Question

If the trace of matrix $A=\left[\begin{array}{ccc}\sin x& 0& 2\\ 1& -4\cos y& 3\\ \sin x& \sin y& 3\sin z\end{array}\right]$ is $8$ for $x,y,z\in [0,\pi ],$ then trace of matrix $B=\left[\begin{array}{ccc}11\sin y& 0& 0\\ 9\sin y& 6\cos x& -3\\ \sin y& -\sin x& \sin z\end{array}\right]$ is

Answer 1

Given : $Tr\left(A\right)=8$
$\sin x-4\cos y+3\sin z=8$
which is possible only when
$\sin x=1,\cos y=-1$ and $\sin z=1$
$\Rightarrow x=\frac{\pi }{2},y=\pi$ and $z=\frac{\pi }{2}$ in $[0,\pi ]$

Now, $B=\left[\begin{array}{ccc}11\sin y& 0& 0\\ 9\sin y& 6\cos x& -3\\ \sin y& -\sin x& \sin z\end{array}\right]$
$\Rightarrow Tr\left(B\right)=11\sin y+6\cos x+\sin z$
$\Rightarrow Tr\left(B\right)=0-0+1$
$\therefore Tr\left(B\right)=1$