Let A, B, C be matrices such that A is symmetric and B and C are skew-symmetric.
Consider the statements
(S1) is symmetric
(S2) is symmetric
Then,
Consider the statements
(S1) is symmetric
(S2) is symmetric
Then,
- A.Only S2 is true
- B.Both S1 and S2 are false
- C.Both S1 and S2 are true
- D.Only S1 is true
Show answer and solution
Correct option: A
Solution
For any matrix M, if M is symmetric (Mᵀ = M) or skew-symmetric (Mᵀ = -M), then (Mⁿ)ᵀ = (Mᵀ)ⁿ.
Key facts:
- A symmetric: Aᵀ = A, so (An)ᵀ = Aⁿ (always symmetric for any n)
- B skew-symmetric: Bᵀ = -B, so (Bn)ᵀ = (-1)ⁿ Bⁿ. For B²⁶: (B²⁶)ᵀ = B²⁶ (symmetric, even power)
- C skew-symmetric: (C¹³)ᵀ = (-1)¹³ C¹³ = -C¹³ (skew-symmetric, odd power)
For S1: Let P = A¹³B²⁶ - B²⁶A¹³. Take transpose:
Pᵀ = (B²⁶)ᵀ(A¹³)ᵀ - (A¹³)ᵀ(B²⁶)ᵀ = B²⁶·A¹³ - A¹³·B²⁶ = -P.
So P is skew-symmetric, not symmetric. S1 is false.
For S2: Let Q = A²⁶C¹³ - C¹³A²⁶. Take transpose
Key facts:
- A symmetric: Aᵀ = A, so (An)ᵀ = Aⁿ (always symmetric for any n)
- B skew-symmetric: Bᵀ = -B, so (Bn)ᵀ = (-1)ⁿ Bⁿ. For B²⁶: (B²⁶)ᵀ = B²⁶ (symmetric, even power)
- C skew-symmetric: (C¹³)ᵀ = (-1)¹³ C¹³ = -C¹³ (skew-symmetric, odd power)
For S1: Let P = A¹³B²⁶ - B²⁶A¹³. Take transpose:
Pᵀ = (B²⁶)ᵀ(A¹³)ᵀ - (A¹³)ᵀ(B²⁶)ᵀ = B²⁶·A¹³ - A¹³·B²⁶ = -P.
So P is skew-symmetric, not symmetric. S1 is false.
For S2: Let Q = A²⁶C¹³ - C¹³A²⁶. Take transpose
[JEE Main 2023 · 25 Jan · Shift 2]