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Main content

Current time:0:00Total duration:14:14

We've learned about matrix
addition, matrix subtraction, matrix multiplication. So you might be wondering,
is there the equivalent of matrix division? And before we get into that,
let me introduce some concepts to you. And then we'll see that there is
something that maybe isn't exactly division, but it's
analogous to it. So before we introduce that, I'm
going to introduce you to the concept of an
identity matrix. So an identity matrix
is a matrix. And I'll denote that
by capital I. When I multiply it times another
matrix-- actually I don't know if I should write
that dot there-- but anyway, when I multiply times
another matrix, I get that other matrix. Or when I multiply that matrix
times the identity matrix, I get the matrix again. And it's important to realize
when we're doing matrix multiplication, that
direction matters. I've actually given you some
information here that-- we can't just assume when we were
doing regular multiplication that, a times b is always
equal to b times a. It's important when we're doing
matrix multiplication, to confirm that it matters
what direction you do the multiplication in. But anyway, and this works
both ways only if we're dealing with square matrices. It can work in one direction or
another if this matrix is non-square, but it won't
work in both. And you can think about that
just in terms of how we learned matrix multiplication,
why that happens. But anyway, I've defined
this matrix. Now what does this matrix
actually look like? It's actually pretty simple. If we have a 2x2 matrix, the
identity matrix is 1, 0, 0, 1. If you want 3x3, it's 1,
0, 0, 0, 1, 0, 0, 0, 1. I think you see the pattern. If you want a 4x4, the identity
matrix is 1, 0, 0, 0 0, 1, 0, 0, 0, 0, 1,
0, 0, 0, 0, 1. So you can see all that any
matrix is, for a given dimension-- I mean we could
extend this to an n by n matrix-- is you just have 1's
along this top left to bottom right diagonals. And everything else is a 0. So I've told you that. Let's prove that it
actually works. Let's take this matrix
and multiply it times another matrix. And confirm that that matrix
doesn't change. So if we take 1, 0, 0, 1. Let's multiply it times-- let's
do a general matrix. Just so you see that this
works for all numbers. a, b, c, d. So what does that equal? We're going to multiply this
row times this column. 1 times a plus 0 times c is a. And that row times
this column. 1 times b plus 0 times d. That's b. Then this row times
this column. 0 times a plus 1 times c is c. Then finally, this row
times this column. 0 times b plus 1 times d. Well, that's just d. There you have it. And it might be a fun exercise
to try it the other way around as well. And actually it's an even better
exercise to try this with a 3x3. And you'll see it
all works out. And a good exercise for you is
to think about why it works. And if you think about it, it's
because you're getting your row information from
here and your column information from here. And essentially, anytime you're
multiplying, let's say this vector times this vector,
you're multiplying the corresponding terms and then
adding them, right? So if you have a 1 and a 0, the
0 is going to cancel out anything but the first term
in this column vector. So that's why you're
just left with a. And that's why it's going to
cancel out everything but the first term in this
column vector. And that's why you're
left with just b. And similarly, this will cancel
out everything but the second term. That's why you're left
with just c there. This times this. You're just left with c. This times this. You're just left with d. And that same thing applies
when you go to 3x3 or n by n vectors. So that's interesting. You have the identity vector. Now if we wanted to complete
our analogy-- so let's think about it. We know in regular mathematics,
if I have 1 times a, I get a. And we also know that 1 over a
times a-- this is just regular math, this has nothing to do
with matrices-- is equal to 1. And you know, we call this
the inverse of a. And that's also the same thing
as dividing by the number a. So is there a matrix analogy? Let me switch colors, because
I've used this green a little bit too much. Is there a matrix, where if I
were to have the matrix a, and I multiply it by this matrix--
and I'll call that the inverse of a-- is there a matrix where
I'm left with, not the number 1, but I'm left with
the 1 equivalent in the matrix world? Where I'm left with the
identity matrix? And it would be extra nice if
I could actually switch this multiplication around. So A times A inverse should
also be equal to the identity matrix. And if you think about it, if
both of these things are true, then actually not only is A
inverse the inverse of A, but A is also the inverse
of A inverse. So they're each other's
inverses. That's all I meant to say. And it turns out there
is such a matrix. It's called the inverse
of A, as I've said three times already. And I will now show you
how to calculate it. So let's do that. And we'll see calculating
it for a 2x2 is fairly straightforward. Although you might think it's a
little mysterious as to how people came up with the
mechanics of it, or the algorithm for it. 3x3 becomes a little hairy. 4x4 will take you all day. 5x5, you're almost definitely
going to do a careless mistake if you did the inverse
