If \(A=\begin{bmatrix}0&1&-3\\-1&0&5\\3&-5&0\end{bmatrix}\) then the value of \(|A^{2025}|\) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 0
The diagonal of A is zero and \(a_{ij} = -a_{ji}\) for every other entry, so
\(\displaystyle A' = \begin{bmatrix}0&-1&3\\1&0&-5\\-3&5&0\end{bmatrix} = -A\)
So A is a skew-symmetric matrix of odd order (3), and such a matrix has determinant 0:
\(|A| = |A'| = |-A| = (-1)^3|A| = -|A|\)
\(2|A| = 0,\quad |A| = 0\)
Directly, expanding along the first row:
\(|A| = 0(0 + 25) - 1(0 - 15) + (-3)(5 - 0) = 15 - 15 = 0\)
Since \(|A^n| = |A|^n\),
\(|A^{2025}| = |A|^{2025} = 0\)