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IIT JEE trace and 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.