations, you're essentially looking for the x and the y values that satisfy both of these equations. This magenta line shows us all the x and y values that satisfy this first linear equation. And this green line shows all of the x's and y's that satisfy the second equation. And of course where they intersect shows us the particular x and y that satisfies both equations. So that's what we did in Algebra 1. We'd solve both of these equations for that. And we'd either do it by substitution, or we'd scale them and add them together, et cetera, et cetera. As you'll see, that's essentially just what we learned in the Gauss-Jordan elimination. It's the exact same thing. It's just when we did the Gauss-Jordan elimination, we just represented it a little different. But I think you know this much. But let's now do it in the matrix world. So how can we represent this problem as a matrix? We could write it like this, and we'll take out a little time to prove to you that it really is the same representation. If you define matrices the way we have defined them in their multiplication. You can define this problem as 3, minus 6, 2, 6. I just took the coefficients, 3, minus 6, 2, 6. And if I were to multiply that by soon. column vector matrix xy. And if I were to set that equal to another column vector matrix, 7, 6. Now you might want to pause it and actually just try to multiply this out, the way that we have learned to multiply matrices. And you will see that you get the same thing. But I will do it now, in case you don't want to do it yourself. So let's just multiply these two matrices. Let's multiply this matrix out and see what happens. So what do you do? You get your row information from the first matrix, column information from the second matrix. And this is of course the product matrix. So this is saying, 3 times x plus 2 times y is equal to 7. Well that's exactly what we wrote up here. 3 times x plus 2 times y is equal to 7. And similarly when you multiply the bottom row, you get minus 6 times x plus 6 times y is equal to 6. So if that was a little confusing to you, go review how we multiply matrices. But if you just multiply this out, you'll get these exact same equations. So hopefully you understand that this is just another way of representing this problem. Although we've gotten rid of the plus signs and the equals signs. But of course you have to know the representation. But why is this useful? Why is this representation useful? Well let's call this matrix a. Let's call this vector x. It's not a variable. It's a vector. So maybe we'll bold it, or we'll put a little vector sign there or something. Whatever. But you'll see it in your textbook. It's bolded real heavy. And then we call this vector b. And the general notation-- if I remember it correctly-- is that anything that's a matrix or a vector is bolded. And matrices that are not vectors, that have more than one dimension in either of the dimensions, they're capital letters. While lower-case letters represent vectors. So these are matrices, but they're also vectors. So that's why they got the lowercase letters. And that's why this one got the uppercase letters. That's just convention. So this equation has the form ax equals b, where a is this matrix, x is this vector-- or this matrix, same thing-- and b is this column vector. So what does that do for us? Well, what happens if we knew a inverse? Well actually, let me take a step back. If these were numbers, what would we do? If I just gave you an algebra equation, ax is equal to b. How do you solve that? Well you would divide both sides of this equation by a. Or another way of saying it, you would multiply both sides of this equation by the inverse of a. So you would essentially say, 1/a times ax is equal to 1/a times b. And then these would cancel out, and you would get x is equal to b/a. That's how we would do it in a traditional simple, linear equation. So how would you do it here? Well what's the matrix analogy to division? And I'm going to give you the answer now. What's the analogy to multiplying by your inverse? Well, it's multiplying by your inverse. So what if we knew the matrix a inverse? We could just multiply both sides of this equation by a inverse. And remember, order matters. So it's not like when you're doing a linear equation you could multiply 1/a on this side. But then you can do it on the right side here. But no. Notice, I put it in front of the numbers in both cases. So you have to do it in front of the numbers in both cases. But if we know a inverse, and if a inverse exists, then we can multiply both sides-- you can say the left side of both sides of this equation by a inverse. a inverse times a, times the vector x is equal to a inverse times b. All I did is I took this expression, and I multiplied both sides by a inverse. And what's a inverse times a? Well that's just the identity matrix. That's the identity matrix, times x is equal to a inverse b. And of course that's just