Given matrix A,
we're asked to find the determinant of A,
then determine if A is invertible,
then find the co-factor matrix of A
and finally find the inverse of A.
Let's begin by finding the determinant.
We will perform co-factor expansion on row two,
because row two has two zero entries.
So starting with the first entry in row two,
the determinant of matrix A is equal to zero times
co-factor two comma one.
Because we have a zero in their product,
we don't have to find co-factor two comma one.
So we move along to the second entry in row two,
which is negative two.
So we have plus negative two times co-factor two comma one,
which is equal to negative one
raised to the power two plus two.
The exponent is two plus two because
the element is in row two, column two.
And then we have times the minor,
the minor is equal to the determinant of the matrix
formed after eliminating row two
and column two from the matrix,
which is the row and column of negative two.
This leaves us with a two by two determinant
where row one is two negative four,
and row two is four six, which we see here.
And now I move along to the third entry in row two,
which is another zero,
which gives us plus zero times co-factor three comma one,
but again, because we have a factor of zero,
we could stop the product of zero.
So simplifying, the first product
and third product are both zero.
Simplifying, we have negative two times
the fourth power of negative one is equal to positive one.
And the determinant is equal to two times six,
which is 12, minus negative four times four,
which is negative 16.
Simplifying, 12 minus negative 16
is equal to 12 plus 16, or 28.
The determinant is equal to negative two times 28,
which is equal to negative 56.
Number two, is A invertible?
And since the determinant does not equal zero,
matrix A is an invertible,
which means matrix A does have an inverse.
For number three, we're asked to find the co-factor matrix.
Because we have a three by three matrix,
the co-factor matrix is defined here
in the upper right-hand corner.
Notice how we have to find nine co-factors.
And I've already set all this up to save some time.
Let's go through the first row.
Co-factor of one comma one is equal to negative one,
raise to power of one plus one,
times the corresponding minor.
The minor is equal to the determinant of the matrix
after eliminating row one and column one from the matrix.
If we eliminate row one and column one,
we can see we have a two-by-two matrix.
The minor is equal to the determinant of this
two by two matrix where row one is negative two zero,
and row two is negative two six.
The next entry in the co-factor matrix
is co-factor one comma two,
which is equal to negative one,
raised to the power of one plus two,
times the corresponding minor.
The minor is equal to the determinant of the matrix
after eliminating row one and now column two.
We have a two by two determinant where row one
is zero zero, and row two is four six.
The entry in row one column three
is co-factor one comma three,
which is equal to negative one,
raised to the power of a one plus three
times the corresponding minor.
To find the minor, we eliminate row one column three.
The minor is equal to the determinant of
the remaining two by two determinant,
where row one is zero negative two,
and row two is four negative two.
Now I'm going to go ahead and stop here,
but the remaining co-factors are already listed,
and then we need to simplify them
to find the co-factor matrix.
Again, let's at least simplify the first row.
In row one column one,
we have the square of negative one, which is positive one.
And the determinant is equal to negative 12 minus zero
or negative 12.
One times negative 12, gives us negative 12
in row one column one of the co-factor matrix.
For the entry in row one column two,
we got the cube of negative one, which is negative one.
And the determinant is zero minus zero,
which of course is zero.
Giving us an entry of zero in row one column two
of the co-factor matrix.
Let's go and take a look at one more.
In row one column three,
we have the fourth power of negative one,
which is positive one.
And the determinant is zero minus negative eight,
which simplifies to one times positive eight,
which is eight, giving us an entry of eight
in row one column three of the co-factor matrix.
So you may want to pause the video
and verify the remaining entries for the co-factor matrix.
For number four, we're asked to find
the inverse of matrix A,
and we'll do this using the adjoint method,
where A inverse is equal to
one divided by the determinant of matrix A,
times the adjoint of A, where adjoint A is equal to
the transpose of the co-factor matrix.
So before we find the inverse matrix,
we do have to find the adjoint of A
by transposing the co-factor matrix that we just found.
So we just found this co-factor matrix,
and therefore the adjoint of A is equal to
the transpose of this matrix.
To transpose the matrix,
the first column of the co-factor matrix
becomes the first row in the transpose,
or the first row of the adjoint of A.
The second column of the co-factor matrix
becomes the second row of the transpose
or of the adjoint of A,
and third column of the co-factor matrix
becomes the third row of the transpose
or the third row of the adjoint matrix.
So now that we have the adjoint of A,
and we already had the determinant,
we can use the formula to find A inverse.
Let's do this on the next slide.
The determinant of A is equal to negative 56,
giving us one divided by negative 56.
And then we have times the adjoint of A,
which again is the transpose of the co-factor matrix.
Next we perform the scalar multiplication,
and then simplify the fractions.
Simplifying, the first role of A inverse
is three fourteenths, two sevenths and one seventh.
The second row is zero, negative one half, zero.
And the third row is negative one seventh,
negative five fourteenths and one fourteenth.
I hope you found this helpful.