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We're nearing the home
stretch of our quest
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to find the inverse of this
three-by-three matrix here.
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And the next thing
that we can do
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is find the determinant
of it, which we already
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have a good bit
of practice doing.
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So the determinant
of C, of our matrix--
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I'll do that same
color-- C, there
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are several ways
that you could do it.
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You could take this
top row of the matrix
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and take the value of
each of those terms
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times the cofactor-- times
the corresponding cofactor--
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and take the sum there.
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That's one technique.
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Or you could do
the technique where
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you rewrite these
first two columns,
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and then you take the product
of the top to left diagonals,
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sum those up, and
then subtract out
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the top right to
the bottom left.
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I'll do the second
one just so that you
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can see that you
get the same result.
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So let's see.
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The determinant is going
to be equal to-- I'll
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rewrite all of these
things-- so negative
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1, negative 2, 2,
2, 1, 1, 3, 4, 5.
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And let me now, just to make
it a little bit simpler,
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rewrite these first two columns.
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So negative 1,
negative 2, 2, 1, 3, 4.
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So the determinant
is going to be
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equal to-- so let
me write this down.
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So you have negative
1 times 1 times 5.
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Well that's just going
to be negative 5,
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taking that product.
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Then you have negative
2 times 1 times 3.
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Well that's negative 6.
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So we'll have negative 6.
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Or you could say plus
negative 6 there.
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And then you have
2 times 2 times 4.
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Well that's just 4 times
4, which is just 16.
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So we have plus 16.
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And then we do the top
right to the bottom left.
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So you have negative
2 times 2 times 5.
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Well that's negative 4 times 5.
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So that is negative 20.
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But we're going to
subtract negative 20.
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So that's negative 4
times 5, negative 20,
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but we're going to
subtract negative 20.
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Obviously that's going to
turn into adding positive 20.
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Then you have negative 1 times
1 times 4, which is negative 4.
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But we're going to
subtract these products.
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We're going to
subtract negative 4.
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And then you have 2 times
1 times 3, which is 6.
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But we have to subtract it.
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So we have subtracting 6.
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And so this simplifies
to negative 5 minus 6
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is negative 11, plus 16
gets us to positive 5.
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So all of this
simplifies to positive 5.
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And then we have plus 20 plus 4.
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Actually, let me do
that green color,
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so we don't get confused.
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So we have plus
20 plus 4 minus 6.
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So what does this get us?
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5 plus 20 is 25, plus 4 is
29, minus 6 gets us to 23.
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So our determinant right
over here is equal to 23.
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So now we are really
in the home stretch.
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The inverse of this
matrix is going
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to be 1 over our determinant
times the transpose
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of this cofactor matrix.
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And the transpose of
the cofactor matrix
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is called the adjugate.
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So let's do that.
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So let's write
the adjugate here.
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This is the drum roll.
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We're really in
the home stretch.
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C inverse is equal to
1 over the determinant,
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so it's equal to 1/23,
times the adjugate of C.
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And so this is going to be equal
to 1/23 times the transpose
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of our cofactor matrix.
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So we have our cofactor
matrix right over here.
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So each row now
becomes a column.
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So this row now
becomes a column.
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So it becomes 1, negative 7,
5 becomes the first column.
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The second row becomes
the second column-- 18,
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negative 11, negative 2.
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And then finally, the third
row becomes the third column.
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You have negative 4, 5, and 3.
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And now we just
have to multiply,
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or you could say divide, each of
these by 23, and we are there.
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So this is the inverse of
our original matrix C, home
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stretch.
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1 divided by 23 is just 1/23.
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Then you have 18/23.
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Actually, let me give myself
a little bit more real estate
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to do this in.
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So there we go.
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So 1 divided by 23-- 1/23,
18/23, negative 4/23,
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negative 7/23, negative 11/23,
5/23, 5/23, negative 2/23.
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And then finally,
assuming we haven't made
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any careless mistakes,
which would shock me
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if we haven't, we get to 3/23.
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And we are done.
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We have successfully inverted
a three-by-three matrix.
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Once again, something I
strongly believe better done
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by a computer and
probably should not
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be part of a typical Algebra
2 curriculum, because it tends
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to be displayed in a,
non-contextual way.
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