Question:

Step 1: Interpret the given condition to identify the type of matrix. The problem states that for an odd order determinant, $a_{ij} + a_{ji} = 0$ for all $i, j$. This condition can be rewritten as $a_{ji} = -a_{ij}$. If $A$ is the matrix associated with this determinant, then the element at row $j$ and column $i$ is the negative of the element at row $i$ and column $j$. This is the definition of a skew-symmetric matrix, i.e., $A^T = -A$. Step 2: Recall fundamental properties of determinants. For any square matrix $A$, its determinant is equal to the determinant of its transpose. $$ \det(A) = \det(A^T) $$ Also, for a scalar $k$ and an $n \times n$ matrix $A$, the determinant of $kA$ is $k^n$ times the determinant of $A$. $$ \det(kA) = k^n \det(A) $$ where $n$ is the order of the matrix. Step 3: Apply the determinant properties to the skew-symmetric matrix. Since $A$ is a skew-symmetric matrix, we have $A^T = -A$. Using the property from Step 2, we can write: $$ \det(A) = \det(A^T) $$ Substitute $A^T = -A$ into the equation: $$ \det(A) = \det(-A) $$ Now, using the property for scalar multiplication with $k = -1$: $$ \det(-A) = (-1)^n \det(A) $$ where $n$ is the order of the matrix. Step 4: Use the information about the odd order of the determinant. The problem specifies that the determinant is of odd order. This means $n$ is an odd integer. For any odd integer $n$, $(-1)^n$ evaluates to $-1$. Substituting this into the equation from Step 3: $$ \det(A) = (-1)^n \det(A) $$ $$ \det(A) = -1 \cdot \det(A) $$ $$ \det(A) = -\det(A) $$ Step 5: Solve the resulting equation for the determinant. From the previous step, we have: $$ \det(A) = -\det(A) $$ To find the value of $\det(A)$, we rearrange the terms: $$ \det(A) + \det(A) = 0 $$ $$ 2 \det(A) = 0 $$ Dividing by 2, we get: $$ \det(A) = 0 $$ Step 6: State the final answer. The value of the odd order determinant, where $a_{ij} + a_{ji} = 0$, is $0$. The final answer is $\boxed{\text{0}}$.