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# IIT JEE symmetric and skew-symmetric matrices

Video transcript

Let p be an odd
prime number and T sub p be the following
set of 2 by 2 matrices. So it's a set of matrices
where each member, we're just calling
it A where A could be made of lowercase,
a, b, c and a. And then these could take any
values between 0 and p minus 1. And remember, p was
an odd prime number. And frankly, all the prime
numbers greater than 2 are odd. So the number of A in Tp,
so the number of matrices such that A is either symmetric
or skew-symmetric or both, and the determinant of A is
divisible by p is-- so let's just make sure we
understand what symmetric and
skew-symmetric mean. So let's think about
symmetric first. So a matrix is symmetric
if it equals its transpose. So in this case, a matrix
a b, c, a-- and that's what the members of
the set look like. This is symmetric if this
is equal to its transpose. And a transpose of
a matrix, you just swap the rows and the columns. So this first row
here, a, b, will become the first column, a, b. And the second row, c, a,
becomes the second column, c, a. So that's what symmetric means. Now if you just
look at this, that means that a is equal
to a. a is equal to a. This essentially
means that b will have to be equal to c in
any of these matrices that are symmetric. Because that would have to
be equal to that and that would have to be equal to that. Fair enough. Now let's think about
what skew-symmetric means. Skew-symmetric just
means that a matrix is equal to the negative
of its transpose. So skew-symmetric
would mean that a, b, c, a is equal to the
negative of its transpose. This is its transpose, so
the negative of its transpose is negative a, negative
c, negative b, negative a. And so the two
results from this mean that a would have to be equal
to negative a, which essentially means that a has
to be equal to 0. The only number,
especially in this set, where it equals the negative
of each other-- frankly, the only number where you equal
the negative of yourself is 0. So that means a equals 0. And it also means that b
must be equal to negative c, or that c must be
equal to negative b. So it also means that b
is equal to negative c. So that's what
either of these mean. Now, either of these
are going to be true. We want to find the number
of members of the set, such that either of these are
true and the determinant of the matrix is divisible by p. And remember, the
determinant-- let me just write this over
here as a reminder. The determinant of a matrix
A is equal to a, b, c, a. And it's just equal to
a times a, or a squared, minus b times c. So in this situation over
here, the symmetric matrix, the determinant here is going
to be equal to a squared minus b times c. But we know that b is equal
to c because it's symmetric. So it's going to be minus--
we could say minus b squared. And over here, what's
the determinant going to be equal to? The determinant of a is going
to be equal to a squared. Well, we know a is 0, so
that's just going to be 0. Minus b times c. If we substitute for c, we know
that c is equal to negative b. We could write negative
b is equal to c. So this is the same thing as
negative b times negative b, which is equal to b squared. So if the matrix
is symmetric, we know that its
determinant is going to be a squared minus b squared. If it's skew-symmetric,
its determinant is going to be equal to b
squared, which is frankly going to be the same
thing as c squared. Now let's think about how
many matrices satisfy this where this determinant is
going to be divisible by p. So this needs to
be divisible by p. So first let's just factor this. This is the same thing, just
because it looks tempting, as a plus b times a minus b. So this is the determinant
in the symmetric case. And it needs to be divisible
by p, which essentially means it needs to equal some
integer constant times p. And remember, p was
an odd prime number. And it's an odd prime number
larger than any of the possible a's or b's. The a's and b's can
only go up to p minus 1. They start at 0 and they
can only go up to p minus 1. So this is larger than
any of these numbers. Now let's think about
how this would work. In order for this to
be divisible by p, that means either the first
term is divisible by p or the second term
is divisible by p. Let's think about that. Either the first term or the
second term is divisible by p, or I guess in theory,
both terms divisible by p. So we could say either-- let
me just write it like this. So one of those are going
to be divisible by p. So let's think about how
these could be divisible by p. So the only way that
this term right over here could be divisible by p, so
a minus b divisible by p So let's think about that. How can this be divisible by p? a and b are both less than p. So their difference is always
going to be smaller than p. Their difference, by definition,
is going to be smaller than p. So how can it be divisible by p? Well, the only way it's going to
be divisible by p is if it's 0. 0 is the only number smaller
then p that is divisible by p. So the only way that this is
