Main content

## IIT JEE

Current time:0:00Total duration:9:19

# IIT JEE trace andÂ determinant

## Video transcript

So we have the same
setup here as we had in the previous problem,
in problem number 42, where p is an odd prime number. And they say, "Let T sub p
be the following set of 2 by 2 matrices," where
each of those matrices have the form a's along the
diagonals and then a b there and a c there. And each of those
numbers are between 0 and p minus 1 inclusive. So they can be 0, 1, 2, 3,
all the way up to p minus 1. And then in number
43, they ask us, "The number of A in the set
such that the trace of A is not divisible by p,
but the determinant of A is divisible by p is?" And then, they tell us,
"The trace of a matrix is the sum of its
diagonal entries." It's actually the sum
of its main diagonal. It's not going to
be that diagonal. It's only the sum of this. So if you sum this, you're going
to get the trace of the matrix. So let's use this first
piece of information. The trace of A is
not divisible by p. So the trace of A is going
to be a plus a, or 2a. So 2a not divisible by p. So let's think of a situation
where 2a is divisible by p. So 2a could be equal
to 0, because 0 is a multiple of p, in which
case a would be equal to 0. So this is a possibility. So if a is equal to 0,
then 2a is divisible by p. So one of the constraints,
from this sentence right here, is that a cannot be equal to 0. Because if a is 0, then the
trace is divisible by p. So a cannot be equal to 0. And let's think about it. Can 2a become any
other multiple of p? Can 2a equal p? Well, if 2a equaled p, then 2
and a would be factors of p, and then p would not be prime. And so we can cross that out. And then 2a can't
be equal to 2p. That would make a equal
to p, which is impossible, because a can only be p minus 1. So that's not possible. And then you can keep going
higher, higher multiples of p, not possible, because a is
going to be smaller than p. So the only constraint that we
get from this first sentence, the trace of a is
not divisible by p, tells us that a
does not equal 0. So now let's look at
the second constraint. The determinant of
A is divisible by p. So the determinant of
A-- let me write this. The determinant of A
is equal to a times a-- it's equal to a squared
minus b times c, or c times b, so minus bc. So this is going to
be divisible by p. This is going to be equal
to some integer times p. Now one thing just
looking at this, there's no obvious
simplification here. But one thing I'm
starting to appreciate, not having taken the
IIT Joint Entrance Exam when I was in high school,
not growing up in India, is that when they have
multiple choices, sometimes it might be faster to just
do the brute force method. And when I say the
brute force method, not just proving mathematically
which of these expressions are true, but just
let's try a simple p. Let's try a simple
p, and then let's find the members of the set
that would satisfy it for that p and then see which of these
expressions equal that p. Let's see what we get. So let's just simplify this. Let's just say p is equal to 3. And I pick p is equal to 3
because that's the smallest odd prime number. It's the smallest number that
satisfies this constraint up here. So let's set p is equal to 3. So if p is equal to 3,
our possibilities for a are that a could be equal
to-- a can't be equal to 0. We know that already. a cannot be equal to 0. So a could be equal to 1
or a could be equal to 2. And if a is equal to 1,
then the determinant of A is going to be equal
to 1 minus bc, which needs to be equal to some
integer multiple of 3, because we set our
p is equal to 3. It must be equal to 3k. If a is equal to 2,
then the determinant is going to be 4,
because it's a squared. So it's going to
be 4 minus bc, must be equal to some
integer multiple of 3. So let's see how many
combinations of b and c we can find in each
of these constraints. So 1 minus bc. So one possibility--
let's think about it. So one possibility is 1 minus
bc is equal to 3 times 0. That's one possibility. Maybe another possibility
is 1 minus bc is equal to 3. Actually, let's think
about this a little bit. Let me try this
possibility here, first. So if 1 minus bc is
equal to 0-- this is pretty straightforward-- bc
would have to be equal to 1. And so b would have
to be equal to 1 and c would have
to be equal to 1. So we have one combination
here, a could be equal to 1, b is equal to 1,
and c is equal to 1. That is one possibility
right over there. And then another
possibility-- now, no matter what bc we pick
here, the maximum value here is going to be 1. If bc are both 0, then this
expression is going to be 1. If they're both 1, this
expression is going to be 0. If they are any other value--
and the only other values that they could be are
some combination of 1 and 2 or 2 and 2-- then we're going
to get a negative number. So the question is, is
a negative multiple of 3 considered to be
a multiple of 3? Now let's just figure it
out assuming that it is, and then we can see if maybe
one of these expressions makes sense. And I'm going to do that in
white just because it depends. If you considered
negative 3 divisible by 3, then we should go with that. So maybe we should
say 1 minus bc could be equal to
negative 3, which is the same thing
as bc minus 1 is equal to 3, which is the
same-- I just multiplied both sides by negative 1--
which is the same thing as bc is equal to 4. And we have to be careful here. You might be tempted to say b
is equal to 1, c is equal to 4, or b is equal to
4, c is equal to 1, but they can't be equal to 4. 4 is bigger than our p. They can only be up to 2. So this constraint is
only, b is equal to 2, and c is equal to 2. And of course, a is equal to 1. So this is, if you do
consider negative 3 to be divisible by 3,
which it actually is. It's negative 1 times 3. It is a multiple. So this would be
one of the cases. But I'll put that
there just in case. You never know what the test
writers are actually thinking. So I'll put that there as
kind of a contingent case. And then this last situation,
4 minus bc is equal to 3k. So 4 minus bc could
be equal to 0. Or 4 minus bc--
let's think about it. 4 minus bc could be equal to 0. 4 minus bc could be equal to 3. That's definitely a possibility,
because this value right here will always be less than 4. And you can't get
anything less than 0 here, because the highest
value for b and c are both 2. 2 times 2 is going to be 4,
so you can't get anything less than 0 here. So you won't have
a case like this. So-- and I just said it. So if you have this equation,
that means 4 is equal to bc. The only possibility with
b and c being less than 3, being 0, 1, or 2, is
that b is equal to 2, and c is equal to 2. Neither one of those
could be equal to 4, because that's
bigger than our p. And so here, we have a is
equal to 2, b is equal to 2, and c is equal to 2. And then if 4 minus
bc is equal to 3, that means that this, that
bc, has to be equal to 1, bc is equal to 1. And so the only
possibility is that-- let me scroll to the
right-- b is equal to 1, and c is equal to 1, and
of course, a is equal to 2. So we've been able to list four
distinct possibilities, four distinct matrices, for
our p is equal to 3. Now let's see if any
of these evaluate to 4 when p is equal to 3. Let me put this down here. So if we have 3 minus 2--
sorry, 3 minus 1, that's 2, 3 squared, 9, minus
3 is 6, plus 1 is 7. So this is 2 times 7. That's 14. That doesn't evaluate to 4. This is 27 minus 4, this is
23, also doesn't evaluate to 4. 3 minus 1 squared,
that's 2 squared. This evaluates to 4. So this is looking pretty good. We just have to hope that
something else doesn't evaluate to 4. This is 2, 3 minus
1 is 2, times 7. Once again, this is also 14. Not there. So I would go with C. And so the exam writers
really did want negative 3 to be a possible-- or they
really do view negative 3 as being divisible
by 3, which it is. But, you know,
sometimes people think of it only as kind of
positive multiples. But it clearly is divisible
by 3, so our answer is C.