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.