Mathematics - Matrices Question with Solution | TestHub

Given,

$x\sin \theta =y\sin \left(\theta +\frac{2\pi }{3}\right)=z\sin \left(\theta +\frac{4\pi }{3}\right)\ne 0$, 

$\Rightarrow y=\frac{x\sin \theta }{\sin \left(\theta +{\displaystyle \frac{2\mathrm{\pi }}{3}}\right)}, z=\frac{x\sin \theta }{\sin \left(\theta +{\displaystyle \frac{4\mathrm{\pi }}{3}}\right)}$

Now, finding the trace of matrix we get,

$x+y+z=x+\frac{x\sin \theta }{\sin \left(\theta +{\displaystyle \frac{2\mathrm{\pi }}{3}}\right)}+\frac{x\sin \theta }{\sin \left(\theta +{\displaystyle \frac{4\mathrm{\pi }}{3}}\right)}$

$\Rightarrow x+y+z=\frac{x\sin \left(\theta +\frac{2\mathrm{\pi }}{3}\right)\sin \left(\theta +\frac{4\mathrm{\pi }}{3}\right)+x\sin \theta \sin \left(\theta +\frac{4\mathrm{\pi }}{3}\right)+x\sin \theta \sin \left(\theta +\frac{2\mathrm{\pi }}{3}\right)}{\sin \left(\theta +\frac{2\mathrm{\pi }}{3}\right)\sin \left(\theta +\frac{4\mathrm{\pi }}{3}\right)}$

$\Rightarrow x+y+z=\frac{x\sin \left(\theta +\frac{2\mathrm{\pi }}{3}\right)\sin \left(\theta +\frac{4\mathrm{\pi }}{3}\right)+x\sin \theta \left(\sin \left(\mathrm{\theta }+\frac{4\mathrm{\pi }}{3}\right)+\sin \left(\mathrm{\theta }+\frac{2\mathrm{\pi }}{3}\right)\right)}{\sin \left(\theta +\frac{2\mathrm{\pi }}{3}\right)\sin \left(\theta +\frac{4\mathrm{\pi }}{3}\right)}$

$\Rightarrow x+y+z=\frac{{\displaystyle \frac{x}{2}}·2\sin \left(\theta +\frac{2\mathrm{\pi }}{3}\right)\sin \left(\theta +\frac{4\mathrm{\pi }}{3}\right)+x\sin \theta \left(2\sin \left(\mathrm{\theta }+\mathrm{\pi }\right)\cos \left(\frac{\mathrm{\pi }}{3}\right)\right)}{\sin \left(\theta +\frac{2\mathrm{\pi }}{3}\right)\sin \left(\theta +\frac{4\mathrm{\pi }}{3}\right)}$

$\Rightarrow x+y+z=\frac{{\displaystyle \frac{x}{2}·\left(\cos \left(\frac{2\pi }{3}\right)-\cos \left(2\theta +2\pi \right)\right)+x\sin \theta \left(-2\sin \theta ·\frac{1}{2}\right)}}{\sin \left(\theta +\frac{2\pi }{3}\right)\sin \left(\theta +\frac{4\pi }{3}\right)}$

$\Rightarrow x+y+z=\frac{{\displaystyle \frac{x}{2}·\left(\frac{-1}{2}-\cos 2\theta \right)-x{\sin }^2\theta }}{\sin \left(\theta +\frac{2\mathrm{\pi }}{3}\right)\sin \left(\theta +\frac{4\mathrm{\pi }}{3}\right)}$

$\Rightarrow x+y+z=\frac{{\displaystyle -x-2x\cos 2\theta -4x{\sin }^2\theta }}{4\sin \left(\theta +\frac{2\mathrm{\pi }}{3}\right)\sin \left(\theta +\frac{4\mathrm{\pi }}{3}\right)}$

$\Rightarrow x+y+z=\frac{{\displaystyle -x-2x\left(1-2{\sin }^2\theta \right)-4x{\sin }^2\theta }}{4\sin \left(\theta +\frac{2\mathrm{\pi }}{3}\right)\sin \left(\theta +\frac{4\mathrm{\pi }}{3}\right)}$

$\Rightarrow x+y+z=\frac{{\displaystyle -3x+4x{\sin }^2\theta -4x{\sin }^2\theta }}{4\sin \left(\theta +\frac{2\mathrm{\pi }}{3}\right)\sin \left(\theta +\frac{4\mathrm{\pi }}{3}\right)}$

$\Rightarrow x+y+z=\frac{{\displaystyle -3x}}{4\sin \left(\theta +\frac{2\mathrm{\pi }}{3}\right)\sin \left(\theta +\frac{4\mathrm{\pi }}{3}\right)}\ne 0$

$\left(\mathrm{I}\right)$ Trace $\left(\mathrm{R}\right)=x+y+z\ne 0$

 $\Rightarrow$ Statement (i) is False

Now, finding $Adj(Adj(R\left)\right)$

Now, using the property $Adj(Adj(R\left)\right)=R\left|R\right|$ we get,

$\mathrm{Trace} \left(\mathrm{Adj}\right(\mathrm{Adj}\left(R\right)\left)\right)=xyz(x+y+z)\ne 0$

{As $x+y+z\ne 0$ proved above and $x,y \& z$ are non zero}

Hence, $\mathrm{Trace}(adj(adj\left(R\right))=0$ is a false statement,

So, neither $\left(\mathrm{I}\right)$ nor $\left(\mathrm{II}\right)$ are true