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Current time:0:00Total duration:13:36

I will now show you my preferred
way of finding an inverse of a 3 by 3 matrix. And I actually think it's
a lot more fun. And you're less likely to
make careless mistakes. But if I remember correctly from
Algebra 2, they didn't teach it this way
in Algebra 2. And that's why I taught the
other way initially. But let's go through this. And in a future video, I will
teach you why it works. Because that's always
important. But in linear algebra, this is
one of the few subjects where I think it's very important
learn how to do the operations first. And then later,
we'll learn the why. Because the how is
very mechanical. And it really just involves
some basic arithmetic for the most part. But the why tends to
be quite deep. So I'll leave that
to later videos. And you can often think about
the depth of things when you have confidence that you at
least understand the hows. So anyway, let's go back
to our original matrix. And what was that original
matrix that I did in the last video? It was 1, 0, 1, 0,
2, 1, 1, 1, 1. And we wanted to find the
inverse of this matrix. So this is what we're
going to do. It's called Gauss-Jordan
elimination, to find the inverse of the matrix. And the way you do it-- and it
might seem a little bit like magic, it might seem a little
bit like voodoo, but I think you'll see in future videos that
it makes a lot of sense. What we do is we augment
this matrix. What does augment mean? It means we just add
something to it. So I draw a dividing line. Some people don't. So if I put a dividing
line here. And what do I put on the other
side of the dividing line? I put the identity matrix
of the same size. This is 3 by 3, so I put a
3 by 3 identity matrix. So that's 1, 0, 0,
0, 1, 0, 0, 0, 1. All right, so what are
we going to do? What I'm going to do is perform
a series of elementary row operations. And I'm about to tell you what
are valid elementary row operations on this matrix. But whatever I do to any of
these rows here, I have to do to the corresponding
rows here. And my goal is essentially to
perform a bunch of operations on the left hand side. And of course, the same
operations will be applied to the right hand side, so that I
eventually end up with the identity matrix on the
left hand side. And then when I have the
identity matrix on the left hand side, what I have left on
the right hand side will be the inverse of this
original matrix. And when this becomes an
identity matrix, that's actually called reduced
row echelon form. And I'll talk more about that. There's a lot of names and
labels in linear algebra. But they're really just fairly
simple concepts. But anyway, let's get started
and this should become a little clear. At least the process
will become clear. Maybe not why it works. So first of all, I said I'm
going to perform a bunch of operations here. What are legitimate
operations? They're called elementary
row operations. So there's a couple
things I can do. I can replace any row
with that row multiplied by some number. So I could do that. I can swap any two rows. And of course if I swap say the
first and second row, I'd have to do it here as well. And I can add or subtract one
row from another row. So when I do that-- so for
example, I could take this row and replace it with this
row added to this row. And you'll see what I
mean in the second. And you know, if you combine it,
you could you could say, well I'm going to multiple this
row times negative 1, and add it to this row, and replace
this row with that. So if you start to feel like
this is something like what you learned when you learned
solving systems of linear equations, that's
no coincidence. Because matrices are actually
a very good way to represent that, and I will show
you that soon. But anyway, let's do some
elementary row operations to get this left hand side into
reduced row echelon form. Which is really just a fancy way
of saying, let's turn it into the identity matrix. So let's see what
we want to do. We want to have 1's
all across here. We want these to be 0's. Let's see how we can do
this efficiently. Let me draw the matrix again. So let's get a 0 here. That would be convenient. So I'm going to keep the
top two rows the same. 1, 0, 1. I have my dividing line. 1, 0, 0. I didn't do anything there. I'm not doing anything
to the second row. 0, 2, 1. 0, 1, 0. And what I'm going to do, I'm
going to replace this row-- And just so you know my
motivation, my goal is to get a 0 here. So I'm a little bit closer
to having the identity matrix here. So how do I get a 0 here? What I could do is I can replace
this row with this row minus this row. So I can replace the third
row with the third row minus the first row. So what's the third row
minus the first row? 1 minus 1 is 0. 1 minus 0 is 1. 1 minus 1 is 0. Well I did it on the left hand
side, so I have to do it on the right hand side. I have to replace this
with this minus this. So 0 minus 1 is minus 1. 0 minus 0 is 0. And 1 minus 0 is 1. Fair enough. Now what can I do? Well this row right here, this
third row, it has 0 and 0-- it looks a lot like what I want
for my second row in the identity matrix. So why don't I just swap
these two rows? Why don't I just swap the
first and second rows? So let's do that. I'm going to swap the first
and second rows. So the first row
stays the same. 1, 0, 1. And then the other side stays
the same as well. And I'm swapping the second
and third rows. So now my second row
is now 0, 1, 0. And I have to swap it on
the right hand side. So it's minus 1, 0, 1. I'm just swapping these two. So then my third row now
becomes what the second row was here. 0, 2, 1. And 0, 1, 0. Fair enough. Now what do I want to do? Well it would be nice if
