Theorem-1 :-  If A be any given square matrix of order n, then A(adj A) = (adj A) A = |A|I , where I is the identity matrix of order n

Solution :-
  Proof :

Theorem-2 :-  A square matrix A is invertible if and only if A is nonsingular matrix.

Solution :-
  Proof : Let A be invertible matrix of order n and I be the identity matrix of order n. 
          Then, there exists a square matrix B of order n such that AB = BA = I 
          Now AB = I. So |AB| =|I| or |A||B| = 1    (since |I|=1, |AB|= |A||B|)
          This gives |A| ≠ 0. Hence A is nonsingular. 
          Conversely, let A be nonsingular. Then |A| ≠ 0
          Now A (adj A) = (adj A) A = |A| I (Theorem 1)
          A[adj(A)/|A|] = [adj(A)/|A|]A = I
          or AB = BA = I, where B = adj(A)/|A|
          Thus A is invertible and A-1 = 1/|A| X adj(A).

Connect with us:

Copyright © 2015-20 by a1classes. All Rights Reserved.