### Video Transcript

Properties of Determinants

In this video, we will learn how to
identify the properties of determinants and use these properties to solve
problems. We will focus on only four
properties, and all of the properties listed will hold on square matrices of any
order. However, we will only focus on
matrices of order two by two and three by three. Before we begin with these
properties, letβs start by recalling how we calculate the determinants of a
two-by-two matrix.

We recall the determinant of a
two-by-two matrix is the difference in the product of the diagonals. In this matrix, thatβs π one one
times π two two minus π one two multiplied by π two one. We can use this definition to
consider our first property. Letβs start by calling this matrix
π΄. We want to see if we can construct
a formula for the determinant of a scalar multiple of π΄. And of course, in this case, weβre
only working with two-by-two matrices. We can do this directly from the
definition of a determinant and scalar multiplication. Remember, when we multiply a matrix
by a scalar, we multiply every entry of our matrix by our scalar.

We can then evaluate the
determinant of this matrix by taking the difference in the products of the
diagonals. This then gives us the determinant
of matrix π times π΄ is equal to π times π one one multiplied by π times π two
two minus π times π one two multiplied by π times π two one. And we can simplify this
expression. We can notice the first term has a
factor of π squared and the second term also has a factor of π squared. Taking out this factor of π
squared gives us the following expression. We can see itβs π squared
multiplied by the difference in the product of the diagonals. Itβs the determinant of π΄. Therefore, weβve shown for any
scalar multiple π and two-by-two matrix π΄, the determinant of π times π΄ is π
squared multiplied by the determinant of π΄.

We could show a very similar result
for three-by-three matrices by using the definition of a determinant, the biggest
difference being when calculating a determinant of a three-by-three matrix, we have
three factors. So, weβll have factors of π cubed
instead of π squared. And in fact, this property is true
in general, although weβve only proven it for two-by-two matrices. We have for any square matrix π΄ of
order π by π and any scalar value of π, the determinant of π times π΄ is equal
to π to the πth power multiplied by the determinant of π΄. When calculating the determinant of
a matrix, we can take out a scalar factor of π. We just need to raise it to the
πth power where π is the order of matrix π΄. There are other operations we can
apply to square matrices. So, letβs see how these affect the
determinant of our matrix.

One of these operations is the
transpose of a matrix. Remember, this means we write the
rows as the columns of the matrix and the columns as the row of the matrix. Letβs see how the determinant of π΄
transposed compares of the determinant of π΄ for our two-by-two matrix. First, itβs worth reiterating when
we take the transpose of a matrix, we write the rows of the matrix as the columns of
our new matrix. We write the first row of matrix π΄
as the first column of matrix π΄ transpose. And we can see the only difference
between matrix π΄ and matrix π΄ transpose is the element in row one, column two is
switched with the element in row two, column one. We can calculate the determinant of
this matrix by taking the difference in the products of the diagonals. We get π one one times π two two
minus π two one multiplied by π one two. Then, we can write this in terms of
the determinant of π΄ by using the commutativity of scalar multiplication. This expression is just equal to
the determinant of π΄.

Therefore, weβve proven for any
two-by-two matrix the determinant of π΄ transpose is equal the determinant of
π΄. We can prove the exact same result
is true for three-by-three matrices or, in fact, any square matrix. The easiest way to do this is to
expand the determinant of π΄ over the first row and then expand the determinant of
π΄ transpose over the first column. This then gives us our second
property. For any square matrix π΄, the
determinant of π΄ transpose is just equal to the determinant of π΄. Letβs now move on to properties
involving the determinants of different ways of combining matrices.

The first way we might think to
combine matrices is to add them together. If we have two two-by-two matrices
π΄ and π΅, what is the determinant of their sum? However, it turns out this is very
difficult to do. And for any order matrices, this
becomes impossible. So instead, letβs consider the
product of these two matrices. What is the determinant of matrix
π΄ times matrix π΅?

