Exams Joint Entrance Examination Mathematics Matrices and Determinants Question 212 Question 212 of 472 Medium MCQ
If A = 1 2 ( 1 3 − 3 1 ) A=\dfrac12\begin{pmatrix}1&\sqrt{3}\\-\sqrt{3}&1\end{pmatrix} A = 2 1 ( 1 − 3 3 1 ) , then:
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A = ( cos 60 ∘ sin 60 ∘ − sin 60 ∘ cos 60 ∘ ) A=\begin{pmatrix}\cos60^\circ & \sin60^\circ\\-\sin60^\circ&\cos60^\circ\end{pmatrix} A = ( cos 6 0 ∘ − sin 6 0 ∘ sin 6 0 ∘ cos 6 0 ∘ ) , a rotation matrix by − 60 ∘ -60^\circ − 6 0 ∘ (i.e. α = 60 ∘ = π / 3 \alpha=60^\circ=\pi/3 α = 6 0 ∘ = π /3 ), so A n A^n A n is rotation by − n × 60 ∘ -n\times60^\circ − n × 6 0 ∘ .
A 30 A^{30} A 30 : rotation by − 1800 ∘ ≡ 0 ∘ ( m o d 360 ∘ ) -1800^\circ\equiv0^\circ \pmod{360^\circ} − 180 0 ∘ ≡ 0 ∘ ( mod 36 0 ∘ ) , so A 30 = I A^{30}=I A 30 = I .
A 25 A^{25} A 25 : rotation by − 1500 ∘ ≡ − 60 ∘ ( m o d 360 ∘ ) -1500^\circ\equiv-60^\circ\pmod{360^\circ} − 150 0 ∘ ≡ − 6 0 ∘ ( mod 36 0 ∘ ) , so A 25 = A A^{25}=A A 25 = A .
A 30 + A 25 − A = I + A − A = I A^{30}+A^{25}-A=I+A-A=I A 30 + A 25 − A = I + A − A = I . Answer the question to unlock