Video Transcript
Find the value of the determinant
of the two-by-two matrix negative one divided by the sin of 𝜃, one, one plus the
cot squared of 𝜃, negative one divided by the sin of 𝜃.
In this question we’re asked to
evaluate the determinant of a two-by-two matrix. And this is a very
complicated-looking matrix. The first thing we need to do is
recall what we mean by the determinant of a two-by-two matrix.
We need to recall to find the
determinant of a two-by-two matrix, we find the difference between the products of
the diagonals. In other words, the determinant of
the two-by-two matrix 𝑎, 𝑏, 𝑐, 𝑑 is equal to 𝑎𝑑 minus 𝑏𝑐, the difference in
the products of the diagonals.
We need to apply this to find the
determinant of the two-by-two matrix given to us in the question. So the first thing we’re going to
need to do is find the product of 𝑎 and 𝑑. And of course that’s the product of
the entry in row one column one and the entry in row two column two. And we can see in our matrix both
of these entries are negative one divided by the sin of 𝜃. So we multiply these together. We get negative one over the sin of
𝜃 multiplied by negative one over the sin of 𝜃.
Next, we need to subtract the value
of 𝑏 times 𝑐. And remember, that’s the product of
the entry in row one column two and the entry in row two column one. And we can see from our matrix this
is one multiplied by one plus the cot squared of 𝜃. And it’s important to remember we
need to subtract this value because we subtract these in our formula.
Now we need to start simplifying
this expression. Let’s start by simplifying negative
one over the sin of 𝜃 multiplied by negative one over the sin of 𝜃. This is a negative multiplied by a
negative. So this simplifies to give us a
positive. And one over the sin of 𝜃
multiplied by one over the sin of 𝜃 is one over sin squared of 𝜃. So our first term simplifies to
give us one over the sin squared of 𝜃.
There’s several different ways we
could simplify the second term in this expression. And they’ll all lead to the correct
result. We’ll just write this as
subtracting one plus the cot squared of 𝜃. And at first, it might be hard to
see how we’re going to simplify this expression any further. And there’s several different ways
of doing this. All of them involve the Pythagorean
identity.
Recall, the Pythagorean identity
tells us the sin squared of 𝜃 plus the cos squared of 𝜃 is equivalent to one. In other words, this is true for
all values of 𝜃. And to use this, we’re going to
need to include one divided by the sin squared of 𝜃. So we’re going to want to divide
both sides of our identity through by the sin squared of 𝜃.
We’ll do this term by term. Let’s start with the sin squared of
𝜃. The sin squared of 𝜃 divided by
the sin squared of 𝜃 is equal to one. Next, the cos squared of 𝜃 divided
by the sin squared of 𝜃 can be written as cos squared of 𝜃 divided by sin squared
of 𝜃. Finally, one divided by the sin
squared of 𝜃 is just one over sin squared 𝜃. And this gives us another version
of the Pythagorean identity.
However, this is not quite in a
form which we can use yet. We need to notice something about
the cos squared of 𝜃 divided by the sin squared of 𝜃. Since this is the quotient of two
squares, we could write this as the cos of 𝜃 divided by the sin of 𝜃 all
squared. But remember, the cos of 𝜃 divided
by the sin of 𝜃 is the cot of 𝜃. In other words, this term is just
equivalent to the cot squared of 𝜃.
So by writing this into our
identity, we’ve shown one plus the cot squared of 𝜃 is equivalent to one over the
sin squared of 𝜃. But what does this mean? We have one divided by the sin
squared of 𝜃 minus one plus the cot squared of 𝜃. And according to our identity,
these are equivalent. And if they are equivalent, they’re
equal for all values of 𝜃. So this is equal to zero for all
values of 𝜃. And this gives us our final answer
of zero.
It’s also worth pointing out,
there’s another way we could’ve answered this question. We could’ve directly used the
definition of the csc of 𝜃 to rewrite our determinant. Then in our final line of working,
instead of having one divided by the sin squared of 𝜃, we would just have the csc
squared of 𝜃. And then we could solve this by
using one of our alternate versions of the Pythagorean identity. We would still get the answer of
zero.
Therefore, we were able to show the
determinant of the two-by-two matrix negative one over the sin of 𝜃, one, one plus
the cot squared of 𝜃, negative one divided by the sin of 𝜃 is equal to zero.
