OK, this is lecture twenty.
And this is the final
lecture on determinants.
And it's about the applications.
So we worked hard in
the last two lectures
to get a formula
for the determinant
and the properties
of the determinant.
Now to use the determinant and,
and always this determinant
packs all this information
into a single number.
And that number can
give us formulas
for all sorts of, things that
we've been calculating already
without formulas.
Now what was A inverse?
So, so I'm beginning with
the formula for A inverse.
Two, two by two formula we know,
right?
The two by two formula for A
inverse, the inverse of a b c d
inverse is one over the
determinant times d a -b -c.
Somehow I want to see what's
going on for three by three
and n by n.
And actually maybe you can see
what's going on from this two
by two case.
So there's a formula
for the inverse,
and what did I divide by?
The determinant.
So what I'm looking
for is a formula
where it has one
over the determinant
and, and you remember why
that makes good sense,
because then that's perfect as
long as the determinant isn't
zero, and that's exactly
when there is an inverse.
But now I have to ask can we
recognize any of this stuff?
Do you recognize what that
number d is from the past?
From last, from
the last lecture?
My hint is think cofactors.
Because my formula is going to
be, my formula for the inverse
is going to be one
over the determinant
times a matrix of cofactors.
So you remember that D?
What's that the cofactor of?
Remember cofactors?
If -- that's the
one one cofactor,
because if I strike out row and
column one, I'm left with d.
And what's minus b?
OK.
Which cofactor is that one?
Oh, minus b is the
cofactor of c, right?
If I strike out the c,
I'm left with a b there.
And why the minus sign?
Because this c was in a two one
position, and two plus one is
odd.
So there was a minus went into
the cofactor, and that's it.
OK.
I'll write down next
what my formula is.
Here's the big formula for
the A -- for A inverse.
It's one over the determinant
of A and then some matrix.
And that matrix is the
matrix of cofactors,
c.
Only one thing, it
turns -- you'll see,
I have to, I transpose.
So this is the matrix
of cofactors, the --
what I'll just --
but why don't we just call
it the cofactor matrix.
So the one one entry of, of
c is the cof- is the one one
cofactor, the thing that we get
by throwing away row and column
one.
It's the d.
And, because of the
transpose, what I see up here
is the cofactor of this
guy down here, right?
That's where the
transpose came in.
What I see here, this is the
cofactor not of this one,
because I've transposed.
This is the cofactor of the b.
When I throw away the b,
the b row and the b column,
I'm left with c, and then
I have that minus sign
again.
And of course the two
two entry is the cofactor
of d, and that's this a.
So there's the formula.
OK.
But we got to think why.
I mean, it worked in
this two by two case,
but a lot of other formulas
would have worked just as well.
We, we have to see
why that's true.
In other words, why is it --
so this is what I aim to find.
And, and let's just sort of
look to see what is that telling
us.
That tells us that the -- the
expression for A inverse --
let's look at a three by three.
Can I just write down
a a b c d e f g h i?
And I'm looking for its inverse.
And what kind of a
formula -- do I see there?
I mean, what --
the determinant is a bunch
of products of three factors,
right?
The determinant of this
matrix'll involve a e i,
and b f times g, and c
times d times h, and minus c
e g, and so on.
So things with three
factors go in here.
Things with how many factors do
things in the cofactor matrix
have?
What's a typical cofactor?
What's the cofactor of a?
The cofactor of a, the one one
entry up here in the inverse
is?
I throw away the row
and column containing a
and I take the determinant
of what's left,
that's the cofactor.
And that's e i minus f h.
Products of two things.
Now, I'm just making
the observation
that the determinant of A
involves products of n entries.
And the cofactor matrix involves
products of n minus 1 entries.
And, like, we never
noticed any of this stuff
when we were
computing the inverse
by the Gauss-Jordan
method or whatever.
You remember how we did it?
We took the matrix A, we
tucked the identity next to it,
we did elimination till
A became the identity.
And then that, the identity
suddenly was A inverse.
Well, that was
great numerically.
But we never knew what
was going on, basically.
And now we see
what the formula is
in terms of letters,
what's the algebra instead
of the algorithm.
