Let R=((∣x000y000z))∣ be a non-zero 3×3 matrix, where xsinθ=ysin(θ+32π)=zsin(θ+34π)=0,θ∈(0,2π). For a square matrix M, let trace (M) denote the sum of all the diagonal entries of M. Then, among the statements:
(I) Trace (R)=0
(II) If trace (adj(adj(R))=0, then R has exactly one non-zero entry.
Solutions · 1
Official solution
\[xsinθ=ysin(θ+32π)=zsin(θ+34π)=0⇒x,y,z=0\]
Also,
\[sinθ+sin(θ+32π)+sin(θ+34π)=0∀θ∈R⇒x1+y1+z1=0⇒xy+yz+zx=0\]
(i) \[\quad \operatorname{Trace}(\mathrm{R})=\mathrm{x}+\mathrm{y}+\mathrm{z}\]
If x+y+z=0 and xy+yz+zx=0
\[\Rightarrow \mathrm{x}=\mathrm{y}=\mathrm{z}=0\]
Statement (i) is False
(ii) \[\quad \operatorname{Adj}(\operatorname{Adj}(\mathrm{R}))=|\mathrm{R}| \mathrm{R}\]
Trace \[(\operatorname{Adj}(\operatorname{Adj}(\mathrm{R})))\]
\[=x y z(x+y+z) \neq 0\]
Statement (ii) is also False