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solution assignment 03 ae602

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Department of Aerospace Engineering AE602 Mathematics for Aerospace Engineers Solution to Assignment 3 1. AB =  C ⇒  ABB −1 = C B −1 (Postmultiplying by  B −1 ) ⇒  A(BB −1 ) = C B −1 (Using associative law of multiplying matrices) ⇒  AI  = C B −1 (Since  BB −1 = I ) ⇒  A =  C B −1 (Since  AI  = A) ⇒  A −1 = (C B −1 ) −1 (Taking incerse of both sides) = (B −1 ) −1 C −1 (Since (AB) −1 = B −1 A −1 ) = B C −1 (Using convolution property of matrix inversion) Similarly, P A = LU ⇒  P AA −1 = LU A −1 ⇒  P I  = P  = LU A −1 ⇒  (LU ) −1 P  = ( LU ) −1 (LU )A −1 = I A −1 ⇒  A −1 = U −1 L −1 P 2.  As  B  is the inverse of  A 2 , (A 2 ) −1 = B ⇒  (AA) −1 = B ⇒  A −1 A −1 = B ⇒  A(A −1 A −1 ) = AB  (Premultiplying by A) ⇒  (AA −1 )A −1 = AB  (Associative law) ⇒  I A −1 = A −1 = AB Thus,  A  is invertible if  A 2 is invertible. 1

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