of a 5x5 matrix. And that's better left
to a computer. But anyway, how do we calculate
the matrix? So let's do that, and then we'll
confirm that it really is the inverse. So if I have a matrix A,
and that is a, b, c, d. And I want to calculate
its inverse. Its inverse is actually--
and this is going to seem like voodoo. In future videos, I will give
you a little bit more intuition for why this works, or
I'll actually show you how this came about. But for now it's almost better
just to memorize the steps, just so you have the confidence
that you know that you can calculate an inverse. It's equal to 1 over this number
times this. a times d minus b times c. ad minus bc. And this quantity down here, ad
minus bc, that's called the determinant of the matrix A. And we're going to
multiply that. This is just a number. This is just a scalar
quantity. And we're going to multiply
that by-- you switch the a and the d. You switch the top left
and the bottom right. So you're left with d and a. And you make these two, you make
the bottom left and the top right, you make
them negative. So minus c minus b. And the determinant-- once
again, this is something that you're just going to take a
little bit on faith right now. In future videos, I promise
to give you more tuition. But it's actually kind of
sophisticated to learn what the determinant is. And if you're doing this in your
high school class, you kind of just have to know
how to calculate it. Although I don't like
telling you that. So what is this? This is also call the
determinant of A. So you might see on an
exam, figure out the determinant of A. So let me just tell you that. And that's denoted by A in
absolute value signs. And that's equal
to ad minus bc. So another way of saying this,
this could be 1 over the determinant. So you could write A inverse
is equal to 1 over the determinant of A times
d minus b minus c, a. Anyway you look at it. But let's apply this to a real
problem, and you'll see that it's actually not so bad. So let's change letters, just so
you know it doesn't always have to be an A. Let's say I have a matrix B. And the matrix B is 3-- I'm
just going to pick random numbers-- minus 4, 2 minus 5. Let's calculate B inverse. So B inverse is going to
be equal to 1 over the determinant of B. What's the determinant? It's 3 times minus 5 minus
2 times minus 4. So 3 times minus 5 is minus
15, minus 2 times minus 4. 2 times minus 4 is minus 8. We're going to subtract that. So it's plus 8. And we're going to multiply
that times what? Well, we switched these two
terms. So it's minus 5 and 3. And we just make these
two terms negative. Minus 2 and 4. 4 was minus 4, so now
it becomes 4. And let's see if we can simplify
this a little bit. So B inverse is equal
to minus 15 plus 8. That's minus 7. So this is minus 1/7. So the determinant of B-- we
could write B's determinant-- is equal to minus 7. So that's minus 1/7 times
minus 5, 4, minus 2, 3. Which is equal to-- this is just
a scalar, this is just a number, so we multiply it times
each of the elements-- so that is equal to minus,
minus, plus. That's 5/7. 5/7 minus 4/7. Let's see. Positive 2/7. And then minus 3/7. It's a little hairy. We ended up with fractions
here and things. But let's confirm that this
really is the inverse of the matrix B. Let's multiply them out. So before I do that I have
to create some space. I don't even need
this anymore. There you go. OK. So let's confirm that that
times this, or this times that, is really equal to
the identity matrix. So let's do that. So let me switch colors. So B inverse is 5/7,
if I haven't made any careless mistakes. Minus 4/7. 2/7. And minus 3/7. That's B inverse. And let me multiply that by B. 3 minus 4. 2 minus 5. And this is going to be
the product matrix. I need some space to
do my calculations. Let me switch colors. I'm going to take this row
times this column. So 5/7 times 3 is what? 15/7. Plus minus 4/7 times 2. So minus 4/7 times 2 is minus--
let me make sure that's right-- 5 times
3 is 15/7. Minus 4-- oh right, right--
4 times 2, so minus 8/7. Now we're going to multiply this
row times this column. So 5 times minus 4
is minus 20/7. Plus minus 4/7 times minus 5. That is plus 20/7. My brain is starting to slow
down, having to do matrix multiplications with fractions
with negative numbers. But this is a good exercise
for multiple parts of the brain. But anyway. So let's go down and
do this term. So now we're going to multiply
this row times this column. So 2/7 times 3 is 6/7. Plus minus 3/7 times 2. So that's minus 6/7. One term left. Home stretch. 2/7 times minus 4
is minus 8/7. Plus minus 3/7 times minus 5. So those negatives cancel out,
and we're left with plus 15/7. And if we simplify,
what do we get? 15/7 minus 8/8 is 7/7. Well that's just 1. This is 0, clearly. This is 0. 6/7 minus 6/7 is 0. And then minus 8/7 plus
15/7, that's 7/7. That's 1 again. And there you have it. We've actually managed to
inverse this matrix. And it was actually harder to
prove that it was the inverse by multiplying, just because we
had to do all this fraction and negative number math. But hopefully that
satisfies you. And you could try it the other
way around to confirm that if you multiply it the other
way, you'd also get the identity matrix. But anyway, that is how
you calculate the inverse of a 2x2. And as we'll see in the next
video, calculating by the inverse of a 3x3 matrix
is even more fun. See you soon.