Mathematical Methods for Physics and Engineering : A Comprehensive Guide

MATRICES AND VECTOR SPACES

and may be straightforwardly derived.

(i) (A−^1 )−^1 =A.

(ii) (AT)−^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 non-
singular square matrixA:

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=IT=(A−^1 A)T.

Using the result (8.39) for the transpose of a product of matrices and noting thatIT=I,
we find
(A−^1 )TAT=I=AT(A−^1 )T.

However, from (8.61), this implies (A−^1 )T=(AT)−^1 and hence proves result (ii) above.
Property(iii). This may be proved in an analogous way to property (ii), by replacing the
transposes in (ii) by Hermitian conjugates and using the result (8.40) for the Hermitian
conjugate 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 byA−^1 ,that

A−^1 AB(AB)−^1 =A−^1 I and hence B(AB)−^1 =A−^1.

Now multiplying on the left byB−^1 gives

B−^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 inverse

matrix can be expressed very simply in terms of the determinant|A|of the matrix

itself. Again we start with the fundamental expression (8.61). Then, using the

property (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 arrive

at the useful result

|A−^1 |=

1

|A|

. (8.62)