Weβll start by constructing an
expression for the determinant of π΄ times π΅. Remember, when we find the product
of two matrices, the elements in row π and column π of their products will be the
sum of the products of the entries in row π of matrix π΄ and column π of matrix
π΅. So, for example, the entry in row
one, column one of matrix π΄ times π΅ is π one one times π one one plus π one two
multiplied by π two one. We can apply the same process to
find the matrix π΄π΅. We need to find the determinant of
this matrix. To do this, we need to take the
difference in the products of the diagonals. Doing this, distributing over our
parentheses and then simplifying gives us the following expression. And this is a very
complicated-looking expression.

However, we can notice something
interesting. This expression is actually equal
to the determinant of matrix π΄ multiplied by the determinant of matrix π΅. We can notice this by taking
factors out of our expression or by using the FOIL method to distribute over these
parentheses. Weβve shown for any two-by-two
matrices π΄ and π΅, the determinant of π΄ times π΅ is equal to the determinant of π΄
multiplied by the determinant of π΅. And in fact, we can prove that this
result holds in general. We have the third property of
determinants of matrices. If π΄ and π΅ are any square
matrices of the same order, then the determinant of π΄ times π΅ is equal to the
determinant of π΄ multiplied by the determinant of π΅.

Thereβs one more property we want
to prove about the determinant of matrices. This time, itβs about the
determinant of a specific type of matrix. This time, we want to consider the
determinant of any triangular matrix, for example, the two-by-two upper triangular
matrix π one one, π one two, zero, π two two. We can calculate the determinant of
this matrix; itβs the difference in the products of the diagonal. This gives us π one one times π
two two minus zero times π one two. And the second term has a factor of
zero, so itβs equal to zero. We can notice something
interesting. The determinant of this matrix is
the product of the entries on its main diagonal. We can show the same is true for a
lower triangular matrix of order two by two. Remember, a lower triangular matrix
is one where every entry above the main diagonal is equal to zero. The determinant of the two-by-two
matrix π one one, zero, π two one, π two two is equal to π one one multiplied by
π two two, the product of the entries on its main diagonal.

Therefore, weβve proven the
determinant of any two-by-two triangular matrix is the product of the entries on its
main diagonal. In fact, we can prove this result
for upper or lower triangular matrices of order three by three, just by using the
definition of a determinant. And, in turn, we can prove this for
any higher-order square matrices. This gives us our fourth property
of determinants. The determinant of any square
triangular matrix, upper or lower, is the product of the elements on its main
diagonal. Letβs now see an example of how we
can use some of these properties to help us answer questions involving the
determinants of different matrices.

If π΄ is a square matrix of order
two by two and the determinant of two π΄ is equal to 12, then the determinant of
three times π΄ transpose is equal to what. Is it option (A) 18, option (B) 24,
option (C) 27? Or is it option (D) 36?

In this question, weβre given some
information about the determinant of a two-by-two matrix π΄. Weβre told the determinant of two
π΄ is equal to 12, and we need to use this information to determine the determinant
of three times the transpose of π΄. Since weβre told that π΄ is a
two-by-two matrix, we might be tempted to start by defining π΄ to be a matrix of
four unknowns. We could then substitute our
expression for π΄ into our equation to find an expression for the determinant of π΄
and then try to use this to find an expression for the determinant of three times
the transpose of π΄. And this would work; however, it
would be very complicated. Instead, we need to notice that our
equations involve determinants of matrices.

So instead, weβll start by
simplifying by using the properties of determinants. Weβll start by simplifying the
expression the determinant of two times π΄. And to do this, weβll start by
recalling the following property. For any square matrix π΅ of order
π by π and any scalar value π, the determinant of π times π΅ is equal to π to
the πth power multiplied by the determinant of π΅. In our case, π΄ is a matrix of
order two by two. So, our value of π is two, giving
us the determinant of two π΄ is equal to two squared multiplied by the determinant
of π΄, which is, of course, four times the determinant of π΄. We can then substitute this
expression for the determinant of two π΄ into the equation weβre given in the
question. This then gives us that four times
the determinant of π΄ is equal to 12. And we can solve for the
determinant of π΄. We divide both sides of the
equation by four. This gives us the determinant of π΄
is equal to three.

