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