Inverting 3x3 part 2: Determinant and adjugate of a matrix Video Lecture - Engineering Mathematics

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