### Video Transcript

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.