If matrix A=[ array cc 1-i & i -i & 1-i array ] where i= -1 , then...

A square $M$ matrix is said to be Hermitian (or selfadjoint) if it is equal to its. Own Hermitian conjugate, i.e. $(\bar{M})^T=M$ Given Matrix A=\left[$\begin{array}{cc}1-i & i \\ -i & 1-i\end{array}$\right](\bar{A})^T=\left[$\begin{array}{cc} 1+i & i \\ -i & 1+i \end{array}$\right]$\begin{aligned} & \text{Now, } A+(\bar{A})^T=\left[\begin{array}{cc}1-i & i \\ -i & 1-i\end{array}\right]+\left[\begin{array}{cc}1+i & i \\ -i & 1+i\end{array}\right] \\ & =\left[\begin{array}{cc}2 & 2 i \\ -2 i & 2\end{array}\right]=2\left[\begin{array}{cc}1 & i \\ -i & 1\end{array}\right]\end{aligned}$$Conjugate transpose of$\left(A+(\bar{A})^T\right)=2\left[$\begin{array}{cc}1 & i \\ -i & 1\end{array}$\right] Hence, $\left(A+(\bar{A})^T\right)$ is hermitian.