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).
