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Remember that in the definition of the determinant of an n-by-n matrix
there are n factorial terms.
So you might be wondering, is there a fast way to compute determinants?
There are at least two other ways to do it.
One is using the cofactor formula which I'm not going to discuss here.
The other one is to use row reduction.
And the method of computing determinants using row reduction
depends on the following result: If you have 2 n-by-n matrices A and B,
then the determinant of the product A B
is the same as the product of determinant of A and determinant of B.
This is something that we discussed in earlier videos.
Now the key is, when you row reduce A to R where R is in reduced row-echelon form
you can write down elementary matrices M_1 up to M_k such that
the product is equal to R.
Now, each of these elementary matrices is a square matrix
and so the determinant of A is given by
the determinant of R divided by
the product of the determinant of M_1
up to the determinant of M_k. So with this,
all we need to know is how to compute determinants of elementary matrices,
which will see is quite a simple thing,
and to compute the determinant of a matrix that is in reduced row-echelon form.
Because R is square
and it is in reduced row-echelon form, there are only two possibilities.
The first is that R is equal to the identity matrix
or R contains a row of 0's.
In the first case the determinant is simply 1.
And in the second case,
it is not hard to see that from the definition of the determinant,
the determinant of R is going to be 0 because
in the definition of the determinant, you need to pick it element from each row.
And if there's a row of 0, then each term in the definition of the determinant will
always be 0.
So before we look at determinants of
elementary matrices in general,
let's look an example
to see how this observation here can be used to compute the determinant of A
where A is given by the following matrix.
So let's row reduce A.
And we'll do that by adding -2 times the first row to the second row.
So that will give us this matrix.
And let me call this matrix Q_1.
So the determinant of A is going to be
the determine of Q_1 divided by the determinant of the matrix that
corresponds to this elementary row operation, which is this.
And the determinant of this,
if you compute it, is going to be 1. So this is the same
as the determinant of Q_1.
Now, if we row reduce Q_1, what we can do is we can add row 2 to row 3
and that will give us the following matrix.
And let me call this Q_2.
So the determinant of Q_1 is going to be
the same as the determinant of Q_2
divided by the determinant of the matrix that corresponds to this elementary row operation.
And in this case,
it's going to be this matrix.
And again you compute the determinant of this thing down here
you get one as well. So this is simply the determinant of Q_2.
And now...uh...let's do one more step.
So if we row reduce this like by adding row 3 to row 1,
I'll get down to the identity matrix.
So this will be my final matrix R.
And so this is going to be the determinant of R divided the determinant of
the elementary matrix that corresponds to this elementary row operation
which again will have determinant 1.
So, in the end this is going to have determinant equal to 1.
So one thing that you see here is
for all three operations that I've performed here, they are of the same kind
and they all have determinant equal to 1.
Now, this is actually not a coincidence and we'll now take care of
all the three elementary row operations and see what the determinants are.
So before we begin, let's recall the definition of the determinant of A.
So A is an n-by-n matrix and this script S sub n denotes the set of
all permutations of 1 up to n.
This is -1 to the number of inversions of the permutation sigma
and A_{1, sigma(1)} means the element of A in the first row and the column
and the column indexed by sigma(1), and so on.
So this is the definition of the determinant of A.
Now, let's consider the elementary matrix that corresponds to
swapping row i and row j.
So the matrix looks like this:
1's on the diagonal except in the ith and jth entry.
So suppose this is
column i and this is column j and this is
row i and this is row j.
Then what we'll see is 1's here.
Right? And everything else is 0.
Now if you look at the formula up here, each term here
is a product that contains exactly one element from
each row and each column. So this matrix is full of 0's.
And the only way you can get a non-zero product here
is to pick the nonzero element from
each of these rows and there's only one permutation that does that.
And it's the permutation given by this
where you get it from the identity permutation
with j and i swapped.
And the number of inversions of this thing, if you think about it
is going to be odd.
So, each element here of the term determined by this permutation is 1
but the number inversions is odd. As a result, the determinant is going to be -1.
In other words, if you look at the elementary matrix that corresponds to
swapping any two rows,
the determinant is always going to be -1.
Now, let's look at the elementary matrices
for the other two elementary row operations.
And those are actually easy to see.
If you accept the following result:
the determinant of a lower-triangular or upper-triangular matrix
is given by the product of entries on the diagonal.
So what is a lower-triangular or upper triangular matrix?
A lower-triangular matrix is simply a matrix in which
all the nonzero entries appear on the diagonal or below the diagonal
like this one. So this is a lower triangular matrix.
And upper-triangular, as you might guess, has nonzero entries
on or above the diagonal like this one.
And if the matrix looks like one of these then you can compute its determinant by
just multiplying the diagonal entries. So in both case,
the diagonal contains a 0, so the determinant of either matrix is 0.
If you look at the elementary matrix that correspond to multiply a row
by a constant alpha, then it will look something like this:
say you look at the elementary matrix for multiplying row i by a constant alpha
then this is the matrix. So 0's everywhere else
1's on the diagonal except on the (i,i)-entry where it will be alpha.
And the determinant of this matrix is going to be just alpha.
Now, if you look at the elementary matrix that corresponds to adding
alpha times row i to row j
it will look something like this: so 1's on the diagonal
and alpha either somewhere up here... down here
below the diagonal or above the diagonal
and in either case the matrix is
upper triangular or lower triangular.
So the determinant is going to be the product of the diagonal entries
which is just 1.
And this is exactly what we found in
our example.
So all the elementary matrices that corresponds to
adding a multiple of a row to another row will have determinant equal to 1.
So with these observations
you can now compute determinants via row reduction.