### Video Transcript

In this video, weβll learn how to
find the determinant of a triangular matrix. Weβll see first that the
determinants of square matrices that have a row or column of zeros is actually equal
to zero and then that the determinant of an upper triangular, diagonal, or lower
triangular matrix is just the product of the entries in the leading diagonal.

Weβre going to be considering
three-by-three square matrices, so letβs first remind ourselves how we find the
determinant of a three-by-three square matrix. To do this, we need to recall how
we define a couple of things: the matrix minors and the cofactors of an π-by-π
square matrix. For the matrix minors, if a matrix
π΄ with elements π sub ππ has order π by π, then the minor, denoted uppercase
π΄ sub ππ, is the determinant of the π minus one by π minus one matrix obtained
by removing row π and column π from matrix π΄. The cofactor for matrix element π
sub ππ, denoted uppercase πΆ sub ππ, is equal to negative one raised to the
power π plus π times π΄ sub ππ, where uppercase π΄ sub ππ is the minor of the
matrix element π ππ.

Then the determinant of a
three-by-three matrix using cofactor expansion, for fixed π equals one, two, or
three, is given by det π΄ equals π sub π one times πΆ sub π one plus π sub π
two times πΆ sub π two plus π sub π three times πΆ sub π three, where the πΆ sub
ππβs are the cofactors of the elements π sub ππ. This is called the cofactor
expansion or the Laplace expansion along row π. Alternatively, we can used fixed π
equals one, two, or three, and the determinant is π sub one π times πΆ sub one π
and so on. And this is called the cofactor
expansion down column π.

We can write this in a slightly
more convenient way using two-by-two determinants. So for example, choosing π equals
one, thatβs the first-row expansion, we have det π΄ equal to π sub one one times
the determinant of the two-by-two matrix with elements π two two, π two three, π
three two, π three three minus π sub one two times the determinant of the matrix
with elements π two one, π two three, π three one, π three three plus π sub one
three times the determinant of π two one, π two two, π three one, π three
two.

One thing to note before going
further is that we need to be careful with the signs of our cofactors. Remember, they each contain
negative one raised to the sum of π and π. And if π plus π is even, this is
positive one. And if π plus π is odd, this is
negative one. Hence, we have the signs of the
cofactors for a three-by-three matrix as shown. Thatβs positive, negative,
positive, negative, positive, negative, positive, negative, positive. And we see that our first-row
expansion does follow this pattern.

Itβs worth considering that which
row or column we choose to expand along can make quite a difference in how much work
we have to put in to calculating our determinant. In fact, choosing to expand along a
row or down a column containing one or more zero elements can simplify the
calculation. We can see how this works with an
example. Suppose our matrix π΄ has elements
five, one, seven, zero, zero, three, and two, four, one. If we decide to expand along the
first row to find the determinant, then weβll need to find the determinants of three
two-by-two matrices. This is perfectly manageable using
the formula for a two-by-two determinant for each of these. However, there is an easier
way.

If we instead expand along the
second row, where we have two zero elements, and remembering the signs of our
cofactors, we see that thereβs no need to calculate the first two determinants since
theyβre both multiplied by zero. And therefore, the first two terms
are equal to zero. And we only have to calculate the
third term. This is equal to negative three
multiplied by five times four minus one times two. Thatβs negative three times 18,
which is negative 54. The calculation using the first-row
expansion of course gives us the same answer. But the second-row expansion is
rather less work than this, which also means thereβs less room for error than when
calculating all three determinants.

So weβve seen that choosing a row
or column containing zero elements simplifies the calculation of the determinant of
a three-by-three matrix. Letβs keep this in mind with our
next example.

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

To find this determinant, we recall
that for a three-by-three matrix π΄, we can use cofactor expansion on any row using
the given formula, where we choose π equal to one, two, or three and expand along
that row and where πΆ sub ππ is the cofactor matrix for element ππ of matrix π΄
and uppercase π΄ sub ππ is the minor of the element in row π column π of matrix
π΄. Now although any choice of row or
column expansion will give us the same determinant, the calculation will be simpler
if we choose a row or column with the most zero elements. And we see that in the given
matrix, all the elements in the third row are zero.

So letting π equal three, we
expand along the third row. And our determinant is π sub three
one times πΆ sub three one plus π sub three two times πΆ sub three two plus π sub
three three times πΆ sub three three. But notice that since all of the
elements π sub three one, π sub three two, and π sub three three are equal to
zero, each of our terms is a multiple of zero. Hence, our determinant is equal to
zero. And so, the value of the
determinant of the given matrix is zero.

Weβve demonstrated in this example
that a three-by-three matrix with a row of zeros has determinant equal to zero. But since cofactor expansion can be
applied along any row or column, the same result is true if any whole row or column
is equal to zero. And this can be generalized to
square matrices of any size. Hence, if π΄ is a square π-by-π
matrix where each entry in a particular row or column is zero, then the determinant
of matrix π΄ is zero.

So, whenever weβre asked to find a
determinant, we should always be mindful to check whether any rows or columns are
zero, since we know then immediately that the determinant will be zero. In our next example, we consider
another special case of determinant calculation.

The value of the determinant of the
three-by-three matrix with elements three, zero, negative two, zero, five, seven,
and zero, zero, four equals what.

When weβre asked to find the
determinant of a three-by-three matrix, we recall that we can use the cofactor
expansion along any specific row π using the formula det π΄ equals π sub π one
times πΆ sub π one plus π sub π two times πΆ sub π two plus π sub π three
times πΆ sub π three. And similarly by choosing column
π, where π is either one, two, or three, we have the same idea expanding down a
column. And this is where πΆ sub ππ is
the cofactor of element π sub ππ in matrix π΄, and uppercase π΄ sub ππ is the
matrix minor of element π sub ππ.

