Question

Let P be a skew-symmetric matrix of order 3. If \( \det(P) = \alpha \), then \( (2025)^\alpha \) is

• A. \( 0 \)
• B. \( 1 \)
• C. \( 2025 \)
• D. \( (2025)^3 \)

Hint:

A crucial property to remember is that the determinant of any skew-symmetric matrix of odd order is always zero. This can be proven using the properties of determinants: \( \det(A^T) = \det \) and \( \det(cA) = c^n \det \). For a skew-symmetric matrix \( P \) of odd order \( n \), \( \det(P) = \det(P^T) = \det(-P) = (-1)^n \det(P) = -\det(P) \), which implies \( 2 \det(P) = 0 \), so \( \det(P) = 0 \).

Answer 1

The Correct Option is B

A skew-symmetric matrix \( P \) satisfies the condition \( P^T = -P \). Taking the determinant of both sides, we get: $$\det(P^T) = \det(-P)$$ We know that \( \det(P^T) = \det(P) \). Also, for a matrix of order \( n \), \( \det(kP) = k^n \det(P) \). Here, the order of matrix \( P \) is \( n = 3 \) and \( k = -1 \). Therefore, $$\det(-P) = (-1)^3 \det(P) = -\det(P)$$ So, we have: $$\det(P) = -\det(P)$$ $$2 \det(P) = 0$$ $$\det(P) = 0$$ Given that \( \det(P) = \alpha \), we have \( \alpha = 0 \). Now we need to find the value of \( (2025)^\alpha \): $$(2025)^\alpha = (2025)^0$$ Since any non-zero number raised to the power of 0 is 1, we have: $$(2025)^0 = 1$$ Thus, \( (2025)^\alpha = 1 \).