The correct choice is (c) 2(A+A’)
To explain: Given that, A=\(\begin{bmatrix}a&b\\c&d\end{bmatrix}\)
⇒A’=\(\begin{bmatrix}a&c\\b&d\end{bmatrix}\)
Let B=A-A’=\(\begin{bmatrix}a&b\\c&d\end{bmatrix}\)–\(\begin{bmatrix}a&c\\b&d\end{bmatrix}\)=\(\begin{bmatrix}a-a&b-c\\c-b&d-d\end{bmatrix}\)=\(\begin{bmatrix}0&b-c\\c-b&0\end{bmatrix}\)
B’=\(\begin{bmatrix}0&c-b\\b-c&0\end{bmatrix}\)=B’
Thus, B=A-A’ is a skew – symmetric.