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DAVID JORDAN: Hello, and
welcome back to recitation.

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I'd like to work on a
problem with you, which

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is to compute the volume
of a parallelepiped using 3

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by 3 determinants.

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So here we've got the
parallelepiped drawn.

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It has, one of its vertices
is at the origin, (0, 0, 0),

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and the other three
edges are given to us

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with these coordinates here.

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So why don't you take some
time to work out this problem,

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pause the video, and check
back with me and I'll show you

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how I solved it.

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OK, welcome back.

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Let's get started.

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So the first thing
that we need to do

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is we need to remember
that computing

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volumes of parallelepipeds is
the same thing as computing 3

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by 3 determinants.

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So the volume is just equal
to the determinant, which

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is built out of the vectors,
the row vectors determining

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the edges.

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So we have-- well,
that's almost true.

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This determinant will be either
a positive or negative number

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and we always want to
take the positive number.

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So I'm going to write
plus or minus here,

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and we'll have to
remember at the end

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that we want a positive number.

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So we can compute
this determinant

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using Laplace expansion as
we did in the last video.

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So for Laplace
expansion, we take

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the-- we can do Laplace
expansion on the first row.

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And that will again be nice,
because this 0 here will

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make our computation simpler.

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And so, remember what we do
is we take the first entry

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in the row, and then we need
to multiply by the minor

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that we get by covering up
that row and that column.

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So we have this little
2 by 2 determinant,

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which looks like minus 1.

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And now we need to
subtract the next entry

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in the row times its
minor, which is now

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this matrix [1, 1; 0, 1].

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So taking that determinant,
we get 1 minus 0.

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So altogether, we get 1.

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So what that tells us is that
our determinant is minus 4.

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But then that tells us
that our volume is plus 4.

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And I'll leave it at that.