When calculating a determinant in
this way, to limit the amount of calculation involved, if possible, we choose to
expand along a row or down a column with the most zero entries. Examining the given matrix, we find
that both the first column and the third row would be good candidates, as they both
have two zero entries. Letβs choose to expand along the
third row so that π is equal to three. Since elements π sub three one and
π sub three two are equal to zero, this gives us zero minus zero plus four times
the determinant of the two-by-two matrix with elements three, zero, zero, five. Note that since weβre expanding
along the third row, our cofactor signs are positive, negative, positive. And the two-by-two matrix with
elements three, zero, zero, five is the matrix minor for element π sub three
three.

So now since our first two terms
are zero, the determinant of the given three-by-three matrix is four times the
determinant of the two-by-two matrix three, zero, zero, five. Using the formula for the
determinant of a two-by-two matrix, we have the determinant of the given matrix as
four times three times five, which is 60. Hence, the value of the determinant
of the given three-by-three matrix is 60.

Now if we consider the calculation
of the determinant here, we can see that it was just the multiplication of the three
entries along the main diagonal. And the reason this was so
straightforward is because the matrix itself is an upper triangular matrix. Letβs remind ourselves of how we
define triangular matrices. If the entries below the main
diagonal are zero, the matrix is an upper triangular matrix. If the entries above the main
diagonal are zero, the matrix is defined as lower triangular. A matrix is triangular if it is
upper or lower triangular or both. As weβve seen, the reason finding
determinants of triangular matrices is so simple is that the zeros in one-half of
the matrix remove most of the calculation.

Letβs look at this a little more
closely for a general upper triangular matrix. If we use cofactor expansion on the
third row, the determinant is then zero times the cofactor of the entry in the third
row first column plus zero times the cofactor of the entry in the third row second
column plus π times the cofactor of the third row third column entry. This is equal to π times the
determinant of the two-by-two matrix π, π, zero, π, which is equal to πππ. In other words, the final result is
just the product of the three entries in the main diagonal. Similarly, for lower triangular
matrices, expanding along the first row, we find the result is equivalent. The determinant equals πππ,
which is the product of the three entries on the main diagonal.

This gives us the property the
determinant of a triangular matrix is the product of the entries on the main
diagonal. We note also that this property
applies to the subclass of triangular matrices, diagonal matrices. These are matrices where only the
entries on the main diagonal are nonzero. Because diagonal matrices are both
upper and lower triangular, they exhibit the same property, that the determinant is
the product of the entries in the leading, or main, diagonal. We note further that this also
applies to both the identity and the zero matrices, where the main diagonal elements
are all one or all zero, respectively, and theyβre both diagonal matrices.

So now letβs look at an example
where we can use this property to compare the determinants of two matrices.

True or False: If π΄ is the
three-by-three square matrix with entries one, four, two, zero, three, six, zero,
zero, four and π΅ the three-by-three matrix with entries one, zero, zero, five,
three, zero, and six, seven, four, then the determinant of π΄ equals the determinant
of π΅.

One way to answer this would be to
calculate each of the two determinants using cofactor expansion along the rows or
columns and comparing. However, we donβt really need to do
this, since both π΄ and π΅ are triangular matrices. π΄ is an upper triangular matrix
since all entries below the main diagonal are zero. And π΅ is lower triangular since
all of the elements above the main diagonal are zero. Now we know that for both upper and
lower triangular matrices, the determinant is just the product of the elements on
the main diagonal. And we know that both our matrices
π΄ and π΅ are triangular. But are their determinants
equal?

By our triangular matrix property,
we know that their determinants are the product of the elements on their main
diagonals. And since both matrices have the
same elements in their main diagonals, one, three, and four, we conclude that their
determinants are both equal to 12. Hence, det π΄ equals det π΅ equals
12. And so, the statement is true.

In our final example, weβre going
to see how to solve an equation by finding the determinant of a diagonal matrix.

Consider the equation the
determinant of the matrix with elements π₯ minus one, zero, zero, zero, π₯ squared
plus π₯ plus one, zero, and zero, zero, one equals two. Determine the value of π₯ to the
sixth power.

Now this looks like a complicated
problem. But actually, we can simplify it by
noticing that the given matrix is a diagonal matrix. That is, only the elements on the
main diagonal are nonzero. And we know that for a diagonal
matrix, which is a special kind of triangular matrix, the determinant is equal to
the product of the elements in the main diagonal. In our case then, this means that
the determinant is equal to π₯ minus one times π₯ squared plus π₯ plus one times
one. Distributing the parentheses and
collecting like terms, weβre left with the determinant equal to π₯ cubed minus
one.

Now we want to find π₯ to the sixth
power, and weβre told that the given determinant is equal to two. So equating two with the
determinant weβve just found, we have π₯ cubed minus one equals two. Solving this for π₯ cubed by adding
one to both sides gives π₯ cubed equal to three. And now using the laws of indices,
squaring both sides, since π₯ to the power three squared is equal to π₯ to the sixth
power, we have π₯ to the sixth power is equal to nine.

Letβs now complete this video by
recapping some of the main points weβve covered. If some of the entries in a matrix
are zero, we may be able to simplify the calculation of the determinant. If a matrix has a zero row or
column, then its determinant is zero. And if a matrix is upper
triangular, lower triangular, or diagonal, then its determinant is the product of
the entries on the main diagonal.