Let R=( … ) be a non-zero 3× 3 matrix, where x sinθ =y sin(θ + 2π 3)=z sin(θ + 4π 3) ≠ 0,θ ∈ (0,2π ). For a square matrix M, let Trace(M) denote the…

Given,

$x\mathrm{sin}\theta =y\mathrm{sin}(\theta +\frac{2\pi }{3})=z\mathrm{sin}(\theta +\frac{4\pi }{3})\neq 0$xsinθ=ysin(θ+32π​)=zsin(θ+34π​)=0,

$$\Rightarrow y=\frac{x\mathrm{sin}\theta }{\mathrm{sin}(\theta +\frac{2\pi }{3})},z=\frac{x\mathrm{sin}\theta }{\mathrm{sin}(\theta +\frac{4\pi }{3})}$$⇒y=sin(θ+32π​)xsinθ​,z=sin(θ+34π​)xsinθ​

Now, finding the trace of matrix we get,

$$x+y+z=x+\frac{x\mathrm{sin}\theta }{\mathrm{sin}(\theta +\frac{2\pi }{3})}+\frac{x\mathrm{sin}\theta }{\mathrm{sin}(\theta +\frac{4\pi }{3})}$$x+y+z=x+sin(θ+32π​)xsinθ​+sin(θ+34π​)xsinθ​

$$\Rightarrow x+y+z=\frac{x\mathrm{sin}(\theta +\frac{2\pi }{3})\mathrm{sin}(\theta +\frac{4\pi }{3})+x\mathrm{sin}\theta \mathrm{sin}(\theta +\frac{4\pi }{3})+x\mathrm{sin}\theta \mathrm{sin}(\theta +\frac{2\pi }{3})}{\mathrm{sin}(\theta +\frac{2\pi }{3})\mathrm{sin}(\theta +\frac{4\pi }{3})}$$⇒x+y+z=sin(θ+32π​)sin(θ+34π​)xsin(θ+32π​)sin(θ+34π​)+xsinθsin(θ+34π​)+xsinθsin(θ+32π​)​

$$\Rightarrow x+y+z=\frac{x\mathrm{sin}(\theta +\frac{2\pi }{3})\mathrm{sin}(\theta +\frac{4\pi }{3})+x\mathrm{sin}\theta (\mathrm{sin}(\theta +\frac{4\pi }{3})+\mathrm{sin}(\theta +\frac{2\pi }{3}))}{\mathrm{sin}(\theta +\frac{2\pi }{3})\mathrm{sin}(\theta +\frac{4\pi }{3})}$$⇒x+y+z=sin(θ+32π​)sin(θ+34π​)xsin(θ+32π​)sin(θ+34π​)+xsinθ(sin(θ+34π​)+sin(θ+32π​))​

$$\Rightarrow x+y+z=\frac{\frac{x}{2}\cdot 2\mathrm{sin}(\theta +\frac{2\pi }{3})\mathrm{sin}(\theta +\frac{4\pi }{3})+x\mathrm{sin}\theta (2\mathrm{sin}(\theta +\pi )\mathrm{cos}(\frac{\pi }{3}))}{\mathrm{sin}(\theta +\frac{2\pi }{3})\mathrm{sin}(\theta +\frac{4\pi }{3})}$$⇒x+y+z=sin(θ+32π​)sin(θ+34π​)2x​⋅2sin(θ+32π​)sin(θ+34π​)+xsinθ(2sin(θ+π)cos(3π​))​

$$\Rightarrow x+y+z=\frac{\frac{x}{2}\cdot (\mathrm{cos}(\frac{2\pi }{3})-\mathrm{cos}(2\theta +2\pi ))+x\mathrm{sin}\theta (-2\mathrm{sin}\theta \cdot \frac{1}{2})}{\mathrm{sin}(\theta +\frac{2\pi }{3})\mathrm{sin}(\theta +\frac{4\pi }{3})}$$⇒x+y+z=sin(θ+32π​)sin(θ+34π​)2x​⋅(cos(32π​)−cos(2θ+2π))+xsinθ(−2sinθ⋅21​)​

$$\Rightarrow x+y+z=\frac{\frac{x}{2}\cdot (\frac{-1}{2}-\mathrm{cos}2\theta )-x{\mathrm{sin}}^{2}\theta }{\mathrm{sin}(\theta +\frac{2\pi }{3})\mathrm{sin}(\theta +\frac{4\pi }{3})}$$⇒x+y+z=sin(θ+32π​)sin(θ+34π​)2x​⋅(2−1​−cos2θ)−xsin2θ​

$$\Rightarrow x+y+z=\frac{-x-2x\mathrm{cos}2\theta -4x{\mathrm{sin}}^{2}\theta }{4\mathrm{sin}(\theta +\frac{2\pi }{3})\mathrm{sin}(\theta +\frac{4\pi }{3})}$$⇒x+y+z=4sin(θ+32π​)sin(θ+34π​)−x−2xcos2θ−4xsin2θ​

$$\Rightarrow x+y+z=\frac{-x-2x(1-2{\mathrm{sin}}^{2}\theta )-4x{\mathrm{sin}}^{2}\theta }{4\mathrm{sin}(\theta +\frac{2\pi }{3})\mathrm{sin}(\theta +\frac{4\pi }{3})}$$⇒x+y+z=4sin(θ+32π​)sin(θ+34π​)−x−2x(1−2sin2θ)−4xsin2θ​

$$\Rightarrow x+y+z=\frac{-3x+4x{\mathrm{sin}}^{2}\theta -4x{\mathrm{sin}}^{2}\theta }{4\mathrm{sin}(\theta +\frac{2\pi }{3})\mathrm{sin}(\theta +\frac{4\pi }{3})}$$⇒x+y+z=4sin(θ+32π​)sin(θ+34π​)−3x+4xsin2θ−4xsin2θ​

$$\Rightarrow x+y+z=\frac{-3x}{4\mathrm{sin}(\theta +\frac{2\pi }{3})\mathrm{sin}(\theta +\frac{4\pi }{3})}\neq 0$$⇒x+y+z=4sin(θ+32π​)sin(θ+34π​)−3x​=0

$(I)$(I) Trace $(R)=x+y+z\neq 0$(R)=x+y+z=0

$\Rightarrow$⇒ Statement (i) is False

Now, finding $Adj(Adj(R))$Adj(Adj(R))

Now, using the property $Adj(Adj(R))=R|R|$Adj(Adj(R))=R∣R∣ we get,

$$\mathrm{Trace}(\mathrm{Adj}(\mathrm{Adj}(R)))=xyz(x+y+z)\neq 0$$Trace(Adj(Adj(R)))=xyz(x+y+z)=0

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

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

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