PYQs / GATE DA / 2026 / Set 1 / Q21 GATE DA 2026 Set 1 — Question 21 Go beyond PYQs with Success Tracker AI-powered personalised practice and doubt support. Unlimited practice on eligible plans; AI usage limits apply. MCQ +1 / -0.33 Easy Vector & Matrix Transformations LU Decomposition & Determinants Linear Algebra Matrix Multiplication & Conformability Matrix Algebra
Linear Algebra → LU Decomposition & Determinants → Matrix Multiplication & Conformability
Last updated 5 September 2026
Question Let
M = ( cos θ − sin θ sin θ cos θ ) M = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} M = ( cos θ sin θ − sin θ cos θ ) be a
2 × 2 2 \times 2 2 × 2 matrix, where
θ = 2 π 5 \theta = \frac{2\pi}{5} θ = 5 2 π , and
I 2 = ( 1 0 0 1 ) I_2 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} I 2 = ( 1 0 0 1 ) .
Which of the following options is equal to
M 2026 M^{2026} M 2026 ?
Solution The matrix
M = ( cos θ − sin θ sin θ cos θ ) M = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} M = ( cos θ sin θ − sin θ cos θ ) represents a rotation in
R 2 \mathbb{R}^2 R 2 by an angle
θ \theta θ . It is a property of rotation matrices that
M k M^k M k represents a rotation by angle
k θ k\theta k θ .
Given
θ = 2 π 5 \theta = \frac{2\pi}{5} θ = 5 2 π , we have:
M 5 = ( cos ( 5 θ ) − sin ( 5 θ ) sin ( 5 θ ) cos ( 5 θ ) ) = ( cos ( 2 π ) − sin ( 2 π ) sin ( 2 π ) cos ( 2 π ) ) = ( 1 0 0 1 ) = I 2 M^5 = \begin{pmatrix} \cos(5\theta) & -\sin(5\theta) \\ \sin(5\theta) & \cos(5\theta) \end{pmatrix} = \begin{pmatrix} \cos(2\pi) & -\sin(2\pi) \\ \sin(2\pi) & \cos(2\pi) \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I_2 M 5 = ( cos ( 5 θ ) sin ( 5 θ ) − sin ( 5 θ ) cos ( 5 θ ) ) = ( cos ( 2 π ) sin ( 2 π ) − sin ( 2 π ) cos ( 2 π ) ) = ( 1 0 0 1 ) = I 2 .
We want to find
M 2026 M^{2026} M 2026 . We can write
2026 2026 2026 as
5 × 405 + 1 5 \times 405 + 1 5 × 405 + 1 .
M 2026 = M 5 × 405 + 1 = ( M 5 ) 405 × M 1 = ( I 2 ) 405 × M = I 2 × M = M M^{2026} = M^{5 \times 405 + 1} = (M^5)^{405} \times M^1 = (I_2)^{405} \times M = I_2 \times M = M M 2026 = M 5 × 405 + 1 = ( M 5 ) 405 × M 1 = ( I 2 ) 405 × M = I 2 × M = M .
Thus, option (B) is correct.
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Solution The matrix
M = ( cos θ − sin θ sin θ cos θ ) M = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} M = ( cos θ sin θ − sin θ cos θ ) represents a rotation in
R 2 \mathbb{R}^2 R 2 by an angle
θ \theta θ . It is a property of rotation matrices that
M k M^k M k represents a rotation by angle
k θ k\theta k θ .
Given
θ = 2 π 5 \theta = \frac{2\pi}{5} θ = 5 2 π , we have:
M 5 = ( cos ( 5 θ ) − sin ( 5 θ ) sin ( 5 θ ) cos ( 5 θ ) ) = ( cos ( 2 π ) − sin ( 2 π ) sin ( 2 π ) cos ( 2 π ) ) = ( 1 0 0 1 ) = I 2 M^5 = \begin{pmatrix} \cos(5\theta) & -\sin(5\theta) \\ \sin(5\theta) & \cos(5\theta) \end{pmatrix} = \begin{pmatrix} \cos(2\pi) & -\sin(2\pi) \\ \sin(2\pi) & \cos(2\pi) \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I_2 M 5 = ( cos ( 5 θ ) sin ( 5 θ ) − sin ( 5 θ ) cos ( 5 θ ) ) = ( cos ( 2 π ) sin ( 2 π ) − sin ( 2 π ) cos ( 2 π ) ) = ( 1 0 0 1 ) = I 2 .
We want to find
M 2026 M^{2026} M 2026 . We can write
2026 2026 2026 as
5 × 405 + 1 5 \times 405 + 1 5 × 405 + 1 .
M 2026 = M 5 × 405 + 1 = ( M 5 ) 405 × M 1 = ( I 2 ) 405 × M = I 2 × M = M M^{2026} = M^{5 \times 405 + 1} = (M^5)^{405} \times M^1 = (I_2)^{405} \times M = I_2 \times M = M M 2026 = M 5 × 405 + 1 = ( M 5 ) 405 × M 1 = ( I 2 ) 405 × M = I 2 × M = M .
Thus, option (B) is correct.
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