### Video Transcript

In this video, we will learn how to
evaluate three-by-three determinants using the cofactors, Laplace expansion, or the
SARAS method. Calculating the determinant of a
square matrix is really useful to gather information about the matrix, such as being
able to tell whether or not the matrix is invertible.

Letβs start with a quick reminder
on how to calculate the determinant of two-by-two matrix. Then weβll go on to see how we can
calculate the determinant of a general square matrix with order π by π using what
we know about finding the determinant of a two-by-two matrix. Letβs consider this two-by-two
matrix. The determinant of π΄ is denoted
with straight lines either side of π΄. Itβs the same symbol that we use
for absolute value. For a two-by-two matrix, itβs given
by the formula ππ minus ππ. Remember, weβve got to be really
careful when handling negative entries in a matrix, especially when weβre trying to
find the determinant.

As an example, find the determinant
of the matrix five, one, negative five, five. Letβs call this matrix π΅. Then we calculate the determined of
π΅ to be five times five minus one times negative five. We calculate this to be 25 minus
negative five. And this is where we must be really
careful with the negatives as this is 25 add five, which is 30. But remember, the entries of the
matrix are not always numbers. For instance, find the determinant
of the matrix πΆ equals π₯, π¦, π§, π₯. Using the same rule, this matrix
has determinant π₯ squared minus π¦π§.

Moving on to think about
three-by-three matrices, there are specific methods for finding the determinant of a
three-by-three matrix. But here weβre going to develop a
more general approach so that this method can be applied to larger matrices. The first thing weβre going to look
at in order to be able to calculate three-by-three determinants are matrix
minors. Consider a matrix π΄ with order π
by π, then the matrix minor, π΄ ππ, is the initial matrix π΄ after having removed
the πth row and the πth column. This means that π΄ ππ is a matrix
with order π minus one by π minus one. Recall that the order of a matrix
means the size of the matrix, so an π-by-π matrix is a matrix with π rows and π
columns. The easiest way to understand
matrix minors is through an example.

Suppose we have this matrix π΄. Letβs calculate the matrix minor π΄
two three. This means removing the second row
and the third column from the initial matrix. We can see that the entries that
remain are the entries negative two, three, zero, negative three. So this is the two-by-two matrix
minor π΄ two three. Letβs also calculate the matrix
minor π΄ three one. This means removing the third row
and the first column. And we can see that weβre left with
the two-by-two matrix three, negative three, three, six.

Weβre going to use this concept
with the formal definition of the general form of the determinant of a matrix. Consider a square matrix π΄ with
order π by π, then the determinant of π΄ is calculated in one of two ways, both of
which involve the determinant of particular matrix minors of π΄. We can choose to calculate the
determinant by using a particular row. Alternatively, we can choose one
particular column. This is quite an overwhelming
definition of the determinant. And it makes a lot more sense when
we see it applied to an example.

Find the determinant of the matrix
one, two, three, three, two, two, zero, nine, eight.

Letβs label this matrix as π΄. We can calculate the determinant of
π΄ by first selecting either one row or one column. A good strategy is to choose the
row or column which contains the most number of zeros. For this matrix, this would either
be the third row or the first column, which both contain one zero entry. Letβs choose the first column. And here is the formula for finding
the determinant of a matrix when we use a particular column π.

For this question, as weβre using
the first column, π is equal to one. And because this is a
three-by-three matrix, π runs from one to three. Here is the same formula but
adapted so that π runs from one to three. And as weβre using the first
column, π is equal to one. Now π one one, π two one, and π
three one are the entries in the column that weβre using. These are the entries one, three,
and zero. And these are the determinants of
the matrix minors.

