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 $\left(B\right)$:

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

Answer 1

The correct option is D lies in $(1,2)$

$A=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]$
$B=A+A^4$
$A^2=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]$$\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]$
$=\left[\begin{array}{cc}{\cos }^2\theta -{\sin }^2\theta & 2\sin \theta \cos \theta \\ -2\sin \theta \cos \theta & -{\sin }^2\theta +{\cos }^2\theta \end{array}\right]$
$A^2=\left[\begin{array}{cc}\cos 2\theta & \sin 2\theta \\ -\sin 2\theta & \cos 2\theta \end{array}\right]$
Simmilarly
$A^4=\left[\begin{array}{cc}\cos 4\theta & \sin 4\theta \\ -\sin 4\theta & \cos 4\theta \end{array}\right]$
$B=A^4+A=\left[\begin{array}{cc}\cos 4\theta & \sin 4\theta \\ -\sin 4\theta & \cos 4\theta \end{array}\right]+\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]$
$B=A^4+A=\left[\begin{array}{cc}\cos 4\theta +\cos \theta & \sin 4\theta +\sin \theta \\ -\sin 4\theta -\sin \theta & \cos 4\theta +\cos \theta \end{array}\right]$
$detB=(\cos 4\theta +\cos \theta {)}^2+(\sin 4\theta +\sin \theta {)}^2$
$={\cos }^{2}4\theta +{\cos }^2\theta +2\cos 4\theta cos\theta +{\sin }^{2}4\theta +{\sin }^2\theta +2\sin 4\theta \sin \theta$
$=2+2(\cos 4\theta \cos \theta +\sin 4\theta \sin \theta )$
$=2+2\cos 3\theta$
at $\theta =\frac{\pi }{5}$
$\left|B\right|=2+2\cos \frac{3\pi }{5}=2-2\left(\sin 18\right)$
$\left|B\right|=2(1-\frac{\sqrt{5}-1}{4})=2\left(\frac{5-\sqrt{5}}{4}\right)=\frac{5-\sqrt{5}}{2}$