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In this section we’re going to examine one of the two
methods that we’re going to be looking at for computing the determinant of a
general matrix. We’ll also see how some
of the ideas we’re going to look at in this section can be used to determine
the inverse of an invertible matrix.
So, before we actually give the method of cofactors we need
to get a couple of definitions taken care of.
Let’s take a look at computing some minors and cofactors.
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Example 1 For
the following matrix compute the cofactors  ,
,
and .
Solution
In order to compute the cofactors we’ll first need the
minor associated with each cofactor.
Remember that in order to compute the minor we will remove the ith
row and jth column of A.
So, to compute (which we’ll need for ) we’ll need to compute the
determinate of the matrix we get by removing the 1st row and 2nd
column of A. Here is that work.
We’ve marked out the row and column that we eliminated and
we’ll leave it to you to verify the determinant computation. Now we can get the cofactor.
Let’s now move onto the second cofactor. Here is the work for the minor.
The cofactor in this case is,
Here is the work for the final cofactor.
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Notice that the cofactor is really just depending upon i and j. If the subscripts of the cofactor add to an
even number then we leave the minor alone (i.e.
no “-” sign) when writing down the cofactor.
Likewise, if the subscripts on the cofactor sum to an odd number then we
add a “-” to the minor when writing down the cofactor.
We can use this fact to derive a table that will allow us to
quickly determine whether or not we should add a “-” onto the minor or leave it
alone when writing down the cofactor.
Let’s start with . In this case the subscripts sum to an even
number and so we don’t tack on a minus sign to the minor. Now, let’s move along the first row. The next cofactor would then be and in this case the subscripts add to an odd
number and so we tack on a minus sign to the minor. For the next cofactor, ,
we would leave the minor alone and for the next, ,
we’d tack a minus sign on, etc.
As you can see from this work, if we start at the leftmost
entry of the first row we have a “+” in front of the minor and then as we move
across the row the signs alternate. If
you think about it, this will also happen as we move down the first
column. In fact, this will happen as we
move across any row and down any column.
We can summarize this idea in the following “sign matrix”
that will tell us if we should leave the minor alone (i.e. tack on a “+”) or change its sign (i.e. tack on a “-”) when writing down the cofactor.
Okay, we can now talk about how to use cofactors to compute
the determinant of a general square matrix.
In fact there are two ways we can used cofactors as the following
theorem shows.
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Theorem 1 If A is an matrix.
(a) Choose any row, say row i, then,
(b) Choose any column, say
column j, then,
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What this theorem tells us is that if we pick any row all we
need to do is go across that row and multiply each entry by its cofactor, add
all these products up and we’ll have the determinant for the matrix. It also says that we could do the same thing
only instead of going across any row we could move down any column.
The process of moving across a row or down a column is often
called a cofactor expansion.
Let’s work some examples of this so we can see it in action.
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Example 2 For
the following matrix compute the determinant using the given cofactor
expansions.
(a) Expand
along the first row. [Solution]
(b) Expand
along the third row. [Solution]
(c) Expand
along the second column. [Solution]
Solution
First, notice that according to the theorem we should get
the same result in all three parts.
(a) Expand along the first
row.
Here is the cofactor expansion in terms of symbols for
this part.
Now, let’s plug in for all the quantities. We will just plug in for the entries. For the cofactors we’ll write down the
minor and a “+1” or a “-1” depending on which sign each minor needs. We’ll determine these signs by going to our
“sign matrix” above starting at the first entry in the particular row/column
we’re expanding along and then as we move along that row or column we’ll
write down the appropriate sign.
Here is the work for this expansion.
We’ll leave it to you to verify the determinant computations.
[Return to Problems]
(b) Expand along the third
row.
We’ll do this one without all the explanations.
So, the same answer as the first part which is good since
that was supposed to happen.
Notice that the signs for the cofactors in this case were
the same as the signs in the first case.
This is because the first and third row of our “sign matrix” are
identical. Also, notice that we didn’t
really need to compute the third cofactor since the third entry was
zero. We did it here just to get one
more example of a cofactor into the notes.
[Return to Problems]
(c) Expand along the
second column.
Let’s take a look at the final expansion. In this one we’re going down a column and
notice that from our “sign matrix” that this time we’ll be starting the
cofactor signs off with a “-1” unlike the first two expansions.
Again, the same as the first two as we expected.
[Return to Problems]
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There was another point to the previous problem apart from
showing that the row or column we choose to expand along won’t matter. Because we are allowed to expand along any
row that means unless the problem statement forces to use a particular row or
column we will get to choose the row/column to expand along.
When choosing we should choose a row/column that will reduce
the amount of work we’ve got to do if possible.
Comparing the parts of the previous example should suggest to us
something we should be looking for in making this choice. In part (b)
it was pointed out that we didn’t really need to compute the third cofactor
since the third entry in that row was zero.
Choosing to expand along a row/column with zeroes in it will
instantly cut back on the number of cofactors that we’ll need to compute. So, when allowed to choose which row/column
to expand along we should look for the one with the most zeroes. In the case of the previous example that
means that the quickest expansions would be either the 3rd row or
the 3rd column since both of those have a zero in them and none of
the other rows/columns do.
So, let’s take a look at a couple more examples.
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Example 3 Using
a cofactor expansion compute the determinant of,
Solution
Since the row or column to use for the cofactor expansion
was not given in the problem statement we get to choose which one we want to
use. Recalling the brief discussion
after the last example we know that we want to choose the row/column with the
most zeroes in it since that will mean we won’t have to compute cofactors for
each entry that is a zero.
So, it looks like the second row would be a good choice
for the expansion since it has two zeroes in it. Here is the expansion for this row. As with the previous expansions we’ll
explicitly give the “+1” or “-1” for the cofactors and the minors as well so
you can see where everything in the expansion is coming from.
We didn’t bother to write down the minors and because of the zero entry. How we choose to compute the determinants
for the first and last entry is up to us at this point. We could use a cofactor expansion on each
of them or we could use the technique we learned in the first section of this
chapter. Either way will get the same
answer and we’ll leave it to you to verify these determinants.
The determinant for this matrix is,
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Example 4 Using
a cofactor expansion compute the determinant of,
Solution
This is a large matrix, but if you check out the third
column we’ll see that there is only one non-zero entry in that column and so
that looks like a good column to do a cofactor expansion on. Here’s the cofactor expansion for this
matrix. Again, we explicitly added in
the “+1” and “-1” and won’t bother to write down the minors for the zero
entries.
Now, in order to complete this problem we’ll need to take
the determinant of a matrix and the only way that we’ve got to do
that is to once again do a cofactor expansion on it. In this case it looks like the third row
will be the best option since it’s got more zero entries than any other row
or column.
This time we’ll just put in the terms that come from
non-zero entries. Here is the
remainder of this problem. Also don’t
forget that there is still a coefficient of 3 in front of this determinant!
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This last example has shown one of the drawbacks to this
method. Once the size of the matrix gets
large there can be a lot of work involved in the method. Also, for anything larger than a matrix you are almost assured of having to do
cofactor expansions multiple times until the size of the matrix gets down to and other methods can be used.