Let p be an odd number and Tp be the following set of 2 × 2 matrices.

Linked Question 1

The number of A in T_p such that det(A) is not divisible by p is

p³ – p²

p³ – 5p

p³ – 3p

2p²

$A = [\begin{matrix} a & b \\ c & a \end{matrix}] : a, b, c \in {0, 1, 2, . ....., p - 1$

| A | = a² – bc

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Linked Question 2

The number of A in T_p such that A is either symmetric or skew-symmetric or both, and det(A) divisible by p is

(p – 1)² + 1

(p – 1)²

2(p – 1)

2p – 1

|A| = a² – bc

If A is symmetric b = c

Then |A| = (a + b) (a + b)

So a & b can attain 2(p – 1) solution

It A is skew symmetric then a = b = c = 0

So total no. of solution = 2p – 2 + 1 = 2p – 1

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Linked Question 3

The number of A in T_p such that the trace of A is not divisible by p but det (A) is divisible by p is

[Note: The trace of a matrix is the sum of its diagonal entries.]

p³ – (p – 1)²

(p – 1)(p² – p + 1)

(p – 1)²

(p – 1)(p² – 2)

Trace = 2a

trace is not divisible by 'P' $\Rightarrow$ a $\neq$ 0

so a = 1, 2, ......., (p – 1)

for each value of 'a' then all (p – 1) option so that a² – bc is divisible by p

e.g. a = 1, p = 7

a² – bc = 1 – bc

so bc = 1 $\equiv$ (1, 1)

= 8 $\equiv$ (2, 4), (4, 2)

= 15 $\equiv$ (3, 5), (5, 3)

= 22 $\equiv$ $ϕ$

= 29 $\equiv$ $ϕ$

= 36 $\equiv$ (6, 6)

Total = 6 = p – 1

so required = (p – 1)(p – 1) = (p – 1)²

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