Video Transcript
Consider the determinant 𝑥, 𝑧,
𝑦, 𝑦, 𝑥, 𝑧, 𝑧, 𝑦, 𝑥. Given that 𝑥 cubed add 𝑦 cubed
add 𝑧 cubed equals negative 73 and 𝑥𝑦𝑧 equals negative eight, determine the
determinant’s 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, 𝑥,
𝑦, and 𝑧, let’s just choose the top row. And here is the general formula to
find the determinant of an 𝑛-by-𝑛 matrix, using a particular row, 𝑖. As we’re using a three-by-three
matrix, 𝑗 runs from one to three. And as we’re using the top row, 𝑖
is equal to one. So here is the particular version
of this general formula applied for our question. Let’s clear some space before we
continue.
I’ve kept the formula that we’re
using on the screen. Let’s call the matrix that we’re
using 𝐴. The straight lines either side of
the matrix denote the determinant. Let’s begin by finding the
different components in the formula that we’re using. The lowercase 𝑎 one one, 𝑎 one
two, and 𝑎 one three are the entries in the row that we’re using. So 𝑎 one one is 𝑥, 𝑎 one two is
𝑧, and 𝑎 one three is 𝑦.
The uppercase, 𝐴 one one, 𝐴 one
two, and 𝐴 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 𝐴 one one by
removing the first row and the first column, like so. And we’re left with the two-by-two
matrix 𝑥, 𝑧, 𝑦, 𝑥. We then do the same for 𝐴 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 𝑦, 𝑧, 𝑧, 𝑥. And finally, we get the matrix
minor 𝐴 one three by removing the first row and the third column. This leaves us with the two-by-two
matrix 𝑦, 𝑥, 𝑧, 𝑦.
But what we actually need for our
formula is the determinant of each of these matrices. Let’s recall a method to find the
determinant of a two-by-two matrix. So the determinant of the matrix
minor 𝐴 one one is 𝑥 times 𝑥 minus 𝑧 times 𝑦. We can equivalently write this as
𝑥 squared minus 𝑦𝑧. In the same way, we find the
determinant of the matrix minor 𝐴 one two to be 𝑦 times 𝑥 minus 𝑧 times 𝑧, or
equivalently 𝑥𝑦 minus 𝑧 squared. And we find the determinant of the
matrix minor 𝐴 one three to be 𝑦 times 𝑦 minus 𝑥 times 𝑧, which is equivalently
𝑦 squared minus 𝑥𝑧.
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 let’s not write out the
determinants of this matrix with the components that we found. From here, let’s just try and
simplify a little bit. Let’s begin by multiplying together
the brackets with algebraic terms. For the first one, we have 𝑥
multiplied by 𝑥 squared minus 𝑦𝑧. So this term becomes one times 𝑥
cubed minus 𝑥𝑦𝑧. We then do the same with the next
term. This gives us negative one times
𝑥𝑦𝑧 minus 𝑧 cubed. And repeating the same with the
last term, we end up with one times 𝑦 cubed minus 𝑥𝑦𝑧.
So now, we’ve multiplied together
the brackets containing algebraic terms. Let’s multiply each term by the
value at the front. The first term is one times 𝑥
cubed minus 𝑥𝑦𝑧, which is of course going to give us 𝑥 cubed minus 𝑥𝑦𝑧. The next term is negative one times
𝑥𝑦𝑧 minus 𝑧 cubed. So we can multiply the second
bracket through by negative one. But we’ve got to be careful because
the 𝑧 cubed already has a minus at the front of it, so that’s going to become a
positive. So we’re gonna end up with minus
𝑥𝑦𝑧 add 𝑧 cubed. Finally, the last term is just one
times 𝑦 cubed minus 𝑥𝑦𝑧. So this gives us 𝑦 cubed minus
𝑥𝑦𝑧.
We can now simplify this a little
bit. We can gather together the 𝑥𝑦𝑧
terms. Bringing these together gives us
minus three 𝑥𝑦𝑧. From here, we notice that we have
𝑥 cubed add 𝑧 cubed add 𝑦 cubed. We’re told in the question that 𝑥
cubed add 𝑦 cubed add 𝑧 cubed gives us negative 73. Additionally, we have negative
three 𝑥𝑦𝑧, and we’re told in the question that 𝑥𝑦𝑧 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 don’t have numerical values, but we can still find the determinant in exactly
the same way.
