(A) A is a real skew-symmetric matrix such that A^2 + I = 0. Then

Question

Match the following:

Column I	Column II
(A) A is a real skew-symmetric matrix such that A2 + I = 0. Then	(p) BA - AB
(B) A is a matrix such that A2 =A. If (I +A) n = I + λA, then λ equals (n ∈ N)	(q) A is of even order
(C) If for a matrix A, A2 = A, and B = I - A, then AB + BA + I - (I - A)2 equals	(r) A
(D) A is a matrix with complex entries and A* stands for transpose of complex conjugate of A. If A* = A and B* = B, then (AB - BA)* equals	(s) 2n - 1
	(t) nC1 + nC2 ... + nCn

Answer 1

Answer is (A) → (q), (B) → (s, t), (C) → (r), (D) → (p)

(A) A² = -I, therefore A is of even order

(B) (I + A)ⁿ = C₀Iⁿ + C₁ IA + C₂IA² + … + C_nIAⁿ

= C_oI + C₁A + C₂A + … + C_nA = I + (2ⁿ - 1)A

Therefore, λ = 2ⁿ - 1

(C) A² = A and B = I - A 

AB + BA + I - (I + A² - 2A) = AB + BA - A + 2A

= AB + BA + A

= A(I - A) + (I - A)A + A = A - A + A - A + A= A

(D) A* = A, B* = B

(AB - BA)* = B* A* - A*B* = BA - AB