GATE 2026 DA Q.21: solution and answer

GATE 2026 · DAQ.21 · 1 mark · MCQ

Let $M = (cos θ sin θ - sin θ cos θ)$ be a $2 \times 2$ matrix, where $θ = \frac{2 π}{5}$ , and $I_{2} = (1001)$ . Which of the following options is equal to $M^{2026}$ ?

(A)
$M^{2}$
(B)
$M$
(C)
$M^{- 1}$
(D)
$I_{2}$

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Answer and solution

ANSWER

$M^{2026} = M$ , so option (B).

Official answer: BSource: Official answer key, GATE 2026 DA (IIT Guwahati)

$M$ is the rotation matrix by angle $θ$ . By the angle-addition formulas, $[cos a sin a - sin a cos a] [cos b sin b - sin b cos b] = [cos (a + b) sin (a + b) - sin (a + b) cos (a + b)]$ , so $M^{n} = [cos n θ sin n θ - sin n θ cos n θ]$ .
Here $θ = \frac{2 π}{5}$ , so $M^{5}$ is a rotation by $5 θ = 2 π$ , giving $M^{5} = I_{2}$ . Hence $M^{n}$ depends only on $n mod 5$ .
Reduce the exponent: $2026 = 5 \times 405 + 1$ , so $2026 \equiv 1 (mod 5)$ .
Therefore $M^{2026} = (M^{5})^{405} M = I_{2}^{405} M = M$ .
Option (B) is correct. The other options are different rotations: $M^{2}$ is a rotation by $\frac{4 π}{5}$ , $M^{- 1} = M^{4}$ is a rotation by $\frac{8 π}{5}$ (equivalently $- \frac{2 π}{5}$ ), and $I_{2}$ is a rotation by $0$ . None equals the rotation by $\frac{2 π}{5}$ .

KEY IDEA

A rotation matrix by $θ = \frac{2 π}{5}$ has order 5, since $M^{5} = I_{2}$ . Reduce the exponent modulo 5: $2026 \equiv 1$ .

WATCH OUT FOR

Reducing the exponent incorrectly, for example reading $2026 \equiv 2$ or $2026 \equiv 0 (mod 5)$ . That gives $M^{2}$ or $I_{2}$ . The remainder of $2026$ on division by 5 is 1.

From the GATE 2026 DA paper. Every question of this paper · Download the full paper · All 7 Special matrices questions, with notes

GATE 2026 DA Q.21: solution and answer

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