OK.
But I have to say why
this is right, right?
I still -- that's a
pretty magic formula.
Where does it come from?
Well, I'll just check it.
Having, having got it
up there, let me --
I'll say, how can we check --
what do I want to check?
I want to check that A times
its inverse gives the identity.
So I want, I want to check
that A times this thing,
A times this -- now I'm going
to write in the inverse --
gives the identity.
So I check that A
times C transpose --
let me bring the
determinant up here.
Determinant of A
times the identity.
That's my job.
That's it, that if this is
true, and it is, then, then I've
correctly identified A
inverse as C transpose divided
by the determinant.
OK.
But why is this true?
Why is that true?
Let me, let me put down
what I'm doing here.
I have A --
here, here's A, here's a11 --
I'm doing this multiplication
-- along to a1n.
And then down in this last row
will be an an1 along to ann.
And I'm multiplying by the
cofactor matrix transposed.
So when I transpose, it'll
be c11 c12 down to c1n.
Notice usually that
one coming first
would mean I'm in row
one, but I've transposed,
so that's, those
are the cofactors.
This first column are the
cofactors from row one.
And then the last column would
be the cofactors from row n.
And why should that come
out to be anything good?
In fact, why should it come
out to be as good as this?
Well, you can tell me what the
one one entry in the product
is.
This is like you're seeing
the main point if you just
tell me one entry.
What do I get up there
in the one one entry
when I do this row of
this row from A times
this column of cofactors?
What, what will I get there?
Because we have seen this.
I mean, we're, right,
building exactly
on what the last
lecture reached.
So this is a11 times c11,
a12 times c12, a1n times c1n.
What does that what
does that sum up to?
That's the cofactor formula
for the determinant.
That's the, this
cofactor formula,
which I wrote, which
we got last time.
The determinant of A is,
if I use row one, let,
let I equal one,
then I have a11 times
its cofactor, a12 times
its cofactor, and so on.
And that gives me
the determinant.
And it worked in
this, in this case.
This row times this thing
is, sure enough, ad minus bc.
But this formula
says it always works.
So up here in this,
in this position,
I'm getting determinant of A.
What about in the
two two position?
Row two times column two
there, what, what is that?
That's just the cofactors,
that's just row two
times its cofactors.
So of course I get
the determinant again.
And in the last here,
this is the last row
times its cofactors.
It's exactly -- you see, we're
realizing that the cofactor
formula is just this
sum of products,
so of course we think,
hey, we've got a matrix
multiplication there.
And we get determinant of A.
But there's one more
idea here, right?
Great.
What else, what have I not --
so I haven't got that formula
completely proved yet, because
I've still got to do all
the off-diagonal stuff,
which I want to be zero,
right?
I just want this
to be determinant
of A times the identity, and
then I'm, I'm a happy person.
So why should that be?
Why should it be that one
row times the cofactors
from a different row, which
become a column because I
transpose, give zero?
In other words, the cofactor
formula gives the determinant
if the row and the, and
the cofactors -- you know,
if the entries of A and the
cofactors are for the same row.
But for some reason, if I
take the cofactors from the --
entries from the first row and
the cofactors from the second
row, for some reason
I automatically
get zero.
And it's sort of
like interesting
to say, why does that happen?
And can I just check that --
of course, we know it happens,
in this case.
Here are the
numbers from row one
and here are the cofactors
from row two, right?
Those are the
numbers in row one.
And th- these are the
cofactors from row two,
because the cofactor
of c is minus b
and the cofactor of d is
a.
And sure enough, that row
times this column gives --
please say it.
Zero, right.
OK.
So now how come?
How come?
Can we even see it in
this two by two case?
Why did -- well, I mean,
I guess we, you know,
in one way we certainly do see
it, because it's right here.
I mean, do we just do
it, and then we get zero.
But we want to
think of some reason
why the answer's zero, some
reason that we can use in the n
by n case.
So let -- here,
here is my thinking.
We must be, if we're
getting the answer's zero,
we suspect that what
we're doing somehow,
we're taking the determinant
of some matrix that
has two equal rows.
