Explanation

(b)
Let, $A = \frac{1}{\sqrt{3}}\begin{bmatrix}
1 & 1 + i \\
1 - i & 1
\end{bmatrix}$
$\therefore A^{T} = \frac{1}{\sqrt{3}}\begin{bmatrix}
1 & 1 - i \\
1 - i & 1
\end{bmatrix}$
$\Rightarrow ({\overline{A}}^{T})\ = \frac{1}{\sqrt{3}}\begin{bmatrix}
1 & 1 + i \\
1 - i & - 1
\end{bmatrix}$
$\therefore A\left( A^{- T} \right) = \frac{1}{\sqrt{3}}\begin{bmatrix}
1 & 1 + i \\
1 - i & - 1
\end{bmatrix} \times \frac{1}{\sqrt{3}}\begin{bmatrix}
1 & 1 + i \\
1 - i & - 1
\end{bmatrix}$
$= \frac{1}{3}\begin{bmatrix}
3 & 0 \\
0 & 3
\end{bmatrix} = \begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix} = I$