Question

Let $\theta =\frac{\pi }{5}$ and $A=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]$. If $B=A+A^4$, then $det\left(A\right)$:

• A. is one
• B. lies in $\left(1,2\right)$
• C. lies in $\left(2,3\right)$
• D. is zero

Answer 1

The correct option is B

lies in $\left(1,2\right)$

Explanation for the correct option:

Finding the value of $det\left(B\right)$ using the given condition:

Given the matrix $A=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]$

Finding the value of $B$ using the condition $B=A+A^4$.

We know that $A^n=\left[\begin{array}{cc}\cos n\theta & \sin n\theta \\ -\sin n\theta & \cos n\theta \end{array}\right]$,

$\Rightarrow A^4=\left[\begin{array}{cc}\cos 4\theta & \sin 4\theta \\ -\sin 4\theta & \cos 4\theta \end{array}\right]$

Then the value of $B$ is determined as,

$\begin{array}{rcl}B& =& A+A^4\\ & =& \left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]+\left[\begin{array}{cc}\cos 4\theta & \sin 4\theta \\ -\sin 4\theta & \cos 4\theta \end{array}\right]\\ & =& \left[\begin{array}{cc}\cos \theta +\cos 4\theta & \sin \theta +\sin 4\theta \\ -\left(\sin \theta +\sin 4\theta \right)& \cos \theta +\cos 4\theta \end{array}\right]\end{array}$

Now finding the determinant of $B$,

$\begin{array}{rcl}det\left(B\right)& =& \left|\begin{array}{cc}\cos \theta +\cos 4\theta & \sin \theta +\sin 4\theta \\ -\left(\sin \theta +\sin 4\theta \right)& \cos \theta +\cos 4\theta \end{array}\right|\\ & =& {\left(\cos \theta +\cos 4\theta \right)}^2-\left(-{\left(\sin \theta +\sin 4\theta \right)}^2\right)\\ & =& {\left(\cos \theta +\cos 4\theta \right)}^2+{\left(\sin \theta +\sin 4\theta \right)}^2\\ & =& {\cos }^2\theta +{\cos }^{2}4\theta +2\cos \theta \cos 4\theta +{\sin }^2\theta +{\sin }^{2}4\theta +2\sin \theta \sin 4\theta \\ & =& \left({\cos }^2\theta +{\sin }^2\theta \right)+\left({\cos }^{2}4\theta +{\sin }^{2}4\theta \right)+2\left(\cos \theta \cos 4\theta +\sin \theta \sin 4\theta \right)\\ & =& 1+1+2\left(\cos \theta \cos 4\theta +\sin \theta \sin 4\theta \right)\end{array}$

We know that $\cos \left(a-b\right)=\cos a\cos b+\sin a\sin b$ so, by replacing $a$ by $4\theta$ and $b$ by $\theta$ in the above equation we get,

$\begin{array}{rcl}det\left(B\right)& =& 2+2\cos \left(4\theta -\theta \right)\\ & =& 2+2\cos 3\theta \end{array}$

Since it is given that $\theta =\frac{\pi }{5}$, substitiute $\theta$ as $\frac{\pi }{5}$ in the above equation.

$\begin{array}{rcl}det\left(B\right)& =& 2+2\cos 3\left(\frac{\pi }{5}\right)\\ & =& 2+2\cos \left(\frac{3\pi }{5}\right)\\ & =& 2+2\cos \left(108\right)\\ & =& 2+2\left(\frac{1-\sqrt{5}}{4}\right)\end{array}$

Taking lcm as $4$ and common factor as $2$ we get,

$\begin{array}{rcl}det\left(B\right)& =& \frac{8+2\left(1-\sqrt{5}\right)}{4}\\ & =& \frac{4+1-\sqrt{5}}{2}\\ & =& \frac{5-\sqrt{5}}{2}\\ & =& 1.381966\end{array}$

Therefore, the value of $det\left(B\right)$ is $1.381966$ which lies between $\left(1,2\right)$.

Hence, option (B) is the correct answer.