UPSC NDA I 2015 Mathematics Paper

Concept:A matrix is skew-Hermitian if its transpose equals the negative of its conjugate transpose, i.e., $A^{T} = -\overline{A}$.Explanation:Let $A = \begin{bmatrix} 0 & -4+i \\ 4+i & 0 \end{bmatrix}$.First, check transpose: $A^{T} = \begin{bmatrix} 0 & 4+i \\ -4+i & 0 \end{bmatrix}$.This is not equal to $A$ or $-A$, so $A$ is not symmetric or skew-symmetric.Now compute the conjugate transpose (Hermitian conjugate): $\overline{A} = \begin{bmatrix} 0 & \overline{-4+i} \\ \overline{4+i} & 0 \end{bmatrix} = \begin{bmatrix} 0 & -4-i \\ 4-i & 0 \end{bmatrix}$.Compare $A^{T}$ and $-\overline{A}$: $-\overline{A} = -\begin{bmatrix} 0 & -4-i \\ 4-i & 0 \end{bmatrix} = \begin{bmatrix} 0 & 4+i \\ -4+i & 0 \end{bmatrix} = A^{T}$.Thus, $A^{T} = -\overline{A}$, satisfying the condition for a skew-Hermitian matrix.Answer:D. Skew-Hermitian