I had a 0 right here. That would get me that much
closer to the identity matrix. So how could I get as 0 here? Well what if I subtracted 2
times row two from row one? Because this would be,
1 times 2 is 2. And if I subtracted that from
this, I'll get a 0 here. So let's do that. So the first row has
been very lucky. It hasn't had to do anything. It's just sitting there. 1, 0, 1, 1, 0, 0. And the second row's not
changing for now. Minus 1, 0, 1. Now what did I say I
was going to do? I'm going to subtract 2 times
row two from row three. So this is 0 minus
2 times 0 is 0. 2 minus 2 times 1,
well that's 0. 1 minus 2 times 0 is 1. 0 minus 2 times negative 1 is--
so let's remember 0 minus 2 times negative 1. So that's 0 minus negative
2, so that's positive 2. 1 minus 2 times 0. Well that's just still 1. 0 minus 2 times 1. So that's minus 2. Have I done that right? I just want to make sure. 0 minus 2 times-- right, 2
times minus 1 is minus 2. And I'm subtracting
it, so it's plus. OK, so I'm close. This almost looks like the
identity matrix or reduced row echelon form. Except for this 1 right here. So I'm finally going to have
to touch the top row. And what can I do? well how about I replace the top
row with the top row minus the bottom row? Because if I subtract
this from that, this'll get a 0 there. So let's do that. So I'm replacing the top
row with the top row minus the third row. So 1 minus 0 is 1. 0 minus 0 is 0. 1 minus 1 is 0. That was our whole goal. And then 1 minus 2
is negative 1. 0 minus 1 is negative 1. 0 minus negative 2., well
that's positive 2. And then the other rows
stay the same. 0, 1, 0, minus 1, 0, 1. And then 0, 0, 1, 2,
1, negative 2. And there you have it. We have performed a series
of operations on the left hand side. And we've performed the
same operations on the right hand side. This became the identity
matrix, or reduced row echelon form. And we did this using
Gauss-Jordan elimination. And what is this? Well this is the inverse of
this original matrix. This times this will equal
the identity matrix. So if this is a, than
this is a inverse. And that's all you have to do. And as you could see, this took
me half the amount of time, and required a lot less
hairy mathematics than when I did it using the adjoint and
the cofactors and the determinant. And if you think about it, I'll
give you a little hint of why this worked. Every one of these operations
I did on the left hand side, you could kind of view them as
multiplying-- you know, to get from here to here,
I multiplied. You can kind of say that
there's a matrix. That if I multiplied by that
matrix, it would have performed this operation. And then I would have had to
multiply by another matrix to do this operation. So essentially what we did is
we multiplied by a series of matrices to get here. And if you multiplied all
of those, what we call elimination matrices, together,
you essentially multiply this times
the inverse. So what am I saying? So if we have a, to go from
here to here, we have to multiply a times the
elimination matrix. And this might be completely
confusing for you, so ignore it if it is, but it might
be insightful. So what did we eliminate
in this? We eliminated 3, 1. We multiplied by the
elimination matrix 3, 1, to get here. And then, to go from
here to here, we've multiplied by some matrix. And I'll tell you more. I'll show you how
we can construct these elimination matrices. We multiply by an elimination
matrix. Well actually, we had
a row swap here. I don't know what you
want to call that. You could call that
the swap matrix. We swapped row two for three. And then here, we multiplied
by elimination matrix-- what did we do? We eliminated this, so
this was row three, column two, 3, 2. And then finally, to get here,
we had to multiply by elimination matrix. We had to eliminate
this right here. So we eliminated row
one, column three. And I want you to know right
now that it's not important what these matrices are. I'll show you how we can
construct these matrices. But I just want you to have kind
of a leap of faith that each of these operations could
have been done by multiplying by some matrix. But what we do know is by
multiplying by all of these matrices, we essentially got
the identity matrix. Back here. So the combination of all of
these matrices, when you multiply them by each
other, this must be the inverse matrix. If I were to multiply each of
these elimination and row swap matrices, this must be the
inverse matrix of a. Because if you multiply
all them times a, you get the inverse. Well what happened? If these matrices are
collectively the inverse matrix, if I do them, if I
multiply the identity matrix times them-- the elimination
matrix, this one times that equals that. This one times that
equals that. This one times that
equals that. And so forth. I'm essentially multiplying--
when you combine all of these-- a inverse times
the identity matrix. So if you think about it just
very big picture-- and I don't want to confuse you. It's good enough at this
point if you just understood what I did. But what I'm doing from all of
these steps, I'm essentially multiplying both sides of this
augmented matrix, you could call it, by a inverse. So I multiplied this by a
inverse, to get to the identity matrix. But of course, if I multiplied
the inverse matrix times the identity matrix, I'll get
the inverse matrix. But anyway, I don't want
to confuse you. Hopefully that'll give you
a little intuition. I'll do this later with some
more concrete examples. But hopefully you see that this
is a lot less hairy than the way we did it with the
adjoint and the cofactors and the minor matrices and the
determinants, et cetera. Anyway, I'll see you
in the next video.