If A= [ sin α -cos α cos α sin α ] and A+A^-1=I , then α=

Correct answer: (C) (π)/6. Given, A = [ sin α -cos α cos α sin α ] Adj A = [ sin α cos α -cos α sin α ] |A| =sin^2 α+cos^2 α=1 A^-1 = [ sin α cos α -cos α sin α ] A+A^-1 = [ sin α -cos α cos α sin α ]+ [ sin α cos α -cos α sin α ] = [ 2 sin α 0 0 2 sin α ] A+A^-1 =I, if 2 sin α=1 sin α =1/2 ⇒ α=(π)/6

Q45 / 160 · AP EAMCET 2022 (06 Jul Shift 2) · Mathematics

If $A=\left[\begin{array}{cc}\sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha\end{array}\right]$ and $A+A^{-1}=I$, then $\alpha=$

A
0
B
$\frac{\pi}{3}$
Correct answer
C
$\frac{\pi}{6}$
D
$\frac{\pi}{4}$

Solution

Given, $\begin{aligned} A & =\left[\begin{array}{cc}\sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha\end{array}\right] \\ \text { Adj } A & =\left[\begin{array}{cc}\sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha\end{array}\right] \\ |A| & =\sin ^2 \alpha+\cos ^2 \alpha=1 \\ A^{-1} & =\left[\begin{array}{cc}\sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha\end{array}\right] \\ A+A^{-1} & =\left[\begin{array}{cc}\sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha\end{array}\right]+\left[\begin{array}{cc}\sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha\end{array}\right] \\ & =\left[\begin{array}{cc}2 \sin \alpha & 0 \\ 0 & 2 \sin \alpha\end{array}\right] \\ A+A^{-1} & =I, \text { if } 2 \sin \alpha=1 \\ \sin \alpha & =\frac{1}{2} \Rightarrow \alpha=\frac{\pi}{6}\end{aligned}$

Hence, the correct answer is option: 'C'.

If $A=\left[\begin{array}{cc}\sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha\end{array}\right]$ and $A+A^{-1}=I$, then $\alpha=$

Solution

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