For a determinant Δ of order 3, the element a<sub>ij</sub> is defined as a<sub>ij</sub> = tan<sup>-1</sup>(tan(i - j)) ∀ i, j, then the value of Δ is equal to (where 'i' represents row and 'j' represents column)
Step-by-Step Solution
Key Concept: The determinant of a matrix where rows or columns are linearly dependent is zero. Here, the matrix is skew-symmetric because a_ij = tan⁻^1(tan(i-j)) = -tan⁻^1(tan(j-i)) = -a_ji, and for a 3x3 skew-symmetric matrix, the determinant is always 0.
The matrix A = [a<sub>ij</sub>] is a 3x3 matrix where a<sub>ij</sub> = tan<sup>-1</sup>(tan(i-j)). Since tan<sup>-1</sup>(tan(\theta)) = \theta for \theta \in (-\pi/2, \pi/2), and here i, j \in {1, 2, 3}, the values of (i-j) are in {-2, -1, 0, 1, 2}. Since -\pi/2 \approx -1.57 and \pi/2 \approx 1.57, the values \pm2 fall outside the principal range. However, the property a<sub>ij</sub> = -a<sub>ji</sub> holds because tan<sup>-1</sup>(tan(i-j)) = -tan<sup>-1</sup>(tan(j-i)). Thus, the matrix is skew-symmetric. The determinant of any odd-order skew-symmetric matrix is 0.
Correct Answer: 0