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In this section we’ll take a look at the second method for
computing determinants. The idea in this
section is to use row reduction on a matrix to get it down to a row-echelon
form.
Since we’re computing determinants we know that the matrix, A, we’re working with will be square and
so the row-echelon form of the matrix will be an upper triangular matrix and we
know how to quickly
compute the determinant of a triangular matrix.
So, since we already know how to do row reduction all we need to know
before we can work some problems is how the row operations used in the row
reduction process will affect the determinant.
Before proceeding we should point out that there are a set
of elementary column operations that mirror the elementary row operations. We can multiply a column by a scalar, c, we can interchange two columns and we can add a multiple of one column onto another column. The operations could just as easily be used
as row operations and so all the theorems in this section will make note of
that. We’ll just be using row operations
however in our examples.
Here is the theorem that tells us how row or column
operations will affect the value of the determinant of a matrix.
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Theorem 1 Let A be a square matrix.
(a) If
B is the matrix that results from
multiplying a row or column of A by
a scalar, c, then 
(b) If
B is the matrix that results from
interchanging two rows or two columns of A
then
(c) If
B is the matrix that results from
adding a multiple of one row of A
onto another row of A or adding a
multiple of one column of A onto
another column of A then
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Notice that the row operation that we’ll be using the most
in the row reduction process will not change the determinant at all. The operations that we’re going to need to
worry about are the first two and the second is easy enough to take care
of. If we interchange two rows the
determinant changes by a minus sign. We
are going to have to be a little careful with the first one however. Let’s check out an example of how this method
works in order to see what’s going on.
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Example 1 Use
row reduction to compute the determinant of the following matrix.
Solution
There is of course no real reason to do row reduction on
this matrix in order to compute the determinant. We can find it easily enough at this
point. In fact let’s do that so we can
check the results of our work after we do row reduction on this.
Okay, now let’s do that with row reduction to see what we’ve
got. We need to reduce this down to
row-echelon form and while we could easily use a multiple of the second row to
get a 1 in the first entry of the first row let’s just divide the first row
by 4 since that’s the one operation we’re going to need to be careful with. So, let’s do the first operation and see
what we’ve got.
So, we called the result B and let’s see what the determinant of this matrix is.
So, the results of the theorem are verified for this
step. The next step is then to convert
the -7 into a zero. Let’s do that and
see what we get.
According to the theorem C should have the same determinant as B and it does (you should verify this statement).
The final step is to convert the 26 into a 1.
Now, we’ve got the following,
Once again the theorem is verified.
Now, just how does all of this help us to find the
determinant of the original matrix? We
could work our way backwards from det(D)
and figure out what det(A) is. However, there is a way to modify our work
above that will allow us to also get the answer once we reach row-echelon
form.
To see how we do this let’s go back to the first operation
that we did. After doing the operation we had,
Written in another way this is,
Notice that the determinants, when written in the “matrix”
form, are pretty much what we originally wrote down when doing the row
operation. Therefore, instead of
writing down the row operation as we did above let’s just use this “matrix”
form of the determinant and write the row operation as follows.
In going from the matrix on the left to the matrix on the
right we performed the operation and in the process we changed the value of
the determinant. So, since we’ve got
an equal sign here we need to also modify the determinant of the matrix on
the right so that it will remain equal to the determinant of the matrix on
the left. As shown above, we can do
this by multiplying the matrix on the right by the reciprocal of the scalar
we used in the row operation.
Let’s complete this and notice that in the second step we
aren’t going to change the value of the determinant since we’re adding a
multiple of the second row onto the first row so we’ll not change the value
of the determinant on the right. In
the final operation we divided the second row by 26 and so we’ll need to
multiply the determinant on the right by 26 to preserve the equality of the
determinants.
Here is the complete work for this problem using these
ideas.
Okay, we’re down to row-echelon form so let’s strip out
all the intermediate steps out and see what we’ve got.
The matrix on the right is triangular and we know that
determinants of triangular matrices are just the product of the main diagonal
entries and so the determinant of A
is,
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Now, that was a lot of work to compute the determinant and
in general we wouldn’t use this method on a matrix, but by doing it on one here it allowed
us to investigate the method in a detail without having to deal with a lot of
steps.
There are a couple of issues to point out before we move
into another more complicated problem.
First, we didn’t do any row interchanges in the above example, but the
theorem tells us that will only change the sign on the determinant. So, if we do a row interchange in our work
we’ll just tack a minus sign onto the determinant.
