We are given that,

We need to find the det(Aⁿ).

To find det(Aⁿ),

First we need to find Aⁿ, and then take determinant of Aⁿ.

Let us find A².

A² = A.A

Let,

For z₁₁: Dot multiply the first row of the first matrix and first column of the second matrix, then sum up.

That is,

(cos θ, sin θ).(cos θ, -sin θ) = cos θ × cos θ + sin θ × (-sin θ)

⇒ (cos θ, sin θ).(cos θ, -sin θ) = cos² θ – sin² θ

By algebraic identity,

cos 2θ = cos² θ – sin² θ

⇒ (cos θ, sin θ).(cos θ, -sin θ) = cos 2θ

For z₁₂: Dot multiply the first row of the first matrix and second column of the second matrix, then sum up.

That is,

(cos θ, sin θ)(sin θ, cos θ) = cos θ × sin θ + sin θ × cos θ

⇒ (cos θ, sin θ)(sin θ, cos θ) = sin θ cos θ + sin θ cos θ

⇒ (cos θ, sin θ)(sin θ, cos θ) = 2 sin θ cos θ

By algebraic identity,

sin 2θ = 2 sin θ cos θ

⇒ (cos θ, sin θ)(sin θ, cos θ) = sin 2θ

Similarly,

If and , then

Now, taking determinant of Aⁿ,

Determinant of 2 × 2 matrix is found as,

So,

Det(Aⁿ) = cos nθ × cos nθ – sin nθ × (-sin nθ)

⇒ Det(Aⁿ) = cos² nθ + sin² nθ

Using the algebraic identity,

sin² A + cos² A = 1

⇒ Det(Aⁿ) = 1

Thus, Det(Aⁿ) is 1.