However, weβre not asked to find
the determinant of π΄; weβre asked to find the determinant of three times the
transpose of π΄. To do this, letβs try simplifying
this expression by using the properties of determinants. First, we recall when we take the
transpose of a matrix, we switch the rows with the columns. So, the transpose of matrix π΄ is
also a matrix of order two by two. This means we can once again apply
the same property. π΄ transpose is a two-by-two
matrix. Therefore, the determinant of three
multiplied by the transpose of π΄ is equal to three squared multiplied by the
determinant of π΄ transpose. And we can simplify this to get
nine multiplied by the transpose of π΄, but we can simplify this expression even
further by using another one of our properties of determinants.

We recall for any square matrix π΅,
the determinant of π΅ transpose is just equal to the determinant of π΅. And we know π΄ transpose is a
square matrix, so we can replace this with the determinant of π΄ to get nine
multiplied by the determinant of π΄. And we know what the determinant of
matrix π΄ is. Itβs equal to three. Therefore, we can just substitute
three for the determinants of π΄ to get nine times three, which is equal to 27,
which we can see is given as option (C).

Therefore, weβve shown if π΄ is a
square matrix of order two by two and the determinant of two π΄ is equal to 12, then
the determinant of three times the transpose of π΄ is equal to 27.

Letβs now go for another example
where we use the properties of the determinant of the product of two matrices.

If the determinant of π΄ times π΅
is equal to 18 and the determinant of π΄ is equal to two, find the determinant of
π΅.

In this question, weβre given some
information about two matrices, π΄ and π΅. Weβre told the determinant of
matrix π΄ multiplied by matrix π΅ is 18 and the determinant of matrix π΄ is equal to
two. We need to use this information to
determine the determinant of π΅. To answer this question, letβs
start by recalling the property of determinants, which links the products of
matrices with the determinant of each individual matrix.

We recall if π΄ and π΅ are square
matrices of the same order, then the determinant of π΄ times π΅ is equal to the
determinant of π΄ multiplied by the determinant of π΅. We might be tempted to apply this
property directly to our question. However, thereβs a problem; weβre
not told the orders of matrices π΄ and π΅. And to apply this property, we do
need the matrices π΄ and π΅ are square matrices of the same order. However, we can show that this must
be the case from the information given. First, weβre told the determinant
of matrix π΄ is equal to two. And we recall we can only find the
determinant of square matrices, so π΄ is a square matrix. Similarly, the determinant of π΄π΅
is equal to 18, so π΄ times π΅ is also a square matrix.

We can use this information to find
an expression for the order of matrix π΅. First, since π΄ is a square matrix,
letβs start by saying its order is of the form π by π. Next, we donβt know the order of
matrix π΅. Letβs say its order is π by
π. Then, there are two different ways
of finding an expression for the order of π΄ times π΅. First, remember, whenever we
multiply two matrices together, its order will be the number of rows of our first
matrix by the number of columns of our second matrix. So, π΄π΅ must have order π by
π. However, weβve already shown that
π΄π΅ is a square matrix. So, the number of rows must be
equal to the number of columns. In other words, π is equal to
π. So, we can replace π with π.

Finally, to determine the value of
π, we notice weβre allowed to multiply π΄ on the right by matrix π΅. And recall for matrix
multiplication to be well defined, the number of columns of our first matrix must be
equal to the number of rows of our second matrix. Therefore, π must be equal to
π. Therefore, weβve shown π΄ is a
square matrix, π΅ is a square matrix, and the orders of matrices π΄ and π΅ are
equal. So, we can just apply our property
to the question. We have the determinant of π΄π΅ is
equal to the determinant of π΄ multiplied by the determinant of π΅. Next, weβre told in the question
the determinant of π΄π΅ is 18 and the determinant of π΄ is two. So, 18 is equal to two times the
determinant of π΅. Finally, we can divide both sides
of the equation through by two to get the determinant of π΅ is nine.

Therefore, we were able to show if
the determinant of π΄π΅ is 18 and the determinant of π΄ is two, then the determinant
of π΅ must be equal to nine.

Letβs now go through a question
which asks us to find the value of the determinant of a triangular matrix.

