Let p be an odd prime number and $T_{p}$ be the following set of $2 \times 2$ matrices: $$T_{p}=\left\{A=\left[\begin{array}{ll} a & b \\ c & a \end{array}\right]: a, b, c \in\{0,1,2, \ldots, p-1\}\right\}$$ The number of A in $T_{p}$ such that $\det(A)$ is not divisible by p is

JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Determinants

Let p be an odd prime number and

T_{p}

be the following set of

2 \times 2

matrices:

T_{p} = {A = [a c b a] : a, b, c \in {0, 1, 2, \dots, p - 1}}

The number of A in

T_{p}

such that

det (A)

is not divisible by p is

Answer: (D)

Total matrices in

T_{p}

p^{3}

(each of

a, b, c

has

p

choices).

For

det (A) = a^{2} - b c

to be divisible by

p

, we need

a^{2} \equiv b c (mod p)

.

**Count matrices with

det (A) \equiv 0 (mod p)

:**

If $a = 0$ : $b c \equiv 0 (mod p)$ , so at least one of $b, c$ is 0. This gives $p^{2} - (p - 1)^{2} = 2 p - 1$ pairs, but actually $2 p - 1$ is wrong. If $a = 0$ , then $b c \equiv 0$ means $b = 0$ or $c = 0$ : $p + p - 1 = 2 p - 1$ pairs. Total: $2 p - 1$ matrices.
If $a \neq = 0$ : for each of $p - 1$ values of $a$ , the pairs $(b, c)$ with $b c \equiv a^{2} (mod p)$ number $p - 1$ (as computed in Q43). Total: $(p - 1)^{2}$ matrices.

Total with divisible determinant:

(2 p - 1) + (p - 1)^{2} = 2 p - 1 + p^{2} - 2 p + 1 = p^{2}

.

Matrices with determinant NOT divisible by

p

p^{3} - p^{2}

JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Single correct option