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LINAN CHEN: Hello.
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Welcome back to recitation.
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I'm sure you are
becoming more and more
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familiar with the
determinants of matrices.
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In the lecture, we also learned
the geometric interpretation
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of the determinant.
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The absolute value of the
determinant of a matrix
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is simply equal to the volume
of the parallelepiped spanned
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by the row vectors
of that matrix.
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So today, we're going
to apply this fact
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to solve the following problem.
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I have a tetrahedron,
T, in this 3D space.
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And the vertices of T are given
by O, which is the origin, A_1,
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A_2, and A_3.
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So I have highlighted
this tetrahedron
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using the blue chalk.
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So this is T.
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And our first goal is to
compute the volume of T
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using the determinant.
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And the second part is:
if I fix A_1 and A_2,
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but move A_3 to another
point, A_3 prime, which
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is given by this
coordinate, I ask you
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to compute the volume again.
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OK.
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So since we want to use the fact
that the determinant is related
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to the volume, we have to
figure out which volume
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we should be looking at.
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We know that the
determinant is related
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to the volume of
a parallelepiped.
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But here, we only
have a tetrahedron.
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So the first goal should be to
find out which parallelepiped
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you should be working with.
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OK, why don't you hit pause and
try to work it out yourself.
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You can sketch the
parallelepiped on this picture.
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And I will return in a while
and continue working with you.
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All right.
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How did your computation go?
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Now let's complete
this picture together.
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As we were saying, we
need a parallelepiped
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so that we can use the fact
that the determinant is
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related to the volume.
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Here, I have a tetrahedron.
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And let's look at these three
edges, OA_1, OA_2, and OA_3.
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All of them meet at the origin.
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So why don't we just consider
the parallelepiped spanned
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by those same three edges?
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It's a natural choice,
because at least it
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shares three edges with T.
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OK, now let's move on
to this picture here.
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As you can see, I have
drawn this parallelepiped
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in red chalk.
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So the blue part
is my original T.
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And the red part,
let me call it P.
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It's the parallelepiped spanned
by edges OA_1, OA_2, and OA_3.
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So that's the parallelepiped
that I'm going to work with.
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Now the next step is to
relate the volume of T
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to the volume of P. OK,
let's recall together
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what is the volume
of a tetrahedron.
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We note that the
volume of a tetrahedron
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is going to be equal to
1/3 of the area of the base
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times the height, right?
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Of course, you can choose any
side, any face to be the base.
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But for convenience, we're going
to choose the triangle OA_1A_2
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to be the base of T.
So the volume of T
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is going to be equal to 1/3
times the area of triangle
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OA_1A_2.
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So I use this A to
indicate the area.
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Then times the height.
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Well, if I choose
this to be my base,
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then this A_3 becomes the apex.
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Then the height is simply
equal to the distance from A_3
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to the triangle OA_1A_2.
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Let me use letter h to
denote this quantity.
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So that's the height.
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And the volume of
T is equal to 1/3
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of the product of the area
of the base times the height.
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OK, this is the volume of T. Now
let's see what the volume of P
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is.
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So P is a parallelepiped.
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The volume of a
parallelepiped is simply
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equal to the area of the
base times its height.
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This time, which face would
you choose to be the base?
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Well, of course, you would like
to choose this parallelogram
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to be the base because it
contains the base of T.
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If we do so, so we want to
choose this parallelogram
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to be the base, then
what is the area of this?
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Well, it clearly contains two
copies of the triangle OA_1A_2.
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So the area of the
parallelogram is simply
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equal to twice the area
of the triangle OA_1A_2.
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Then what is the height of P?
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Again, if you choose
this face to be the base,
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then A_3 becomes the apex again.
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Then the height of
the parallelepiped
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is equal to the
distance from A_3
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to the base, which is the
same as the distance from A_3
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to triangle OA_1A_2.
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So here, the height
is also equal to h.
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Now you can compare
these two formulae.
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You see that volume
of T is simply
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equal to 1/6 of
volume of P. That's
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the connection
between the volume
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of the tetrahedron with the
volume of the parallelepiped.
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In order to compute
the volume of T,
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we only need to compute
the volume of P.
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Now let's compute the volume
of this parallelepiped.
