Question

If ${\displaystyle \cos \theta \left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]+\sin \theta \left[\begin{array}{cc}x& -\cos \theta \\ \cos \theta & x\end{array}\right]}$ = I₂, find x.

Answer 1

${\displaystyle \left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]\times \left[\begin{array}{cc}\cos \theta & -x\\ x& \cos \theta \end{array}\right]}$ = I₂ 

${\displaystyle \left[\begin{array}{cc}{\cos }^2\theta +x\sin \theta & -x\sin \theta +\sin \theta \cos \theta \\ -\sin \theta \cos \theta +x\cos \theta & x\sin \theta +{\cos }^2\theta \end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]}$

${\displaystyle \left[\begin{array}{cc}{\cos }^2\theta +x\sin \theta & 0\\ 0& x\sin \theta +{\cos }^2\theta \end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]}$

cos²θ + x sinθ = 1

x sinθ = 1 – cos²θ

x sinθ = sin²θ

x = ${\displaystyle \frac{{\sin }^2\theta }{\sin \theta }}$

x = sinθ

– x cosθ + sinθ cosθ = 0

sinθ cosθ = x cosθ

⇒ sinθ = x ...(÷ cosθ) 

∴ x = sinθ

The value of x = sinθ