If $A = [a_{ij}]_{3×3}$, where $a_{ij} =\left\{\begin{matrix}(-1)^{i+j}-1,& i=j\\(-1)^{i+j},&i≠j\end{matrix}\right.$ then the value of $A + A^T$ is:

2A

The correct answer is Option (3) → 2A

Given $A=[a_{ij}]_{3\times3}$ with

$a_{ij}=\begin{cases}(-1)^{i+j}-1, & i=j\\(-1)^{i+j}, & i\ne j\end{cases}$

For $i\ne j$:

$a_{ij}=(-1)^{i+j}=(-1)^{j+i}=a_{ji}$

For $i=j$:

$a_{ii}=(-1)^{2i}-1=1-1=0$

Hence $a_{ij}=a_{ji}$ for all $i,j$, so $A$ is symmetric.

Therefore

$A^T=A$

$A+A^T=A+A=2A$