Question

Let \[A=\left[ \begin{matrix}
   \cos \alpha & -\sin \alpha \\
   \sin \alpha & \cos \alpha \\
\end{matrix} \right],\left( \alpha \in R \right)\] such that ${{A}^{32}}=\left[ \begin{matrix}
   0 & -1 \\
   1 & 0 \\
\end{matrix} \right]$. Then the value of $\alpha $ is
A. $\dfrac{\pi }{16}$
B. $0$
C. $\dfrac{\pi }{32}$
D. $\dfrac{\pi }{64}$

Answer 1

Hint: We will first try to figure out the value of ${{A}^{32}}$, this will be done by applying the principle of mathematical induction. After that we will compare the values with the ones that are given in the question and subsequently find the value $\alpha $.

Complete step by step answer:
We have the given matrix: $A=\left[ \begin{matrix}
   \cos \alpha & -\sin \alpha \\
   \sin \alpha & \cos \alpha \\
\end{matrix} \right]$
if $P\left( n \right)$ : $A=\left[ \begin{matrix}
   \cos \alpha & -\sin \alpha \\
   \sin \alpha & \cos \alpha \\
\end{matrix} \right]$then\[{{A}^{n}}=\left[ \begin{matrix}
   \cos n\alpha & -\sin n\alpha \\
   \sin n\alpha & \cos n\alpha \\
\end{matrix} \right],n\in \mathbb{N}\]

We can prove by the principle of mathematical induction. It is a mathematical technique which is used to prove a statement, a formula or a theorem is true for every natural number. The technique involves two steps to prove a statement, as stated below
− Step 1(Base step) − It proves that a statement is true for the initial value.
Step 2(Inductive step) − It proves that if the statement is true for the nth iteration , then it is also true for (n+1)th iteration.
Now let’s find out for $n=2$
\[\begin{align}
  & A=\left[ \begin{matrix}
   \cos \alpha & -\sin \alpha \\
   \sin \alpha & \cos \alpha \\
\end{matrix} \right] \\
 & {{A}^{2}}=\left[ \begin{matrix}
   \cos \alpha & -\sin \alpha \\
   \sin \alpha & \cos \alpha \\
\end{matrix} \right]\left[ \begin{matrix}
   \cos \alpha & -\sin \alpha \\
   \sin \alpha & \cos \alpha \\
\end{matrix} \right] \\
 & {{A}^{2}}=\left[ \begin{matrix}
   {{\cos }^{2}}\alpha -{{\sin }^{2}}\alpha & -\cos \alpha \sin \alpha -\sin \alpha \cos \alpha \\
   \sin \alpha \cos \alpha & {{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha \\
\end{matrix} \right] \\
 & {{A}^{2}}=\left[ \begin{matrix}
   \cos 2\alpha & -\sin 2\alpha \\
   \sin 2\alpha & \cos 2\alpha \\
\end{matrix} \right] \\
\end{align}\]
Therefore P(n) is true for n=2.
Now, assume P(k) to be true and then prove P(k+1) is true:
 $P\left( k \right)$ : $A=\left[ \begin{matrix}
   \cos \alpha & -\sin \alpha \\
   \sin \alpha & \cos \alpha \\
\end{matrix} \right]$then\[{{A}^{k}}=\left[ \begin{matrix}
   \cos k\alpha & -\sin k\alpha \\
   \sin k\alpha & \cos k\alpha \\
\end{matrix} \right],k\in \mathbb{N}\]
We will have to prove that P(k+1) is true: for that we have to show that: \[{{A}^{k+1}}=\left[ \begin{matrix}
   \cos \left( k+1 \right)\alpha & -\sin \left( k+1 \right)\alpha \\
   \sin \left( k+1 \right)\alpha & \cos \left( k+1 \right)\alpha \\
\end{matrix} \right],k\in \mathbb{N}\]
Taking LHS:
$\begin{align}
  & {{A}^{k+1}}={{A}^{k}}.A \\
 & {{A}^{k+1}}=\left[ \begin{matrix}
   \cos k\alpha & -\sin k\alpha \\
   \sin k\alpha & \cos k\alpha \\
\end{matrix} \right].\left[ \begin{matrix}
   \cos \alpha & -\sin \alpha \\
   \sin \alpha & \cos \alpha \\
\end{matrix} \right] \\
 & {{A}^{k+1}}=\left[ \begin{matrix}
   \cos \left( k+1 \right)\alpha & -\sin \left( k+1 \right)\alpha \\
   \sin \left( k+1 \right)\alpha & \cos \left( k+1 \right)\alpha \\
\end{matrix} \right]=RHS \\
\end{align}$
Therefore by the principal of mathematical induction, P(n) is true for $n\in \mathbb{N}$

We will now take n = 32:
\[{{A}^{32}}=\left[ \begin{matrix}
   \cos \left( 32\alpha \right) & -\sin \left( 32\alpha \right) \\
   \sin \left( 32\alpha \right) & \cos \left( 32\alpha \right) \\
\end{matrix} \right],n\in \mathbb{N}\]
Now we will compare this value of ${{A}^{32}}$ to the one given in the given that is: $\left[ \begin{matrix}
   0 & -1 \\
   1 & 0 \\
\end{matrix} \right]$
\[\Rightarrow {{A}^{32}}=\left[ \begin{matrix}
   \cos \left( 32\alpha \right) & -\sin \left( 32\alpha \right) \\
   \sin \left( 32\alpha \right) & \cos \left( 32\alpha \right) \\
\end{matrix} \right]=\left[ \begin{matrix}
   0 & -1 \\
   1 & 0 \\
\end{matrix} \right]\]
So on comparing we get :
$\begin{align}
  & \cos \left( 32\alpha \right)=0 \\
 & \sin \left( 32\alpha \right)=1 \\
\end{align}$
Taking inverse we get
 $\begin{align}
  & \Rightarrow 32\alpha =\dfrac{\pi }{2} \\
 & \Rightarrow \alpha =\dfrac{\pi }{64} \\
\end{align}$
Therefore the correct answer is Option D.

Note:
There is no need to prove and derive the whole principle of mathematical induction, students can directly apply it just by showing it for one value of n. Also take care while comparing the matrices as chances of silly mistakes are higher there.