UPSC NDA I and II 2020 Math Solved Paper

Concept:A Hermitian matrix is a square matrix that equals its own conjugate transpose. Its diagonal elements are real.Explanation:Given matrix $A = \begin{bmatrix} 1-i & i \\ -i & 1-i \end{bmatrix}$ where $i = \sqrt{-1}$.First, compute the conjugate of $A$, denoted $\overline{A}$: $\overline{A} = \begin{bmatrix} 1+i & -i \\ i & 1+i \end{bmatrix}$.Next, take the transpose of $\overline{A}$ to get the conjugate transpose: $(\overline{A})^{T} = \begin{bmatrix} 1+i & i \\ -i & 1+i \end{bmatrix}$.Now add $(\overline{A})^{T}$ and $A$: $(\overline{A})^{T} + A = \begin{bmatrix} 1+i & i \\ -i & 1+i \end{bmatrix} + \begin{bmatrix} 1-i & i \\ -i & 1-i \end{bmatrix} = \begin{bmatrix} 2 & 2i \\ -2i & 2 \end{bmatrix}$.The resulting matrix has real diagonal entries $2$ and $2$, and the off-diagonal elements are complex conjugates: $2i$ and $-2i$. This satisfies the condition for a Hermitian matrix.Answer:Thus, $(\overline{A})^{T} + A$ is Hermitian. The correct option is C.