Question

The value of determinant $\Delta =\left|\begin{array}{ccc}\sin \left(n\pi \right)& \tan \left(2n\pi \right)& \cos \left(\right(2n+1\left)\frac{\pi }{2}\right)\\ 1& 0& -1\\ \ln \left(7\right)& \cos \left(\frac{\pi }{4}\right)& -2\end{array}\right|$ is

Answer 1

For, $n\in I$
$\sin \left(n\pi \right)=0$
$\tan \left(2n\pi \right)=0\text{and}\cos \left(\right(2n+1\left)\frac{\pi }{2}\right)=0$
So, $\Delta$ will become
$\Delta =\left|\begin{array}{ccc}0& 0& 0\\ 1& 0& -1\\ \ln \left(7\right)& \cos \left(\frac{\pi }{4}\right)& -2\end{array}\right|$
As all the elements of $R_1$ is $0.$
Hence, the value of $\Delta =0$