Match of the column.
Column –I | Column –II |
(i) A is a real skew symmetric Matrix | [A] BA – AB such that A2 + I = 0. Then |
(ii) A is a matrix such that A2 = A. | [B]A is of If (I +A)n = I + λA, then λ equals even order |
(iii) If for a matrix A, A2 = A, and B = I – A, then AB + BA + I – (I – A)2 equals | [C] A |
(iv) A is a matrix with complex entries and A* stands for transpose of complex conjugate of A. If A*= A & B* = B, then (AB – BA)* equals | [D] 2n – 1 |
Text Solution
Verified by Experts(i) [B]; (ii) [D]; (iii) [C]; (iv) [A]
Ans.
(i) [B]
(ii) [D]
(iii) [C]
(iv) [A]
Sol.
(i) A2 = – I ∴ A is of even order
(ii) (I + A)n = C0 In + C1 IA + C2 IA2 + ....... + Cn IAn
= C0 I + C1 A + C2A + ..... + can
= I + (2n – 1) A
∴ λ = 2n – 1
(iii) 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
(iv) A* = A, B* = B
(AB – BA)* = B* A* – A* B* = BA – AB
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