In this example, we are going to compute the adjugate matrix of a matrix.
So given a matrix, you can compute its adjugate matrix.
So the matrix we're going to work with is the matrix A, which is a minus 2, 5, 1, 4,
1, 0, and a 3, 5, 5.
So that's the matrix A. And what we're asked to do is to compute the
adjugate matrix of the matrix A. So first, let's talk about the general definition of
an adjugate matrix.
In general, when we're working with a square matrix, so, again, this has to be for a square
matrix, the adjugate of the matrix A can be written in this general form C1,1, C2,1, all
the way to Cn1, C1,2, C2,2, all the way down to Cn2.
And then down this column.
So this is what the adjugate matrix form looks like in general, C1,1 is just a number, C2,1
is just a number, C1,2 is just a number.
These are all just scalar numbers.
And these numbers are actually the cofactors of the matrix A.
For our particular problem, since we're working with a matrix A that has n equal to 3, we
don't have to go to end here.
We only have to go to 3.
So if we want to fill this definition, which is a general definition for our specific case,
it turns into this 3 by 3 matrix that we're actually looking to compute in this problem.
So there's actually nine numbers here that I need to know to be able to write down the
adjugate matrix.
Notice one thing here, the subscripts written here are correct C1,1, C2,1, C3,1, et cetera,
across that first row.
It looks a little backwards, right?
Normally, we do rows and columns, so you might have guessed that should have been C1,2, and
this should have been C1,3 because these are in the first row.
But the adjugate matrix is actually transposed in this definition in terms of the cofactors.
So this is written down correctly.
And just be careful, when you populate your matrix not to get things out of order.
So what do we need to do to actually compute the adjugate matrix for our particular example
we're working?
We need to compute these nine numbers.
So let's go about computing those cofactors.
So C1,1 I can get by blanking out the first row in the first column of matrix A. What
I'm left with is the 2 by 2 matrix 1, 0, 5, 5, and I can compute the determinant of that.
And I get the number 5.
So C1,1 is equal to 5.
What about C2,1-- C1,2?
I'm sorry.
C1,2, very similar thing, I need to cross out the first row, second column.
And then, what am I going to be left with?
4, 0, negative 3, 5.
The other tricky part is, remember when we do cofactors the sign toggles back and forth.
So I'm actually going to have a negative sign here and then the matrix 4, 0, a negative
3, 5, like I said.
So I need to take the negative of this determinate quantity.
So I've got the negative of 20 minus 0, which is a negative 20.
And then we can just keep going.
We need to compute C1,3, et cetera.
So let's just keep going.
C1,3 is going to be positive, the determinant of some 2 by 2 matrix.
To know what that 2 by 2 matrix is, I went ahead and wrote down the A matrix again, so
row 1, column 3 I'm going to strike out.
So what am I left with?
I'm left with this 2 by 2 matrix.
That's the 2 by 2 matrix I need to fill in here.
So that's what I've written down right there.
If I do my little computation I get 20 minus a negative 3, which is 23.
And we just keep going.
C2,1, I'm just going to change signs, so it's a negative.
I get the 2 by 2 matrix by crossing out the second row, first column, and I'm left with
5, 1, 5, 5.
So that's what goes right there.
I get a negative 25 minus 5, which is a negative 20.
C2,2 will be plus the determinant of this 2 by 2 matrix.
So that is a negative 10 minus a minus 3, which gives me a negative 7.
And Three more to go now.
C2,3 will be minus the determinant of this 2 by 2 matrix, which is a negative, a negative
10 minus a minus 15, which is a negative 5.
C3,1 I get by striking out the third row and first column.
So I'm left with this submatrix 5, 1, 1, 0.
5, 1, 1, 0.
C3,2, and actually I just noticed a little error.
Originally I had a minus sign here, but this should definitely be a plus sign right there,
right?
We've been toggling plus, minus, plus, minus as we go due to the co-factor definition.
So this should definitely be a plus there, which means it's a plus here, which is indeed
a negative 1.
So This final negative 1 is correct.
But just a second ago that was minus and that should have been plus.
Anyway, continuing our computation.
C3,2 we get by striking out the third row of A and the second column of A. We're left
with negative 2, 1, 4, 0.
We do this computation and we get plus 4.
And then, finally, C3,3 we're going to have plus this 2 by 2 determinant computation,
which is equal to a negative 22.
So we've now computed all nine of these numbers.
If we go back to the definition of the adjugate matrix, we can now fill them in.
And if we fill them in, this is what we get for our final adjugate matrix computation.