So I believe that if we multiply
these by the cofactors from
some other row, we're taking
the determinant -- ye,
what matrix are we taking
the determinant of?
Here it's, this is it.
We're, when we do that times
this, we're really taking --
can I put this in little
letters down here?
I'm taking -- let me look
at the matrix a b a b.
Let me call that matrix
AS, A screwed up.
OK.
All right.
So now that matrix is
certainly singular.
So if we find its determinant,
we're going to get zero.
But I claim that if we find
its determinant by the cofactor
rule, go along the
first row, we would
take a times the cofactor of a.
And what is the --
see, how -- oh no -- let
me go along the second row.
OK.
So let's see, which --
if I take --
I know I've got a
singular matrix here.
And I believe that when I
do this multiplication, what
I'm doing is using the cofactor
formula for the determinant.
And I know I'm
going to get zero.
Let me try this again.
So the cofactor formula
for the determinant
says I should take a times
its cofactor, which is this b,
plus b times its cofactor,
which is this minus a.
OK.
That's what we're doing,
apart from a sign here.
Oh yeah, so you know, there
might be a minus multiplying
everything.
So if I take this
determinant, it's A --
the determinant of this, the
determinant of A screwed up is
a times its
cofactor, which is b,
plus the second guy times its
cofactor, which is minus a.
And of course I get
the answer zero,
and this is exactly what's
happening in that, in that,
row times this wrong column.
OK.
That's the two by two picture,
and it's just the same here.
That the reason I get a zero
up in there is, the reason
I get a zero is that when I
multiply the first row of A
and the last row of
the cofactor matrix,
it's as if I'm taking the
determinant of this screwed up
matrix that has first
and last rows identical.
The book pins this
down more specific --
and more carefully than
I can do in the lecture.
I hope you're seeing the point.
That this is an identity.
That it's a beautiful
identity and it tells us what
the inverse of the matrix is.
So it gives us the inverse,
the formula for the inverse.
OK.
So that's the first goal of
my lecture, was to find this
formula.
It's done.
OK.
And of course I could
invert, now, I can,
I sort of like I can see what --
I can answer
questions like this.
Suppose I have a matrix,
and let me move the one one
entry.
What happens to the inverse?
Just, just think
about that question.
Suppose I have some
matrix, I just write down
some nice, non-singular
matrix that's got an inverse,
and then I move the one
one entry a little bit.
I add one to it, for example.
What happens to
the inverse matrix?
Well, this formula
should tell me.
I have to look to see what
happens to the determinant
and what happens to
all the cofactors.
And, the picture,
it's all there.
It's all there.
We can really understand
how the inverse changes
when the matrix changes.
OK.
Now my second application is to
-- let me put that over here --
is to Ax=b.
Well, the -- course, the
solution is A inverse b.
But now I have a
formula for A inverse.
A inverse is one
over the determinant
times this C transpose times B.
I now know what A inverse is.
So now I just have to
say, what have I got here?
Is there any way to, to make
this formula, this answer,
which is the one
and only answer --
it's the very same answer we got
on the first day of the class
by elimination.
Now I'm -- now I've got
a formula for the answer.
Can I play with it further
to see what's going on?
And Cramer's, this Cramer's
Rule is exactly, that --
a way of looking
at this formula.
OK.
So this is a formula for x.
Here's my formula.
Well, of course.
The first thing I
see from the formula
is that the answer x always
has that in the determinant.
I'm not surprised.
There's a division
by the determinant.
But then I have to say a little
more carefully what's going on
And let me tell you what
Cramer's Rule is. up here.
Let, let me take x1,
the first component.
So this is the first
component of the answer.
Then there'll be a
second component and a,
all the other components.
Can I take just the first
component of this formula?
Well, I certainly have
determinant of A down under.
And what the heck
is the first --
so what do I get
in C transpose b?
What's the first entry
of C transpose b?
That's what I have
to answer myself.
Well, what's the first
entry of C transpose b?
This B is -- let me
tell you what it is.
OK.
Somehow I'm
multiplying cofactors
by the entries of B,
right, in this product.
Cofactors from the matrix
times entries of b.
