Video Transcript
Properties of Determinants
In this video, we will learn how to
identify different properties of determinants and how to use these properties of
determinants to simplify problems. Before we start listing the
properties of determinants, we can recall we already know several different methods
of calculating the determinant of different matrices.
For example, we know we can
calculate the determinant of a square matrix by adding and subtracting the products
of the diagonals. Or alternatively, we can calculate
the determinant of a matrix by expanding over the first row. In fact, we can also show that we
can expand over any row or column of the matrix. We will still get the same value
for the determinant. What this means is we can choose
any row or column of the matrix to expand over to calculate its determinant; we can
choose the easiest row or column. We can use this result to prove a
very useful property of determinants.
If we had a square matrix with an
entire row or column filled with zeros, then we could calculate the determinant of
this matrix by expanding over the zero row or column. In this calculation, every term
would have a factor of zero. So we would be able to show the
determinant of this matrix is zero. And it is worth reiterating here we
can only calculate the determinant of square matrices. So, when we’re talking about the
properties of determinants, we assume all of our matrices are square. Otherwise, we wouldn’t be able to
calculate the determinant of this matrix.
Another useful property we can show
is that taking the transpose of a matrix does not affect its determinant. In other words, for any square
matrix 𝐴, the determinant of 𝐴 transpose is equal to determinant of 𝐴, where we
recall to take the transpose of a matrix, we write the rows of the matrix as the
corresponding columns of our new matrix. And before we continue, it’s worth
pointing out here we can prove all of the properties of the determinant shown in
this video. Most of the proofs of the
properties will just involve using the definition of a determinant and checking the
left-hand side of the equation and the right-hand side of the equation are
equal. However, as we’ll see, there’s far
too many properties of the determinant to prove in this video.
And before we move on to the next
property, there is one useful thing. We can use the results about the
determinant of a matrix being equal to the determinant of its transpose. If we wanted to prove one of our
previous properties involving rows and columns of a matrix, we already know the
determinant of a matrix is equal to the determinant of its transpose. And the rows of matrix 𝐴 are the
columns of matrix 𝐴 transpose. So, we would only need to prove
this property for rows and then use the transpose property to prove it for
columns.
The next property we can show is
that if all entries in a single row or column of a matrix share a common factor,
then we can take out this shared factor outside of the determinant. And we can see an example of
this. Consider the determinant of the
two-by-two matrix two, one, five, 10. In this matrix, we can see the
second row has a shared factor of five. Our determinant properties tell us
we can take the shared factor of five outside of the calculation for the
determinant. It’s equal to five times the
determinant of the two-by-two matrix two, one, one, two. And we could calculate both sides
of this equation separately.
In our first matrix, we take the
difference in the products of the diagonal. That’s two times 10 minus five
times one, which is equal to 15. And we would do the same in our
second matrix. We take the difference in the
products of the diagonals. That’s two times two minus one
times one, which is equal to three. Then, we multiply this by a factor
of five to see the right-hand side of the equation is also 15.
Another useful property we can show
is if a square matrix has a repeated row or column, then its determinant will be
equal to zero. And this result can be particularly
useful for finding the determinants of large matrices. We should always check if there’s a
repeated row or column in any matrix so we can determine that its determinant is
zero.
The next useful property we can
show is if we switch any two rows or columns of our matrix, then this will switch
the sign of the determinant of that matrix. To see an example of this, I’d
switch the two columns in our two-by-two matrix two, one, five, 10. Switching the first column, the
second column gives us the two-by-two matrix one, two, 10, five. And we have notation for this: 𝑐
sub one is the first column of the matrix and 𝑐 sub two is the second column of the
matrix. We use a double-sided arrow to show
that we’re interchanging the two columns. If we then calculate the
determinant of this matrix, we get one multiplied by five minus 10 multiplied by
two, which we can calculate is negative 15, the negative of the original matrix. So, switching any two rows or
columns of our matrix will switch the sign of our determinant.
Another property we can show about
determinants is that we can add or subtract scalar multiples of the rows or columns
of the matrix to other rows or columns of the matrix without affecting the
determinant. And to help us see what this means,
let’s take a look at an example. If we wanted to calculate the
determinant of the two-by-two matrix one, three, zero, two, we could do this
directly, or we could rewrite our matrix by using this property. We can replace the second column of
this matrix by subtracting three times the first column from this matrix. And remember, our property
guarantees that this won’t affect the value of the determinant. We can write this as 𝑐 two is
being replaced with 𝑐 two minus three 𝑐 one. Subtracting three lots of the first
column from the second column, we get zero, two.
