Mathematical Methods for Physics and Engineering - Matematica.NET

MATRICES AND VECTOR SPACESand may be straightforwardly derived.(i) (A −1 ) −1 = A.(ii) (A T ) −1 =(A −1 ) T .(iii) (A † ) −1 =(A −1 ) † .(iv) (AB) −1 = B −1 A −1 .(v) (AB ···G) −1 = G −1 ···B −1 A −1 .◮Prove the properties (i)–(v) stated above.We begin by writing down the fundamental expression defining the inverse of a nonsingularsquare matrix A:AA −1 = I = A −1 A. (8.61)Property (i). This follows immediately from the expression (8.61).Property (ii). Taking the transpose of each expression in (8.61) gives(AA −1 ) T = I T =(A −1 A) T .Using the result (8.39) for the transpose of a product of matrices and noting that I T = I,we find(A −1 ) T A T = I = A T (A −1 ) T .However, from (8.61), this implies (A −1 ) T =(A T ) −1 and hence proves result (ii) above.Property (iii). This may be proved in an analogous way to property (ii), by replacing thetransposes in (ii) by Hermitian conjugates and using the result (8.40) for the Hermitianconjugate of a product of matrices.Property (iv). Using (8.61), we may write(AB)(AB) −1 = I = (AB) −1 (AB),From the left-hand equality it follows, by multiplying on the left by A −1 ,thatA −1 AB(AB) −1 = A −1 I and hence B(AB) −1 = A −1 .Now multiplying on the left by B −1 givesB −1 B(AB) −1 = B −1 A −1 ,and hence the stated result.Property (v). Finally, result (iv) may extended to case (v) in a straightforward manner.For example, using result (iv) twice we find(ABC) −1 = (BC) −1 A −1 = C −1 B −1 A −1 . ◭We conclude this section by noting that the determinant |A −1 | of the inversematrix can be expressed very simply in terms of the determinant |A| of the matrixitself. Again we start with the fundamental expression (8.61). Then, using theproperty (8.52) for the determinant of a product, we find|AA −1 | = |A||A −1 | = |I|.It is straightforward to show by Laplace expansion that |I| = 1, and so we arriveat the useful result|A −1 | = 1|A| . (8.62)266