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

The matrix $A$ is defined where each element follows:

• Diagonal elements (when $i = j$): $a_{ij} = (-1)^{i+j} - 1$
• Non-diagonal elements (when $i \neq j$): $a_{ij} = (-1)^{i+j}$

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For Row 1:

$a_{11} = (-1)^{1+1} - 1$

$= (-1)^2 - 1$

$= 1 - 1$

$= 0$

$a_{12} = (-1)^{1+2} = (-1)^3 = -1$

$a_{13} = (-1)^{1+3} = (-1)^4 = 1$

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For Row 2:

$a_{21} = (-1)^{2+1} = (-1)^3 = -1$

$a_{22} = (-1)^{2+2} - 1$

$= (-1)^4 - 1$

$= 1 - 1$

$= 0$

$a_{23} = (-1)^{2+3} = (-1)^5 = -1$

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For Row 3:

$a_{31} = (-1)^{3+1} = (-1)^4 = 1$

$a_{32} = (-1)^{3+2} = (-1)^5 = -1$

$a_{33} = (-1)^{3+3} - 1$

$= (-1)^6 - 1$

$= 1 - 1$

$= 0$

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$A = \begin{pmatrix} 0 & -1 & 1 \\ -1 & 0 & -1 \\ 1 & -1 & 0 \end{pmatrix}$

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The transpose $A^T$ is obtained by flipping rows and columns:

$A^T = \begin{pmatrix} 0 & -1 & 1 \\ -1 & 0 & -1 \\ 1 & -1 & 0 \end{pmatrix}$

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Comparing $A$ and $A^T$, the matrices are equal: $A = A^T$

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$A + A^T = A + A$

$= 2A$

Therefore, the value of $A + A^T$ is $2A$.