So any time I'm multiplying
cofactors by numbers,
I think, I'm getting the
determinant of something.
And let me call
that something B1.
So this is a matrix, the matrix
whose determinant is coming out
of that.
And we'll, we'll see what it is.
x2 will be the determinant
of some other matrix B2, also
divided by determinant of A.
So now I just --
Cramer just had a good idea.
He realized what matrix it was,
what these B1 and B2 and B3
and so on matrices were.
Let me write them on
the board underneath.
OK.
So what is this B1?
This B1 is the matrix that
has b in its first column
and otherwise the
rest of it is A.
So it otherwise it has the
rest, the, the n-1 columns of A.
It's the matrix with --
it's just the matrix A
with column one replaced
by the right-hand side,
by the right-hand side b.
Because somehow when I take
the determinant of this guy,
it's giving me this
matrix multiplication.
Well, how could that be?
How could -- so what's,
what's the determinant formula
I'll use here?
I'll use cofactors, of course.
And I might as well use
cofactors down column one.
So when I use cofactors
down column one,
I'm taking the first
entry of b times what?
Times the cofactor c11.
Do you see that?
When I, when I use
cofactors here,
I take the first
entry here, B one
let's call it, times
the cofactor there.
But what's the cofactor in --
my little hand-waving is meant
to indicate that it's a
matrix of one size smaller,
the cofactor.
And it's exactly c11.
Well, that's just
what we wanted.
This first entry
is c11 times b1.
And then the next entry is
whatever, is c21 times b2,
and so on.
And sure enough, if
I look here, when
I'm, when I do the
cofactor expansion,
b2 is getting multiplied by
the right thing, and so on.
So there's Cramer's Rule.
And the book gives
another kind of cute proof
without, without building
so much on, on cofactors.
But here we've got
cofactors, so I thought
I'd just give you this proof.
So what is B --
in general, what is Bj?
This is the, this is A with
column j replaced by, by b.
So that's -- the determinant
of that matrix that you divide
by determinant of A to get xj.
So x -- let me change
this general formula.
xj, the j-th one,
is the determinant
of Bj divided by the
determinant of A.
And now we've said what Bj is.
Well, so Cramer found a rule.
And we could ask him, OK,
great, good work, Cramer.
But is your rule any
good in practice?
So he says, well, you couldn't
ask about a rule in mine,
right, because it's just --
all you have to do is find
the determinant of A and these
other determinants, so I guess
-- oh, he just says, well,
all you have to do is
find n+1 determinants,
the, the n Bs and the A.
And actually, I remember
reading -- there was a book,
popular book that, that kids
interested in math read when I
was a kid interested in
math called Mathematics
for the Million or something,
by a guy named Bell.
And it had a little page
about linear algebra.
And it said,-- so it
explained elimination
in a very complicated way.
I certainly didn't
understand it.
And, and it made it, you
know, it sort of said, well,
there is this formula
for elimination,
but look at this great
formula, Cramer's Rule.
So it certainly said Cramer's
Rule was the way to go.
But actually, Cramer's Rule
is a disastrous way to go,
because to compute
these determinants,
it takes, like,
approximately forever.
So actually I now think
of that book title
as being Mathematics
for the Millionaire,
because you'd have to
be able to pay for,
a hopelessly long calculation
where elimination, of course,
produced the x-s, in an instant.
But having a formula allows
you to, with, with letters, you
know, allows you to do algebra
instead of, algorithms.
So the, there's some value
in the Cramer's Rule formula
for x and in the explicit
formula for, for A inverse.
They're nice
formulas, but I just
don't want you to use them.
That'ss what it comes to.
If you had to -- and Matlab
would never, never do it.
I mean, it would
use elimination.
OK.
Now I'm ready for number
three in today's list
of amazing connections coming
through the determinant.
And that number three is the
fact that the determinant gives
a volume.
OK.
So now -- so that's
my final topic for --
among these -- this my
number three application,
that the determinant is actually
equals the volume of something.
Can I use this little space
to consider a special case,
and then I'll use the
far board to think
about the general rule.
So what I going to prove?
Or claim.
I claim that the
determinant of the matrix
is the volume of a box.
