If \(A = [a_{ij}]_{3\times 3}\), where \(a_{ij} = \begin{cases}(-1)^{i+j}-1, & i=j\\ (-1)^{i+j}, & i \ne j\end{cases}\) then the value of \(A + A^T\) is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(2A\)
On the diagonal, i = j, so i + j = 2i is even:
\(a_{ii} = (-1)^{2i} - 1 = 1 - 1 = 0\)
Off the diagonal, the entry depends only on i + j, which does not change when i and j are swapped:
\(a_{ij} = (-1)^{i+j} = (-1)^{j+i} = a_{ji}\)
\( \displaystyle A = \begin{bmatrix}0 & -1 & 1\\ -1 & 0 & -1\\ 1 & -1 & 0\end{bmatrix} \)
So A is a symmetric matrix, \(A^T = A\), and
\(A + A^T = A + A = 2A\)