Exams Joint Entrance Examination Mathematics Matrices and Determinants Question 138 Question 138 of 472 Easy MCQ
If A = 1 2 [ 1 3 − 3 1 ] A=\dfrac12\begin{bmatrix}1 & \sqrt{3}\\-\sqrt{3} & 1\end{bmatrix} A = 2 1 [ 1 − 3 3 1 ] , then:
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A = [ cos 60 ∘ sin 60 ∘ − sin 60 ∘ cos 60 ∘ ] A=\begin{bmatrix}\cos60^\circ & \sin60^\circ\\-\sin60^\circ&\cos60^\circ\end{bmatrix} A = [ cos 6 0 ∘ − sin 6 0 ∘ sin 6 0 ∘ cos 6 0 ∘ ] is the rotation matrix for angle α = π 3 \alpha=\dfrac\pi3 α = 3 π , so A n A^n A n corresponds to angle n α n\alpha n α .
30 α = 1800 ∘ ≡ 0 ∘ ⇒ A 30 = I 30\alpha=1800^\circ\equiv0^\circ\Rightarrow A^{30}=I 30 α = 180 0 ∘ ≡ 0 ∘ ⇒ A 30 = I . 25 α = 1500 ∘ ≡ 60 ∘ = α ⇒ A 25 = A \quad 25\alpha=1500^\circ\equiv60^\circ=\alpha\Rightarrow A^{25}=A 25 α = 150 0 ∘ ≡ 6 0 ∘ = α ⇒ 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