OK, and you say, which box?
Fair enough.
OK.
So let's see.
I'm in -- shall we say we're
in, say three by three?
Shall we suppose -- let's,
let's say three by three.
So, so we can really -- we're,
we're talking about boxes
in three dimensions, and
three by three matrices.
And so all I do -- you
could guess what the box is.
Here is, here is,
three dimensions.
OK.
Now I take the first row of
the matrix, a11, a22, A --
sorry. a11, a12, a13.
That row is a vector.
It goes to some point.
That point will be -- and
that edge going to it,
will be an edge of the box,
and that point will be a corner
of the box.
So here is zero zero
zero, of course.
And here's the first row of
the matrix: a11, a12, a13.
So that's one edge of the box.
Another edge of the box
is to the second row
of the matrix, row two.
Can I just call
it there row two?
And a third row of
the box will be to --
a third row -- a third edge of
the box will be given by row
three.
So, so there's row three.
That, the coordinates,
what are the coordinates
of that corner of the box?
a31, a32, a33.
So I've got that edge of the
box, that edge of the box,
that edge of the box,
and that's all I need.
Now I just finish
out the box, right?
I just -- the proper word,
of course, is parallelepiped.
But for obvious
reasons, I wrote box.
OK.
So, OK.
So there's the, there's
the bottom of the box.
There're the four
edge sides of the box.
There's the top of the box.
Cute, right?
It's the box that
has these three edges
and then it's completed to
a, to a, each, you know,
each side is a, is
a parallelogram.
And it's that box whose volume
is given by the determinant.
That's -- now it's -- possible
that the determinant is
negative.
So we have to just say
what's going on in that case.
If the determinant is
negative, then the volume, we,
we should take the
absolute value really.
So the volume, if we, if we
think of volume as positive,
we should take the absolute
value of the determinant.
But the, the sign, what does
the sign of the determinant --
it always must
tell us something.
And somehow it, it will tell
us whether these three is a --
whether it's a right-handed
box or a left-handed box.
If we, if we reversed
two of the edges,
we would go between
a right-handed box
and a left-handed box.
We wouldn't change the
volume, but we would
change the, the cyclic, order.
So I won't worry about that.
And, so one special
case is what?
A equal identity matrix.
So let's take that special case.
A equal identity matrix.
Is the formula determinant
of identity matrix,
does that equal the
volume of the box?
Well, what is the box?
What's the box?
If A is the identity matrix,
then these three rows are
the three coordinate
vectors, and the box is --
it's a cube.
It's the unit cube.
So if, if A is the
identity matrix, of course
our formula is
Well, actually that
proves property one --
that the volume right.
has property one.
Actually, we could, we could,
we could get this thing if we --
if we can show that the box
volume has the same three
properties that define
the determinant,
then it must be the determinant.
So that's like the, the, the
elegant way to prove this.
To prove this amazing fact
that the determinant equals
the volume, first we'll check
it for the identity matrix.
That's fine.
The box is a cube
and its volume is one
and the determinant is one
and, and one agrees with one.
Now let me take one -- let me
go up one level to an orthogonal
matrix.
Because I'd like to take this
chance to bring in chapter --
the, the previous chapter.
Suppose I have an
orthogonal matrix.
What did that mean?
I always called those things Q.
What was the point
of -- suppose I have,
suppose instead of the identity
matrix I'm now going to take A
equal Q, an orthogonal matrix.
What was Q then?
That was a matrix whose columns
were orthonormal, right?
Those were its columns
were unit vectors,
perpendicular unit vectors.
So what kind of a
box have we got now?
What kind of a box comes
from the rows or the columns,
I don't mind, because
the determinant
is the determinant
of the transpose,
so I'm never worried about
that.
What kind of a box,
what shape box have we
got if the matrix is
an orthogonal matrix?
It's another cube.
It's a cube again.
How is it different
from the identity cube?
It's just rotated.
It's just the orthogonal
matrix Q doesn't
have to be the identity matrix.
It's just the unit cube
but turned in space.
So sure enough, it's the unit
cube, and its volume is one.
Now is the determinant one?
What's the determinant of Q?
