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Grade: 12

                        

Please explain me Passage 1...

5 years ago

Answers : (2)

Latika Leekha
askIITians Faculty
165 Points
							Hello student,
The passage says
p is an odd prime number and TP denotes such a set of 2X2 matrices whose elements are of the form a, b in first row and c, a in second row where a, b, c all elements are from the set {0, 1, 2, …. , p-1). this means that the elements of the matrix can be numbers ranging from 0 till 1 less than that odd prime number p. Infact all prime numbers gretaer than 2 are odd.
Further on the basis of this you are required to answer the following questions.
I am solving the following question too:
The number of A in TP such that A is either symmetric or skew-symmetric or both and det is also divisible by p is 2p-1.
For symmetric , we will have b = c and so det = a2-b2
Similarly, for skew-symmetric, a = 0 and b = -c and so det = b2.
Now for det to be divisible by p, in symmetric we have a2-b2 = (a-b)(a+b) should be divisible by p i.e. = kp
Now for this to happen, either (a-b) or (a+b) shoulod be divisible by p.
(a-b) is divisible only when a-b = 0 i.e. a = b
There are p possibilities in this case.
Next, (a+b) can be divisible when a+b = 0.p
i.e. a and b are negatives – not possible.
Also, if a = 0 then b = p not possible.
LEaving these cases there are a total of (p-1) possibilities.
Hence, total nuber of possibilities are p + p - 1 = 2p -1.
Now, second case i.e. skew-symmetric, det = b2, so b2 should be divisible by p
i.e. b.b should be divisible by p i .e.b should be divisible by p.
but then b = 0 and a = 0 because it is skew-symmetric.
But this case has already been included in the previous case.
Hence, toal number of cases 2p-1.
5 years ago
SHIKHER CHITRANSH SRIVASTAV
34 Points
							thank you it was very helpful...
						
5 years ago
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