Recall that we find the matrix
minor π΄ one one by removing the first row and the first column of the original
matrix. Weβre left with the two-by-two
matrix two, two, nine, eight. We then calculate the determinant
of this matrix using the usual formula for finding the determinant of a two-by-two
matrix. This is two times eight minus two
times nine. This is 16 minus 18, which gives us
negative two. We can then calculate the matrix
minor π΄ two one by removing the second row and the first column. This gives us the matrix two,
three, nine, eight. We can then calculate its
determinant to be two times eight minus three times nine. And we find that this is negative
11. We do the same, calculating the
matrix minor π΄ three one by removing the third row and the first column of the
original matrix. And we find its determinant to be
two times two minus three times two, which gives us negative two.

The final step is to calculate the
values negative one to the power of one add one, negative one to the power of two
add one, and negative one to the power of three add one. We find these to be one, negative
one, and one. Remember, raising negative one to
an even power gives us one and raising negative one to an odd power gives us
negative one. Letβs now substitute into this
formula the values that we found. Substituting in these values
becomes clear why itβs useful to choose a row or column that has the most number of
zeros. Because this last term is being
multiplied by zero, the whole term will be zero, which just means itβs one less
thing that we actually need to calculate.

Tidying things up a bit, we find
that one times one times negative two gives us negative two and negative one times
three times negative 11 gives us 33. Adding these together, we get the
final answer of 31.

So the main things to think about
are which row or column youβre going to be using and carefully calculating each
little bit of the formula. And weβve got to make sure that
weβre always really careful around negatives. Letβs now see another example,
although, this time, weβll expand along a row rather than a column.

Calculate the determinant of π΄
when π΄ equals three, zero, negative one, zero, one, zero, two, two, four.

To make things easier for
ourselves, weβll identify the row or column that contains the most number of entries
which are zero. For this matrix, thatβs going to be
the second row of π΄. Letβs remind ourselves of the
general formula for finding the determinant of an π-by-π matrix. Here is the formula. Remember that π represents the row
number and π represents the column number. So as we have three columns, π
runs from one to three. And weβre expanding along the
second row, so π equals two. So this is the formula that weβre
going to be using.

The entries π two one, π two two,
and π two three are zero, one, and zero, respectively. Because two of these entries are
zero, we find that both of these terms are going to be zero because theyβre both
being multiplied by zero. So thereβs actually only one term
that we need to calculate here. Letβs begin by calculating the
matrix minor π΄ two two. This is going to be the two-by-two
matrix we get by removing the second row and the second column. So we see that the matrix minor π΄
two two is three, negative one, two, four. But what we actually need for our
formula is its determinant. So we calculate this in the usual
way for finding determinants of a two-by-two matrix. This gives us three times four
minus negative one times two. This gives us 12 minus negative
two, which gives us 14.

The final thing we can do is
calculate the value of negative one to the power of two add two. As this is negative one to the
fourth power, which is an even power, this is going to give us one. So we calculate the determinant of
π΄ to be one times one times 14. And this, of course, just gives us
14. So carefully selecting the row or
column that you choose when youβre finding the determinant of a matrix can really
reduce the amount of calculations necessary.

Remember that not all matrices
contain numeric entries. Letβs have a look at this in an
example.

Consider the determinant π₯, π§,
π¦, π¦, π₯, π§, π§, π¦, π₯. Given that π₯ cubed add π¦ cubed
add π§ cubed equals negative 73 and π₯π¦π§ equals negative eight, determine the
determinantβs numerical value.

Remember that we can choose to
calculate this determinant by picking a row or column. We often choose the row or column
with the most amount of zeros. But as we have three unknowns, π₯,
π¦, and π§, letβs just choose the top row. And here is the general formula to
find the determinant of an π-by-π matrix, using a particular row, π. As weβre using a three-by-three
matrix, π runs from one to three. And as weβre using the top row, π
is equal to one. So here is the particular version
of this general formula applied for our question. Letβs clear some space before we
continue.

Iβve kept the formula that weβre
using on the screen. Letβs call the matrix that weβre
using π΄. The straight lines either side of
the matrix denote the determinant. Letβs begin by finding the
different components in the formula that weβre using. The lowercase π one one, π one
two, and π one three are the entries in the row that weβre using. So π one one is π₯, π one two is
π§, and π one three is π¦.

