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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.