Therefore, the determinant of the
two-by-two matrix one, three, zero, two is equal to the determinant of the
two-by-two matrix one, zero, zero, two. This property is particularly
useful for helping us simplify determinants of matrices of order three by three and
higher.
There are many more properties of
determinants we can show. So, let’s clear some space and go
over a few more of these properties. Our next property that we can show
is that if we multiply all of the entries of a row or column by the corresponding
cofactors of a different row or column, then the sum of these values will be
zero. This is quite a complicated
property, so it’s much easier to see what this means in an example. Let’s start by applying this
property to the three-by-three matrix one, two, three, four, five, six, seven,
eight, nine. And we’ll use the first row of this
matrix, but remember, we can use any row or column of this matrix.
We now want to multiply every entry
of this row of the matrix by the corresponding cofactors of another row of this
matrix. Then, when we sum these values, we
should get zero. Let’s multiply these entries by the
corresponding cofactors of the third row of the matrix; that’s the row seven, eight,
nine. Let’s start by finding the matrix
minor we get from the entry seven in our matrix. That’s the determinant of the
two-by-two matrix we get by removing the first column and third row from the
matrix. That’s the two-by-two matrix two,
three, five, six. Remember, we need to check the
parity of the sum of the row and column numbers of this entry, and it’s in the first
column and third row. These sum to give four, which is
even. So, this will be a positive
minor.
Finally, in our property, we’re
multiplying this by the corresponding entry in the first row. So, we multiply this by one. We can then do exactly the same for
the next entry in this row. The matrix minor of this entry will
be the determinant of the two-by-two matrix one, three, four, six. We need to take the negative of
this value. And finally, we need to multiply
this by the corresponding entry in the first row; that’s two.
We then do exactly the same for the
third entry in this row. We get three times the determinant
of the two-by-two matrix one, two, four, five. And our matrix property guarantees
that this sum must be equal to zero. We can calculate this to check that
this is true in this case. And there is another slightly
different way we can think about this property. We could say that we’re calculating
the determinant of this matrix by expanding over the third row of the matrix. However, we’re using the
coefficients of the first row instead.
Another useful property of
determinants we can show is the determinants of any square triangular matrix will be
equal to the products of all of its entries on its main diagonal. And there’s a few things worth
pointing out here. First, this works for both upper
and lower triangular matrices. Similarly, since diagonal matrices
are both upper and lower triangular matrices, this property will also work for
diagonal matrices. Finally, if any of the entries on
the main diagonal of our triangular matrix is zero, then this product will have a
factor of zero. So, its determinant is zero, and
therefore it’s not invertible. Let’s see a use of this property in
an example.
Let’s say we want to calculate the
determinant of the three-by-three matrix one, two, three, zero, four, five, zero,
zero, six. We can see we’re calculating the
determinant of an upper triangular matrix because all of the entries below the
leading diagonal are zero. And therefore, our property tells
us we can calculate the determinant of this matrix by just calculating the products
of the main diagonals. That’s one times four times six,
which is equal to 24.
The next property we’re going to
discuss is true for a square matrix of any order and is true for any row or column
of this matrix. However, if we wanted to include
all of these possibilities in our statement, then the statement itself would get
very complicated. So, we’ll just state this property
for the first row of a three-by-three matrix. However, this does hold for any
row, any column, and any-size square matrix.
The determinant of any square
matrix where one of the rows or columns is written as a sum can be split as
follows. It’s equal to the sum of the
determinants of the following two matrices. And the only difference between
these two matrices is we’ve split the row or column over which we had a sum in our
original matrix. The first matrix took the first
term in each of these sums, and the second matrix took the second term in each of
these sums.
And finally, it’s worth reiterating
although the statement we’ve written is only written for the first row of a
three-by-three matrix, the statement holds for any square matrix and it works for
any row or any column of the matrix. The second thing we might not
notice about this statement is that it also works in the opposite direction. In other words, if we’re
calculating the sum of the determinants of two or more matrices where the only
difference involves one row or one column, then we can add these rows or columns
together so we only need to calculate the determinant of one matrix.