going to be divisible by p, if a minus b is equal to 0. Or that's the same thing as
saying that a is equal to b. Now how many possibilities are
there for a being equal to b? Well, we can just look
at all the possibilities. There are, if we look over
here, there are p possibilities. You have one, two, three. Let me write in
a a darker color. a could be 0. That's one possibility. Two possibilities,
three possibilities. And if you were to just count
all the way up, it would be p. Notice when I'm
counting it, it's one more than the
number I'm counting. So this was p possibilities. And then b has to
be the same thing, so there's not much
freedom in selecting a b once you select the a. So there are p possibilities
where a is equal to b. There are p possibilities
of that happening. And let's just
keep in mind-- so a might be equal to 0,
which is equal to b. So this would include
both of them equalling 0, both of them equaling 1, both
of them equaling 2, all the way to both of them
equaling p minus 1. So there are p
possibilities of this. And let's think about how a
plus b could be divisible by kp. So remember, both a
and b are less than p. Both a and b are less than p. So this is a thought experiment. So a plus b could be
equal to 0 times p. That is a possibility. But frankly, when
you think about it, that means that a would have
to be equal to negative b. But both of these
are positive numbers, so that's not a possibility. So actually, a plus
b can't be 0 times p. I said 0 times p based on just
these are both smaller numbers. But then they would have to
be the negatives of each other and both of these
numbers are not negative. Well, I guess let
me take that back. This would work in the situation
where both a and b are 0. But we are ready included
that in this over here. We already included that
in these p possibilities where a and b are
both equal to 0. So let's not add to that. Now let's think about, because
this isn't a new possibility. a plus b is equal to p. Well, that's
completely possible. How many situations are there
where a plus b is equal to p? So if you pick your
a, it completely determines what your
b is going to be. Your b is going to be p minus a So can a be equal to 0? Well, no. If a was 0, b
would have to be p. And we know that b cannot be
p. b can at most be p minus 1. So a can be 1 and then b would
be p minus 1. a could be 2 and then b could
be p minus 2, all the way to a being p minus
1 and then b would be 1. So over here, there are
p minus 1 possibilities. And notice they don't overlap. And none of these possibilities
is either a or b equal to 0 or is either a or b
equal to each other. So they're never going
to be equal to each other because p over 2-- and you don't
have a situation where a plus b is equal to p, where
they're the same thing. Because if they were the same
thing, you'd go to this one. And you know this is never going
to be the case because this is a prime number. If it was possible that b could
equal a and this equation hold, you would have 2a
is equal to p, a is equal to p over
2, which would mean that p is divisible by 2. But it's prime. It can't be divisible by 2. So these are all p minus
1 distinct possibilities. They don't overlap
with any of these. a and b are never equal to 0. And well frankly, a and
b never equal each other in these possibilities. So so far, just from
the symmetric case, we've counted up minus
1 plus p possibilities. So we've counted up a total
of 2p minus 1 possibilities. I just added up all of
these possibilities. Now let's think
about the situation of being skew-symmetric. When you're skew-symmetric,
the determinant of a is equal to b squared. And we want the
possibilities where the determinant
is divisible by p. So b squared would have
to be equal to k times p. So let's think about this. If b times b is divisible by
p-- and remember, both of these are smaller than
p and p is prime. That means that either one
of these, or essentially it means that b would have
to be divisible by p. It means that b would
have to be divisible by p. And the only way that b is
going to be divisible by p, since b is smaller than
p, is if b is equal to 0. 0 divided by any
number other than 0-- because when you divide
by 0, it's undefined. But 0 divided by any other
number you have 0 remainder, so you can view it as divisible. So you could say, OK, here's
another possibility here. But we have to be very,
very, very careful. We already included that
possibility over here. We already included
the possibility of b equaling 0 in
this combination here where a is equal
to b is equal to 0. And let me be clear. We already said over here--
so you might say, wait. Over here, we're just
saying b is equal to 0. We're not saying that
a and b are equal to 0. But remember, in the
skew-symmetric case, a has to be equal to 0. So in the case where
b is equal to 0, you're saying that a
and b are equaling 0, but that was one of
the possibilities that we considered here. So that would work,
but we already counted that possibility. So there's a total of
2p minus 1 possibilities where a is either
symmetric or skew-symmetric and the determinant of
a is divisible by p. So 2p minus 1, the only
answer choice is D.