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Current time:0:00Total duration:8:14

So in problem 44, once again
we have the exact same setup that we had in
problems 42 and 43. But now they're
asking us the number of A in T sub p such
that the determinant of A is not divisible by p. Well, we've already done
some work in problem 43 in figuring out some
set of the members that are divisible by p. So why don't we just figure
out the total number of members of T sub p, and then subtract
the total members whose determinants are divisible
by p, and then we'll get this answer right over here. So what's the total
members in T sub p? Well, for a, I have p
possible choices, right? If I'm picking between-- up to
p minus 1, but I'm including 0, there's actually p
possible choices. So I have p possible
choices for a, p possible choices for
b by the same logic, and p possible choices for c. So I have p to the
third possibilities. So there are a total of p to
the third members in T sub p. Now, what did the
last problem tell us? This was number 43. It told us the members-- maybe
I should write it this way. The number of members, numbers
of A's, such that the trace of A-- this was
the language I used in the last video or the last
problem-- such that the trace of A not divisible by p,
but the determinant of A is. And we got the answer
as p minus 1 squared. Now, if you remember that
problem, and even just to do that problem,
when we thought about this statement,
such that the trace of A is not divisible by
p, we figured out that this was exactly equal to
saying that A does not equal 0. So one way to view the
result of problem 43 is if you assume that
A is not equal to 0, there are p minus 1 squared
members of T sub p where the determinant
is divisible by p. So we're starting to count some
of the A's whose determinant is divisible by p. But these are the only--
the matrix A's whose a entry is not equal to 0. So if we want to
count all of them, we have to also count the
ones where a does equal 0, so where a does equal 0. So let's think about that case. We can figure out how
many matrices there are where a does equal 0,
and the determinant of A is divisible by p. Add it to that. Then we have the total
number of A's whose determinants are divisible by p. And we could subtract
that from the total number of possibilities, and hopefully,
we'll get one of the answers up here. So let's think a
little bit about what will the determinant
look like over here? So in this situation, the
determinant of our matrix will be-- it's a squared. It's a times a minus b
times c, a squared minus bc. Now, we're assuming
that a is equal to 0. If a is equal to 0, it's
equal to negative bc. And we need to think
about how is this going to be a
multiple, or how is this going to be divisible by p? So this has got to be equal
to some multiple of p. So the first thing, just
think about b's and the c's. They are both non-negative. So this value right
here cannot be negative. So if you put a negative sign
there, it cannot be positive. So it can't be any
positive multiples of p. It could be 0. So we could have negative bc. We could have negative
bc is equal to 0, which is equivalent to saying
that bc is equal to 0. That's one possibility,
where it equals 0 times p. That's still a multiple. Or maybe it could be negative. Maybe we could have negative bc
is equal to negative 1 times p. But when you think about
it here, if this was true, that would mean that bc
is equal to p, so b or c would be factors of p. Now, we know that p
is a prime number. Its factors are only 1 and p. b and c-- I mean, one of
these guys could be 1, but then the other guy
would have to be p. They can't be p. They only get as
high as p minus 1. This is not an option. So the only possibility, if a is
equal to 0, for the determinant of capital A to still
be divisible by p, is for bc to be equal to 0. So let's think about how
many possibilities there are over here. So let's think about
all of the combinations. So there's the one situation,
where b and c are equal to 0. b and c is equal to 0. Now let's think
about the other ones. There's one where b is
equal to 0 and c isn't. bc will still equal
0 in this situation. So how many
possibilities are there? How many possibilities
are there for this? Well, b is equal to 0. C over here can be any
value, although we've already considered the case
where c is equal to 0. So it could be any
value so that we don't double count, any value
where c does not equal 0. So it could be 1
through p minus 1. Or another way to
think about it, there's p minus 1 possibilities. Now let's think about c
is equal to 0 and b isn't. Well, exact same logic. Once again, we don't
want to double count that they're both
being equal to 0. So let me write it. And c isn't-- well,
we're saying it isn't 0, so we are actually already
putting that constraint there. So it has p minus
1 possibilities. So how many total
possibilities do we have where the
determinant of A is equal to 0 and a is equal to 0? So we have one here. We have one possibility here,
then p minus 1 and p minus 1. So our total possibilities
are p minus 1 plus p minus 1 plus 1, which is 2p minus 2
plus 1, which is 2p minus 1. 2p minus 1 incremental
possibilities when a is equal to 0
for the determinant of A to be divisible by p. So that's what we just
got in this video. In the last video, we had p
minus 1 squared possibility when a does not equal 0. So if we want the total
number of A's, where the determinant of
A is divisible by 0, we can add the number that
we got in the last video to the number we just got. So let's do that. So if I were to
rewrite this over here, this is equal to p
squared minus 2p plus 1. And then I want to
add this 2p minus 1 here, so plus 2p minus 1. And what do I get? These cancel out. These cancel with that. And these guys cancel out. So there's a total of p
squared possible A's where the determinant of
A is divisible by p. Now, we're very close, but
this isn't what they're asking. They want the total members
where the determinant of A is not divisible by p. So all we have to do here is
take our total membership, which is p to the third,
and subtract from that the members whose determinants
are divisible by p. And we just figured that out. We subtract p squared. And now we'll get the number
of A's where determinant of A not divisible by p. Now let's see if that's one
of the choices, and it is. D, p cubed minus p squared.