Matrices & Determinants
Matrices and Determinants
Grade Class 12
Question:
Let p be an odd prime number and T<sub>p</sub> be the following set of 2 × 2 matrices : <br>T<sub>p</sub> = {A = <math><mfenced open='[' close=']'><mtable><mtr><mtd><mi>a</mi></mtd><mtd><mi>b</mi></mtd></mtr><mtr><mtd><mi>c</mi></mtd><mtd><mi>a</mi></mtd></mtr></mtable></mfenced></math> : a, b, c ∈ {0, 1, 2, ..., p-1}}. The number of A in T<sub>p</sub> such that A is either symmetric or skew-symmetric or both, and det(A) divisible by p is -
(A) (p - 1)<sup>2</sup>
(B) 2 (p - 1)
(C) (p - 1)<sup>2</sup> + 1
(D) 2p - 1
Step-by-Step Solution
Key Concept: A matrix A = [[a, b], [c, a]] is symmetric if b = c. It is skew-symmetric if a = 0, b = -c (mod p), and a = 0. The determinant is a^2 - bc. We need to count cases where A is symmetric or skew-symmetric and a^2 - bc is divisible by p.
For A = [[a, b], [c, a]] to be symmetric, b = c. Then det(A) = a^2 - b^2 = (a-b)(a+b). For this to be divisible by p, a = b or a = -b (mod p). There are p choices for a and b such that a = b, and p choices for a and b such that a = -b. The case a = b = 0 is counted in both, so total symmetric matrices with det(A) divisible by p is p + p - 1 = 2p - 1. For skew-symmetric, a = 0 and b = -c. Then det(A) = 0 - b(-b) = b^2. For this to be divisible by p, b must be 0. This gives the matrix [[0, 0], [0, 0]], which is already counted in the symmetric case. Thus, the total count is 2p - 1.
Correct Answer: 4