We believe that the determinant
of Q better be one or minus
one, so that our formula
is -- checks out in that --
if we can't check it in these
easy cases where we got a cube,
we're not going to get
it in the general case.
So why is the determinant
of Q equal one or minus one?
What do we know about Q?
What's the one matrix statement
of the properties of Q?
A matrix with orthonormal
columns has --
satisfies a certain equation.
What, what is that?
It's if we have this orthogonal
matrix, then the fact --
the way to say what it, what
its properties are is this.
Q prime, u- u- Q
transpose Q equals I.
Right?
That's what -- those are the
matrices that get the name Q,
the matrices that
Q transpose Q is I.
OK.
Now from that, tell me
why is the determinant one
or minus one.
How do I, out of this fact --
this may even be a
homework problem.
It's there in the, in the
list of exercises in the book,
and let's just do it.
How do I get, how do I discover
that the determinant of Q
is one or maybe minus one?
I take determinants of
both sides, everybody says,
so I won't --
I take determinants
of both sides.
On the right-hand side -- so
I, when I take determinants
of both sides,
let me just do it.
Take the determinant of
-- take determinants.
Determinant of the
identity is one.
What's the determinant
of that product?
Rule nine is paying off now.
The determinant of a product is
the determinant of this guy --
maybe I'll put it, I'll use
that symbol for determinant.
It's the determinant of that
guy times the determinant
of the other guy.
And then what's the
determinant of Q transpose?
It's the same as the
determinant of Q.
Rule ten pays off.
So this is just
this thing squared.
So that determinant squared is
one and sure enough it's one
or minus one.
Great.
So in these special
cases of cubes,
we really do have
determinant equals volume.
Now can I just push
that to non-cubes.
Let me push it first to
rectangles, rectangular boxes,
where I'm just multiplying
the e- the edges are --
let me keep all the
ninety degree angles,
because those are -- that,
that makes my life easy.
And just stretch the edges.
Suppose I stretch that first
edge, suppose this first edge
I double.
Suppose I double
that first edge,
keeping the other
edges the same.
What happens to the volume?
It doubles, right?
We know that the volume
of a cube doubles.
In fact, because we know that
the new cube would sit right
on top --
I mean, the new, the added
cube would sit right on --
would fit --
probably a geometer would
say congruent or something --
would go right in, in the other.
We'd have two.
We have two identical cubes.
Total volume is now two.
OK.
So I want -- if I double an
edge, the volume doubles.
What happens to the determinant?
If I double, the first
row of a matrix, what ch-
ch- what's the effect
on the determinant?
It also doubles, right?
And that was rule number 3a.
Remember rule 3a was
that if I, I could,
if I had a factor in, in row
one, T, I could factor it out.
So if, if I have a factor
two in that row one,
I can factor it out
of the determinant.
It agrees with the -- the volume
of the box has that factor two.
So, so volume satisfies
this property 3a.
And now I really close, but I
-- but to get to the very end
of this proof, I have
to get away from right
angles.
I have to allow the
possibility of, other angles.
And -- or what's
saying the same thing,
I have to check that the
volume also satisfies 3b.
So can I --
This is end of proof
that the -- so I'm --
determinant of A equals volume
of box, and where I right now?
This volume has properties,
properties one, no problem.
If the box is the
cube, everything is --
if the box is the unit
cube, its volume is one.
Property two was if
I reverse two rows,
but that doesn't change the box.
And it doesn't change the
absolute value, so no problem
there.
Property 3a was if I mul-
you remember what 3a was?
So property one was about
the identity matrix.
Property two was about
a plus or minus sign
that I don't care about.
Property 3a was a
factor T in a row.
But now I've got property
three B to deal with.
What was property 3b?
This is a great way to
review these, properties.
So that 3b, the property
3b said -- let's do,
let's do two by two.
So said that if I
had a+a', b+b', c,
d that this equaled what?
So this is property 3b.
This is the linearity
in row one by itself.
So c d is staying the same,
and I can split this into a b
and a' b'.
That's property 3b, at least
in the two by two case.
And what I --
I wanted now to show
that the volume, which
two, two by two, that
means area, has this,
has this property.