Second, we took the matrix all the way down to row-echelon
form, but if you stop to think about it there’s really nothing special about
that in this case. All we need to do is
reduce the matrix to a triangular matrix and then use the fact that can quickly
find the determinant of any triangular matrix.
From this point on we’ll not be going all the way to
row-echelon form. We’ll just make sure
that we reduce the matrix down to a triangular matrix and then stop and compute
the determinant.
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Example 2 Use
row reduction to compute the determinant of the following matrix.
Solution
We’ll do this one with less explanation. Just remember that if we interchange rows
tack a minus sign onto the determinant and if we multiply a row by a scalar
we’ll need to multiply the new determinant by the reciprocal of the scalar.
Okay, we’ve gotten the matrix down to triangular form and
so at this point we can stop and just take the determinant of that and make
sure to keep the scalars that are multiplying it. Here is the final computation for this
problem.
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Example 3 Use
row reduction to compute the determinant of the following matrix.
Solution
Okay, there’s going to be some work here so let’s get
going on it.
Okay, that was a lot of work, but we’ve gotten it into a
form we can deal with. Here’s the
determinant.
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Now, as the previous example has shown us, this method can
be a lot of work and its work that if we aren’t paying attention it will be
easy to make a mistake.
There is a method that we could have used here to
significantly reduce our work and it’s not even a new method. Notice that with this method at each step we
have a new determinant that needs computing.
We continued down until we got a triangular matrix since that would be
easy for us to compute. However, there’s
nothing keeping us from stopping at any step and using some other method for
computing the determinant. In fact, if
you look at our work, after the second step we’ve gotten a column with a 1 in
the first entry and zeroes below it. If
we were in the previous section we’d just do a cofactor expansion along this
column for this determinant. So, let’s do
that. No one ever said we couldn’t mix
the methods from this and the previous section in a problem.
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Example 4 Use
row reduction and a cofactor expansion to compute the determinant of the
matrix in Example 3.
Solution.
Okay, this “new” method says to use row reduction until we
get a matrix that would be easy to do a cofactor expansion on. As noted earlier that means only doing the
first two steps. So, for the sake of
completeness here are those two steps again.
At this point we’ll just do a cofactor expansion along the
first column.
At this point we can use any method to compute the
determinant of the new matrix so we’ll leave it to you to verify
that
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There is one final idea that we need to discuss in this
section before moving on.
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Theorem 2 Suppose that A is a square matrix and that two of
its rows are proportional or two of its columns are proportional. Then .
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When we say that two rows or two columns are proportional
that means that one of the rows (columns) is a scalar times another row (column)
of the matrix.
We’re not going to prove this theorem but if you think about
it, it should make some sense. Let’s
suppose that two rows are proportional.
So we know that one of the rows is a scalar multiple of another
row. This means we can use the third row
operation to make one of the rows all zero.
From Theorem 1 above we know that both of these matrices must have the
same determinant and from Theorem
7 from the Determinant Properties section we know that if a matrix has a
row or column of all zeroes, then that matrix is singular, i.e. its determinant is zero.
Therefore both matrices must have a zero determinant.
Here is a quick example showing this.
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Example 5 Show
that the following matrix is singular.
Solution
We can use Theorem 2 above upon noticing that the third
row is -2 times the first row. Using this theorem is all we really need to do this example.
So, technically we’ve answered the question. However, let’s go through the steps
outlined above to also show that this matrix is singular. To do this we’d do one row reduction step
to get the row of all zeroes into the matrix as follows.
We know by Theorem 1 above that these two matrices have
the same determinant. Then because we
see a row of all zeroes we can invoke Theorem 7 from the
Determinant Properties to say that the determinant on the right must be zero,
and so be singular.
Then, as we pointed out, these two matrices have the same
determinant and so we’ve also got and so A
is singular.
You might want to verify that this matrix is singular by
computing its determinant with one of the other methods we’ve looked at for
the practice.
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We’ve now looked at several methods for computing
determinants and as we’ve seen each can be long and prone to mistakes. On top of that for some matrices one method may
work better than the other. So, when
faced with a determinant you’ll need to look at it and determine which method
to use and unless otherwise specified by the problem statement you should use
the one that you find the easiest to use.
Note that this may not be the method that somebody else chooses to use,
but you shouldn’t worry about that. You
should use the method you are the most comfortable with.