x. The identity matrix times any other matrix is just that matrix. So that's just the matrix x, or the vector x times a inverse b. So, if you're given a linear equation, and if you know the inverse of this matrix, to solve for x and y, we just have to multiply this number times the inverse. And you might say, Sal, that's such a pain. Because this is such a simple linear equation to solve. Why would I go through all the trouble of getting an inverse, and then multiplying the inverse times this number. And I would agree with you to some degree. That for a 2 by 2 system of equations, it is easier to solve it the way that you did it in Algebra 1 or Algebra 2. But if you're doing it for a 3 by 3, well, finding a matrix is still pretty difficult for a 3 by 3, so it's still difficult. But as you get to larger and larger numbers, it's sometimes-- well, finding a matrix can be difficult too-- But actually the real place where it you really, really pays off is, let's say that you have a bunch of linear equations to solve. And the left hand side stays the same. But you keep changing the right hand side. So let's say you have ax equals b. And then you have another one that says, ax equals c, and ax equals d. And these numbers keep changing. And these numbers are the same. Then it really pays off to solve for the inverse. And then every time you need to find a new solution, you just multiply your new right-hand side times your inverse, and you just get the answer. And that'll really pay off when we view this in another way. But anyway, I wanted to show you that this is the same thing. And so let's solve it using what our knowledge is of matrices. Let me erase this here, and I know I'm running over time, but hopefully I'm not completely boring you. So I'm going to keep that there, just because I think it's nice to have that visual representation of what we're doing. Always to remember what's going on. So, what's a inverse? So first of all, the a inverse is equal to 1 over the determinant of a times the adjoint of this matrix. I don't want to get fancy with terminology and all that, but what was that? 2 by 2 is fairly easy. You swap these two terms. You get a 6 and a 3. And then you make these two terms negative. So a minus 6 becomes a 6. And a 2 becomes a minus 2. And what's the determinant of a? The determinant of a is equal to this times this minus this times this. So 3 times 6. 3 times 6 is 18 minus this times this. So 6 times 2 is 12. That's a minus 6. That's minus 12. So minus minus 12. It's equal to plus. So 18 plus 12 is equal to 30. So what does a inverse equal? 1 over 30 times this thing. So a inverse is equal to-- we could even keep the 1/30 on the outside. That might simplify things. Well actually I'll put it-- So a inverse is equal to what? This divided by 30. So that's 1/5, minus-- Actually I do want to keep it on the outside, because that's going to make the later multiplications easier. So anyway, a is equal to 1/30 times 6, minus 2, 6, 3. That's a inverse. So now let's solve for x and y. So we said x and y is equal to a inverse times b. So we could say x-- another way to write x is like this. x is just this vector. x and y. Not to get confused, this x is different than that x, even though I've written them the same. If I was a typographer, I would make this really bold, so that you know that this is a vector. Maybe I should put a vector notation. I don't know. You could do a bunch of things with it. It's equal to a inverse times this. So that's 1/30. I did that just for the matrix addition. I didn't divide everything by 30, just so the matrix multiplication's a little easier. Minus 2, 3, times 7/6. And so what is this equal to? It's equal to 1/30 times-- I know I'm crowding this down here-- let's see. 6 times 7 minus 2 times 6. So 6 times 7 is 42. Minus 2 times 6, so minus 12. So that's equal to the 30. And then 6 times 7 plus 2 times 6. So 6 times 7, once again is 42. Plus 2 times 6. So 42 plus 12 is 50. Is that right? 6 times 7-- oh I'm sorry. This is a 3. That's why I was getting confused. See, it's important to have good penmanship. So it's 6 times 7 is 42, plus 3 times 6. So it's 42 plus 18, which is 60. And of course you divide both of them by 30. So you get the final xy. I'll write it here. I don't want to erase anything. So we get xy is equal to-- divide both of those by 30-- is equal to 1 and 2. And so that tells us that these two linear equations intersect at the point x is equal to 1, y is equal to 2. That might seem a little bit like a lot of work, but that's just because I took the time to explain it and all that. But if you just immediately took that, represented it this way, found the inverse, and multiplied, it wouldn't have taken you that much time. And I encourage you to do that as an exercise. Anyway, I'll see you in the next video. And in the next video, we're going to do this exact same problem, but we're going to see that this data represents a different problem. See