Let me just emphasize that we
have got -- we're getting --
this is a formula, then, for
the area of a parallelogram.
The area of this parallelogram
-- can I just draw it?
OK, here's the, here's
the parallelogram.
I have the row a b.
That's the first row.
That's the point a b.
And I tack on c d.
c d, coming out of here.
And I complete
the parallelogram.
So this is --
well, I better
make it look right.
It's really this one that has
coordinates c d and this has
coordinates -- well,
whatever the sum is.
And of course
starting at zero zero.
So we all know,
this is a+c, b+d.
Rather than --
I'm pausing on that
proof for a minute
just to going back
to our formula.
Because I want you to see
that unlike Cramer's Rule,
that I wasn't that
impressed by, I'm
very impressed by this
formula for the area
of a parallelogram.
And what's our formula?
What, what's the area
of that parallelogram?
If I had asked you
that last year,
you would have said OK,
the area of a parallelogram
is the base times the height,
right?
So you would have figured
out what this base, the --
how long that base was.
It's like the square root
of A squared plus b squared.
And then you would
have figured out
how much is this
height, whatever it is.
It's horrible.
This, I mean, we got square
roots, and in that height
there would be other
revolting stuff.
But now what's the formula
that we now know for the area?
It's the determinant
of our little matrix.
It's just ad-bc.
No square roots.
Totally rememberable, because
it's exactly a formula
that we've been studying the
whole, for three lectures.
OK.
That's, you know, that's
the most important point
I'm making here.
Is that if you know the
coordinates of a box,
of the corners, then
you have a great formula
for the volume,
area or volume, that
doesn't involve any lengths
or any angles or any heights,
but just involves the
coordinates that you've got.
And similarly, what's the
area of this triangle?
Suppose I chop that off
and say what about --
because you might often be
interested in a triangle
instead of a parallelogram.
What's the area
of this triangle?
Now there again,
everybody would have
said the area of a triangle is
half the base times the height.
And in some cases, if you know
the base that a, that's --
and the height, that's fine.
But here, we, what we know is
the coordinates of the corners.
We know the vertices.
And so what's the
area of that triangle?
If I know these, if I know
a b, c d, and zero zero,
what's the area?
It's just half, so
it's just half of this.
So this is, this
is a- a b -- a d -
b c for the parallelogram
and one half of that,
one half of ad-bc
for the triangle.
So I mean, this is a totally
trivial remark, to say, well,
divide by two.
But it's just that you
more often see triangles,
and you feel you know
the formula for the area
but the good formula for
the area is this one.
And I'm just going to --
I'm just going to
say one more thing
about the area of a triangle.
It's just because
it's -- you know,
it's so great to have a
good formula for something.
What if our triangle did
not start at zero zero?
What if our triangle,
what if we had this --
what if we had -- so I'm
coming back to triangles again.
But let me, let me put this
triangle somewhere, it's --
I'm staying with triangles,
I'm just in two dimensions,
but I'm going to allow you
to give me any three corners.
And in -- those six numbers
must determine the area.
And what's the formula?
The area is going
to be, it's going
to be, there'll be that
half of a parallelogram.
I mean, basically this can't
be completely new, right?
We've got the area when -- we,
we know the area when this is
zero zero.
Now we just want to lift our
sight slightly and get the area
when all th- so let me
write down what it, what it
comes out to be.
It turns out that if you do
this, x1 y1 and a 1, x2 y2
and a 1, x3 y3 and a
1, that that works.
That the determinant
symbol, of course.
It's just -- if I gave you
that determinant to find,
you might subtract
this row from this.
It would kill that one.
Subtract this row from this,
it would kill that one.
Then you'd have a simple
determinant to do with
differences, and it would --
this little
subtraction, what I did
was equivalent to
moving the triangle
to start at the origin.
I did it fast,
because time is up.
And I didn't complete
that proof of 3b.
I'll leave -- the book has a
carefully drawn figure to show
why that works.
But I hope you saw
the main point is
that for area and
volume, determinant
gives a great formula.
OK.
And next lectures are
about eigenvalues,
so we're really
into the big stuff.
Thanks.
