Question

If the minor of three-one element (i.e $M_{31}$) in the determinant $\left|\begin{array}{ccc}0& 1& \sec \alpha \\ \tan \alpha & -\sec \alpha & \tan \alpha \\ 1& 0& 1\end{array}\right|$ is $1$ then find the value of $\alpha$. $(0\le \alpha \le \pi )$

Answer 1

$$\left|\begin{array}{ccc}0& 1& \sec \alpha \\ \tan \alpha & -\sec \alpha & \tan \alpha \\ 1& 0& 1\end{array}\right|$$

$M_{31}=\tan \alpha +{\sec }^2\alpha$

$\Rightarrow \tan \alpha +{\sec }^2\alpha$

$\Rightarrow \tan \alpha +1+{\tan }^2\alpha =1$

$\Rightarrow \tan \alpha +{\tan }^2\alpha =0$

$\Rightarrow \tan \alpha \left(1+\tan \alpha \right)=0$

$\Rightarrow \tan \alpha =0or1+\tan \alpha =0$

$\Rightarrow \alpha =0or\tan \alpha =-1$

$\Rightarrow \alpha =0or\alpha =\frac{3\pi }{4}$