Question

If $\left[\begin{array}{cc}1& -\tan \theta \\ \tan \theta & 1\end{array}\right]{\left[\begin{array}{cc}1& -\tan \theta \\ \tan \theta & 1\end{array}\right]}^{-1}={\left[\begin{array}{cc}\cos a& -\sin a\\ \sin a& \cos a\end{array}\right]}^{-1}$ then $a$=

• A. $0$
• B. $\frac{\pi }{2}$
• C. $\frac{\pi }{4}$
• D. $\frac{\pi }{6}$

Answer 1

The correct option is A $0$

We know that $A{A}^{-1}=I$

$\left[\begin{array}{cc}1& -\tan \theta \\ \tan \theta & 1\end{array}\right]{\left[\begin{array}{cc}1& -\tan \theta \\ \tan \theta & 1\end{array}\right]}^{-1}={\left[\begin{array}{cc}\cos a& -\sin x\\ \sin a& \cos a\end{array}\right]}^{-1}$

$\Rightarrow \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]={\left[\begin{array}{cc}\cos a& -\sin x\\ \sin a& \cos a\end{array}\right]}^{-1}$

Post-multiply both sides by $\left[\begin{array}{cc}\cos a& -\sin x\\ \sin a& \cos a\end{array}\right]$ we get

$\Rightarrow \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\left[\begin{array}{cc}\cos a& -\sin x\\ \sin a& \cos a\end{array}\right]={\left[\begin{array}{cc}\cos a& -\sin x\\ \sin a& \cos a\end{array}\right]}^{-1}\left[\begin{array}{cc}\cos a& -\sin x\\ \sin a& \cos a\end{array}\right]$

Using $A{A}^{-1}=I$ and $IA=A$ we have

$\Rightarrow \left[\begin{array}{cc}\cos a& -\sin x\\ \sin a& \cos a\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

Since both the matrices on LHS and RHS are of same order,hence we equating the corresponding rows and columns on both sides, we get

$\Rightarrow \cos a=1,\sin a=0$

$\Rightarrow \cos a=\cos 0,\sin a=\sin 0$

$\therefore a=0$