A is nonsingular. Conversely, let A be nonsingular. Then A ≠ Now A ( adj A) = ( adj A) A = A I (Theorem ) A A I | A | | A | adj adj AB = BA = I, where B = | A | adj Thus A is invertible and A – = | A | adj Example If A = , then verify that A adj A = |A| I. Also find A – .
Solution We have A = ( – ) – ( – ) + ( – ) = ≠ Now A = , A = – , A = – , A = – , A = ,A = , A = – , A = , A = Therefore adj A = DETERMINANTS Now A ( adj A) = −+ −+ −+ −+ −+ −+ = ( ) = A . I Also A – a d j Example If A = and B , then verify that (AB) – = B – A – . Solution We have AB = Since, AB = – ≠ , (AB) – exists and is given by (AB) – = (AB) AB adj =− Further, A = – ≠ and B = ≠ . Therefore, A – and B – both exist and are given by A – = − = ,B Therefore B A − = − = − Hence (AB) – = B – A – Example Show that the matrix A = satisfies the equation A – 4A + I = O, where I is × identity matrix and O is × zero matrix.
Using this equation, find A – . Solution We have A.A Hence 4A I + = O Now A – 4A + I = O Therefore A