There are two more properties of
determinants we need to show. First, the determinant of the
product of two matrices is the product of their determinants, where, of course, both
𝐴 and 𝐵 need to be square matrices. Otherwise, we wouldn’t be able to
calculate their determinants. And we can actually use this
property to show one final useful property. The determinant of 𝐴 times 𝐴 will
be equal to the determinant of 𝐴 multiplied by the determinant of 𝐴. In other words, the determinant of
𝐴 times 𝐴 is equal to the determinant of 𝐴 all squared.
We can continue this to show for
any positive integer exponent 𝑛 the determinant of matrix 𝐴 to the 𝑛th power is
equal to the determinant of matrix 𝐴 all raised to the 𝑛th power. Let’s now see some examples of how
we can use these properties of determinants to simplify and evaluate problems
involving the determinants of different matrices.
Find the value of the determinant
of the three-by-three matrix four, one, negative eight, negative six, three, six,
zero, zero, zero.
In this question, we’re asked to
evaluate the determinant of a three-by-three matrix. And we might be tempted to jump in
straightaway and evaluate this determinant by expanding over the first row. And this would work; we would get
the correct answer. However, whenever we’re asked to
evaluate a determinant, we can always check to see if we can simplify this problem
by using properties of determinants. And in this case, we can notice the
third row of this matrix is all zeros.
We can then recall the following
fact about the determinants of matrices. If all of the entries of a row or
column of a square matrix is zero, then its determinant will also be equal to
zero. And it’s worth pointing out, in
this case, this is exactly the same as saying that we’re going to calculate the
determinant of our matrix by expanding over its third row, since then in our
calculation of the determinant every single term would have a factor of zero. So, the determinant of this matrix
would still be zero. In either case, we were able to
show the determinant of the three-by-three matrix four, one, negative eight,
negative six, three, six, zero, zero, zero is equal to zero because it has a row
entirely made of zeros.
Let’s now see another example of
calculating the determinant of a three-by-three matrix without expanding over the
rows or columns.
Without expanding, find the value
of the determinant of the three-by-three matrix eight, negative three, negative two,
seven, one, negative eight, 24, negative nine, negative six.
In this question, we’re asked to
evaluate the determinant of a three-by-three matrix. We could do this by expanding over
any of its rows or columns. However, the question explicitly
asks us to do this without expanding. There’s a few different methods we
could use to evaluate determinants by using different properties of the
determinant. For example, we can recall that
we’re allowed to add and subtract linear multiples of the rows and columns to other
rows and columns of the matrix. And this won’t affect the
determinant of this matrix.
We could then use these to try and
rewrite our matrix as an upper triangular matrix. Then, the determinant of an upper
triangular matrix is the product of the entries on its leading diagonal. However, this is a complicated
process, so we should always check to see if there’s an easier method first. We can start by recalling that we
can take out shared factors among a row or column of the matrix. In particular, we can notice the
third row of this column all share a factor of three. Taking out the shared factor of
three from the row gives us three multiplied by the determinant of the
three-by-three matrix eight, negative three, negative two, seven, one, negative
eight, eight, negative three, negative two.
At this point, we might also notice
that in the third column of this matrix every single entry shares a factor of two or
negative two. And we could take this factor
outside of our matrix in exactly the same way. However, we can already evaluate
the determinant of this matrix by using a different property. We can instead notice that the
first row of this matrix is equal to its third row. We can then recall if a matrix has
a repeated row or column, then its determinant will be equal to zero. So, because the first and third row
of this matrix is repeated, we can conclude its determinant is zero, giving us three
times zero, which is equal to zero.
And before we finish with this
question, it’s worth noting we’ve shown a useful property about matrices. We’ve shown if one of the rows or
columns of this matrix is a scalar multiple of the other row or column of this
matrix, then its determinant is zero. In either case, we were able to
show the determinant of the three-by-three matrix eight, negative three, negative
two, seven, one, negative eight, 24, negative nine, negative six is equal to
zero.
Let’s now see an example of
combining multiple determinants into one determinant.
Evaluate the determinant of the
matrix negative six, one, one, one plus the determinant of the matrix negative five,
one, one, one plus the determinant of the matrix negative four, one, one, one. And we keep adding determinants of
matrices of this form up to the determinant of the matrix 10, one, one, one.
