IIT JEE divisible determinants

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.