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We know that it's related to
the determinant of a matrix.
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And the row vectors
of that matrix
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are given by these three edges.
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So because all of
them start from 0,
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we only need the coordinate
of A_1, A_2, and A_3.
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So here, volume of P is
equal to the absolute value.
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So don't forget the
absolute value sign.
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Absolute value of the
determinant of a three
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by three matrix.
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So we just need to copy the
coordinates of the vertices
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down here.
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The first one is [2, 2, -1].
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These two are too close.
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-1.
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And A_2 is [1, 3, 0].
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A3 is [-1, 1, 4].
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The absolute value
of this determinant.
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And if you compute this, this
is a three by three matrix.
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The determinant should
be easy to compute.
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And the result should be 12.
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So that's the volume
of P, which means
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the volume of the tetrahedron
T is equal to 12 over 6,
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which is 2.
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Did you get the correct answer?
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OK, so in order to
compute the volume of T,
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we have related to
a parallelepiped, P,
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which contains T. All right.
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Now let's look at
the second part.
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The second part says that if
I keep A_1 and A_2 unchanged,
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but I move A_3 to
a new point-- so A3
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is going to be moved to a
point given by A_3 prime,
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and the coordinate is
-201, -199, and 104.
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And I'm asking you
to compute the volume
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of the new tetrahedron.
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Well as you can see,
this point seems
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to be far away from the origin.
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I'm not even able to
draw this point here.
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But you can imagine, as
this point goes far away
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from the origin,
this spike is going
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to become more and more pointy.
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In other words, the entire
tetrahedron looks more and more
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like a needle.
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But nonetheless, we
can use the same method
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to compute the volume.
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So we follow the same
idea, the volume of T
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is going to be equal to
1/6 of the volume of P.
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And in this case,
that's going to become
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the absolute value of the
determinant of a three
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by three matrix
whose row vectors are
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given by these three edges.
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So in this case, we again
copy down the coordinates
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of three vertices.
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The first one is [2, 2, -1].
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The second one is [1, 3, 0].
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The third one should become
this, so [-201, 199, 104].
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OK.
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That should give me the
volume of the new tetrahedron.
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Let me call it T prime
just to differentiate it
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from the previous tetrahedron.
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And of course, you can compute
this determinant explicitly.
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If you do so, you will see
the answer should be 2 again.
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But in fact, there is
a way that you can just
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read out the answer directly
without any real computation.
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Let's pay attention
to the last row.
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In other words, let's pay
attention to this new A_3,
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well, A_3 prime.
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What do you observe here?
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A_3 prime, if you consider the
difference between A_3 and A_3
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prime, in other words, if you
consider how much you have
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moved your apex, you will
see that's equal to -100*A_1.
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Right?
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So A_3 is [-1, 1, 4].
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A_3 prime is this.
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That's exactly equal
to -100 times A_1.
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What does that mean?
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Well from the point
of view of the matrix,
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you are subtracting from the
original third row 100 times
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the first row.
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But this row operation does
not change the determinant.
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In other words, you know that
this determinant should be
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the same as the previous one.
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So you can write out 2 directly.
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From this picture, we
can also see that fact.
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So from this picture, we
know that this section
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is saying that I need to move
A_3 in the inverse direction
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of A_1 by 100 times A_1.
192
00:12:50,730 --> 00:12:54,500
So you're moving
A_3 parallel to A_1.
193
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But it doesn't matter how
far you've moved your apex.
194
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You're moving in a way
that remains parallel
195
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to the base, which is saying
that this movement does not
196
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change the height.
197
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Since A_1 and A_2 are
fixed, your base is fixed,
198
00:13:12,780 --> 00:13:14,730
and you're not
changing the height.
199
00:13:14,730 --> 00:13:17,660
So of course, the
volume is not changing.
200
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That's also a
reason, another way
201
00:13:20,450 --> 00:13:22,700
to see that the
volume of T prime
202
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is simply equal to
the volume of T.
203
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OK, this completes this problem.
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I hope this example
was helpful to you.
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And we should keep in mind that
the fact that the determinant
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is related to
volume sometimes can
207
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become very handy in
computing the volume
208
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of certain geometric objects.
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Thank you for watching,
and I'm looking forward
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to seeing you soon.