In this question, we’re asked to
evaluate an expression. And we can see in this expression
every term is the determinant of a two-by-two matrix. So, one way to answer this question
would just be to evaluate all of these determinants and add them together. However, we can simplify this
expression first by using the properties of determinants.
To do this, we need to notice
something interesting about all of the matrices we’re given. All of these matrices have exactly
the same second row. And one of our properties of
determinants tells us how to add two matrices together which only have one row or
column which differ. We recall that we can add the
determinants of two matrices together which have all of the same elements except for
the same row or column by combining these into one matrix where we add the
corresponding entries in the row or column that differs.
Now, we could use this property to
just add two of the determinants together. However, we can notice, in this
case, all of our matrices are of the same form. So instead, we’ll just apply this
property to all of the terms. So, to use this property on this
sum, we first know that the second row of this matrix is going to be one, one. Next, the entry in row one, column
one of our matrix is going to be the sum of all of the entries in row one, column
one of our terms. This will, of course, be negative
six plus negative five. And we keep adding integers of this
form up to 10.
We would then do the same for the
entries in row one, column two. However, we can see that all of
these entries are equal to one. So, when we add these together, we
would just get the number of terms. And the number of terms from
negative six to 10 is equal to 17. Therefore, by using the properties
of determinants, we were able to rewrite the sum of determinants as one
determinant.
Now, to evaluate the determinant of
this two-by-two matrix, we’re going to need to evaluate the sum we have in its first
row and first column. And there’s several different ways
we can do this. For example, this is an arithmetic
sequence with first term negative six and common difference one. We could then use the formula for
the sum of a finite arithmetic sequence. However, there is a second way we
can do this. We can notice the term negative six
will cancel with six, negative five will cancel with five, and this will continue
all the way down to negative one canceling with one. And of course, adding zero doesn’t
change its value. So, this simplifies to give us
seven plus eight plus nine plus 10. And if we evaluate this, we see
it’s equal to 34. Therefore, the sum of these
determinants is equal to the determinant of the two-by-two matrix 34, 17, one,
one.
Finally, we can evaluate the
determinant of this matrix by finding the difference in the products of its
diagonals. That’s 34 times one minus 17 times
one, which is equal to 17. Therefore, we were able to evaluate
the sum of these determinants by using our properties of determinants. We were able to show this sum was
equal to 17.
Let’s now see one last example of
using the properties of the determinants to answer questions.
Consider the determinant of the
two-by-two matrix 𝑥, 𝑦, 𝑧, 𝑤 is equal to six. Find the value of the determinant
of the two-by-two matrix 𝑥 minus 10𝑦, 𝑦, 𝑧 minus 10𝑤, 𝑤.
In this question, we’re given the
determinant of a two-by-two matrix and asked to determine the determinant of a
different two-by-two matrix. And both of these matrices contain
four unknowns, so we can’t evaluate these directly. We might be tempted to expand the
determinant in our first matrix and then expand the determinant in our second matrix
and try to rewrite this expression in terms of our first determinant. However, there’s a much simpler
method involving the properties of determinants.
To do this, we need to notice
something interesting about the second matrix we’re given. In the first entry of the first
column, we’re subtracting 10𝑦, and in the second entry of the second column, we’re
subtracting 10𝑤. This is a scalar multiple of the
second column of our first matrix. In fact, we’re subtracting this
directly from the first column of this matrix. In other words, to generate the
second matrix, we’re subtracting 10 lots of the second column from the first column
of our first matrix. And we can recall adding and
subtracting scalar multiples of one row or column to a different row or column will
not affect the determinant. Therefore, the determinant of the
second matrix will be equal to the determinant of the first matrix, which we’re told
is equal to six.
Let’s now go over some of the key
points of this video. First, we showed that we can use
all of the many properties of determinants to simplify expressions involving
determinants. This means whenever we’re given an
expression involving determinants or asked to evaluate a determinant, we should
always check to see if we can simplify this problem by using the properties of
determinants. And there are many properties of
the determinants, so many that we can’t list all of them in the key points.
A few of these are taking the
transpose of a square matrix doesn’t affect its determinant. If all of the entries in a row or
column of a square matrix are zero, then its determinant will be equal to zero. If a square matrix has a repeated
row or column, then its determinant will be equal to zero. And the determinant of any
triangular matrix will be the product of its main diagonals. And of course, this also extends to
diagonal matrices.
