JEE Main · 2024 · 30 Jan 2024 · Shift 2 · Matrices

Since $x\sin \theta =y\sin \left(\theta +\frac{2\pi }{3}\right)=z\sin \left(\theta +\frac{4\pi }{3}\right)=\lambda \ne 0$, each of $x,y,z$ and each of the three sines must individually be non-zero (a product equal to a non-zero number cannot have a zero factor). So $x=\frac{\lambda }{\sin \theta }$, $y=\frac{\lambda }{\sin (\theta +2\pi /3)}$, $z=\frac{\lambda }{\sin (\theta +4\pi /3)}$, all non-zero.

Writing $a=\sin \theta ,b=\sin (\theta +2\pi /3),c=\sin (\theta +4\pi /3)$, the identity $a+b+c=0$ holds for every $\theta$. Then
$x+y+z=\lambda \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\lambda \cdot \frac{ab+bc+ca}{abc}$.

Since $a+b+c=0$, $ab+bc+ca=\frac{(a+b+c{)}^2-(a^2+b^2+c^2)}{2}=-\frac{a^2+b^2+c^2}{2}$.

As $a,b,c$ are all non-zero, $a^2+b^2+c^2>0$, and $abc\ne 0$, so this ratio is never zero — hence $x+y+z\ne 0$ for any valid $\theta$. Trace$(R)=x+y+z\ne 0$, so statement (I) is false.

For a $3\times 3$ matrix, $\mathrm{adj}(\mathrm{adj}(R))=|R|R$, so
Trace$(\mathrm{adj}(\mathrm{adj}(R)))=xyz(x+y+z)$.

Since $x,y,z$ are always non-zero and $x+y+z\ne 0$ always, this trace is never zero — the hypothesis "trace$(\mathrm{adj}(\mathrm{adj}(R)))=0$" is never actually satisfied. A conditional statement with a false hypothesis is vacuously true, so statement (II) is true regardless of what it claims about $R$ having exactly one non-zero entry.

Only statement (II) is true.