Take a matrix:
$\begin{bmatrix} a & b\\ b& a \end{bmatrix}$
Sum of each row$= a+b$ [ matrix is invertible hence determinant is not $0$ i.e. $a$ and $b$ are different]
Now lets prove:
$2\times \begin{bmatrix} a & b\\ b& a \end{bmatrix} = \begin{bmatrix} 2a & 2b\\ 2b& 2a \end{bmatrix}:$ Sum of each row $= 2(a+b)$
$\begin{bmatrix} a & b\\ b& a \end{bmatrix}\times \begin{bmatrix} a & b\\ b& a \end{bmatrix} = \begin{bmatrix} a^{2} +b^{2} & ab +ba\\ ab +ba & a^{2} +b^{2} \end{bmatrix}:$ Sum of each rows = $(a^2+b^2+2ab)= (a+b)^2$
$\frac{1}{a^{2}-b^{2}}\begin{bmatrix} a & -b\\ -b& a \end{bmatrix}= \frac{(a-b)}{a^{2}-b^{2}} = \frac{(a-b)}{(a+b) (a-b)}= \frac{1}{a+b}$
Hence, all are true
Correct Answer: $E$