Matrices & Determinants
Diagonal Matrix — Trace and Adjoint Conditions
nta_pyq_2024_jan
Grade 12
Question:
Let $R=\begin{pmatrix}x&0&0\\0&y&0\\0&0&z\end{pmatrix}$ be a non-zero $3\times3$ matrix, where $x\sin\theta=y\sin\!\left(\theta+\dfrac{2\pi}{3}\right)=z\sin\!\left(\theta+\dfrac{4\pi}{3}\right)\neq0$, $\theta\in(0,2\pi)$. For a square matrix $M$, let trace$(M)$ denote the sum of all diagonal entries of $M$. Then, among the statements: (I) Trace$(R)=0$ (II) If trace$(\text{adj}(\text{adj}(R)))=0$, then $R$ has exactly one non-zero entry.
Both (I) and (II) are true
Neither (I) nor (II) is true
Only (II) is true
Only (I) is true
Step-by-Step Solution
Key Concept: From the common value condition: $\sin\theta+\sin(\theta+2\pi/3)+\sin(\theta+4\pi/3)=0$ for all $\theta$. So $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Rightarrow xy+yz+zx=0$. This does NOT mean $x+y+z=0$. Statement (I) claims trace$(R)=x+y+z=0$, which is not necessarily true. Statement (II): trace(adj(adj(R)))$=xyz(x+y+z)$, and this $=0$ doesn't imply exactly one non-zero entry.
Neither statement is true. Answer: (2).
Correct Answer: 2