The uppercase, π΄ one one, π΄ one
two, and π΄ one three denote the matrix minors. Remember that we get these from
removing a particular row and a particular column from the original matrix. For example, we get π΄ one one by
removing the first row and the first column, like so. And weβre left with the two-by-two
matrix π₯, π§, π¦, π₯. We then do the same for π΄ one
two. We get the matrix minor by removing
the first row and the second column. This leaves us with the two-by-two
matrix π¦, π§, π§, π₯. And finally, we get the matrix
minor π΄ one three by removing the first row and the third column. This leaves us with the two-by-two
matrix π¦, π₯, π§, π¦.

But what we actually need for our
formula is the determinant of each of these matrices. Letβs recall a method to find the
determinant of a two-by-two matrix. So the determinant of the matrix
minor π΄ one one is π₯ times π₯ minus π§ times π¦. We can equivalently write this as
π₯ squared minus π¦π§. In the same way, we find the
determinant of the matrix minor π΄ one two to be π¦ times π₯ minus π§ times π§, or
equivalently π₯π¦ minus π§ squared. And we find the determinant of the
matrix minor π΄ one three to be π¦ times π¦ minus π₯ times π§, which is equivalently
π¦ squared minus π₯π§.

The final thing we need to
calculate for our formula are the values of negative one to the power of one add
one, negative one to the power of one add two, and negative one to the power of one
add three. Remember that raising negative one
to an even power gives us one and raising negative one to an odd power gives us
negative one. So negative one to the power of one
add one is negative one squared, which is one. Negative one to the power of one
add two is negative one cubed, which is negative one. And finally, negative one to the
power of one add three is negative one to the fourth power, which gives us one.

So letβs not write out the
determinants of this matrix with the components that we found. From here, letβs just try and
simplify a little bit. Letβs begin by multiplying together
the brackets with algebraic terms. For the first one, we have π₯
multiplied by π₯ squared minus π¦π§. So this term becomes one times π₯
cubed minus π₯π¦π§. We then do the same with the next
term. This gives us negative one times
π₯π¦π§ minus π§ cubed. And repeating the same with the
last term, we end up with one times π¦ cubed minus π₯π¦π§.

So now, weβve multiplied together
the brackets containing algebraic terms. Letβs multiply each term by the
value at the front. The first term is one times π₯
cubed minus π₯π¦π§, which is of course going to give us π₯ cubed minus π₯π¦π§. The next term is negative one times
π₯π¦π§ minus π§ cubed. So we can multiply the second
bracket through by negative one. But weβve got to be careful because
the π§ cubed already has a minus at the front of it, so thatβs going to become a
positive. So weβre gonna end up with minus
π₯π¦π§ add π§ cubed. Finally, the last term is just one
times π¦ cubed minus π₯π¦π§. So this gives us π¦ cubed minus
π₯π¦π§.

We can now simplify this a little
bit. We can gather together the π₯π¦π§
terms. Bringing these together gives us
minus three π₯π¦π§. From here, we notice that we have
π₯ cubed add π§ cubed add π¦ cubed. Weβre told in the question that π₯
cubed add π¦ cubed add π§ cubed gives us negative 73. Additionally, we have negative
three π₯π¦π§, and weβre told in the question that π₯π¦π§ is equal to negative
eight. So substituting in those values
gives us negative 73 minus three times negative eight, which is negative 73 add
24. And this gives us negative 49. So sometimes, we encounter matrices
which donβt have numerical values, but we can still find the determinant in exactly
the same way.

Letβs now summarize the main points
from this lesson. For a matrix π΄, a matrix minor π΄
ππ is the initial matrix π΄ after having removed the πth row and the πth
column. And here is the general formula for
the determinant of an π-by-π matrix π΄. We can either choose a particular
row or a particular column. But it often helps to choose a row
or column that contains the most number of zeros.