Find the value of the determinant
of the three-by-three matrix five, negative one, negative eight, zero, two, 60,
zero, zero, zero.

In this question, weβre asked to
evaluate the determinant of a given three-by-three matrix. We could do this by using the
definition of a determinant. However, thereβs actually an easier
method if we can just notice the property of this matrix. We need to notice that this is an
upper triangular matrix. This means every entry below the
leading diagonal of this matrix is equal to zero. The leading diagonal of a matrix is
the entries whose row number is equal to the column number. So, for this matrix, thatβs five,
two, and zero. So, this matrix is an upper
triangular matrix.

And we recall the determinant of
any square triangular matrix is the product of all of the entries on its leading
diagonal. In our case, the leading diagonal
has terms five, two, and zero. Therefore, the determinant of this
matrix is five multiplied by two multiplied by zero, which we can evaluate is equal
to zero.

Therefore, the determinants of the
three-by-three matrix five, negative one, negative eight, zero, two, 60, zero, zero,
zero is equal to zero.

Letβs now go through an example
where we need to use the determinant of a diagonal matrix to find the value of a
variable.

Consider the equation the
determinant of the three-by-three matrix π₯ minus one, zero, zero, zero, π₯ squared
plus π₯ plus one, zero, zero, zero, one is equal to two. Determine the value of π₯ to the
sixth power.

In this question, weβre given an
equation involving the determinant of a three-by-three matrix which has a variable
π₯. We need to use this to determine
the value of π₯ to the sixth power. To do this, we need to find an
expression for the determinant of this matrix. We could do this by using the
definition of a determinant, expanding over one of the rows or columns. However, we can also notice that
every element not on the main diagonal of this matrix is equal to zero. In other words, this three-by-three
matrix is a diagonal matrix. And we can use this to find the
determinant of this matrix because every diagonal matrix is an upper and lower
triangular matrix. For example, every entry below the
main diagonal of our matrix is zero. So, this is an example of an upper
triangular matrix.

Now, we just need to recall that
the determinant of any square triangular matrix is the product of the elements on
its main diagonal. And itβs worth noting since all
diagonal matrices are both upper and lower triangular matrices, this property will
also hold for any square diagonal matrix. Therefore, we can evaluate the
determinant of this matrix by finding the product of its leading diagonal, π₯ minus
one times π₯ squared plus π₯ plus one multiplied by one. We can then distribute over our
parentheses. We get π₯ cubed plus π₯ squared
plus π₯ minus π₯ squared minus π₯ minus one. And if we simplify this expression,
we see itβs equal to π₯ cubed minus one. And remember, weβre told in the
question this determinant is equal to two.

Therefore, our expression for the
determinant is equal to two. π₯ cubed minus one is equal to
two. We want to use this to find the
value of π₯ to the sixth power. So, letβs start by adding one to
both sides of the equation to give us that π₯ cubed is equal to three. And then, we can find an expression
for π₯ to the sixth power by squaring both sides of our equation, giving us that π₯
to the sixth power is equal to nine, which is our final answer.

Therefore, weβve shown if the
determinant of the three-by-three matrix π₯ minus one, zero, zero, zero, π₯ squared
plus π₯ plus one, zero, zero, zero, one is equal to two, then the value of π₯ to the
sixth power is nine.

Letβs now go over the key points of
this video. We found and proved four useful
results involving the determinants of matrices. First, we showed for any square
matrix π΄ of order π by π and scalar π, the determinant of ππ΄ is equal to π to
the πth power multiplied by the determinant of π΄. Next, we showed for any square
matrix π΄, the determinant of the transpose of π΄ is just equal to the determinant
of π΄. Next, we showed for any square
matrices π΄ and π΅ of the same order, the determinant of π΄ times π΅ is equal to the
determinant of π΄ multiplied by the determinant of π΅. Finally, we showed the determinant
of any square triangular matrix is the product of all of the terms on its principal
diagonal. And this result holds for both
upper and lower triangular matrices. And it even holds for square
diagonal matrices because all diagonal matrices are both upper and lower triangular
matrices.