Page 248 - Advanced engineering mathematics

228 CHAPTER 7 Matrices and Linear Systems

7. If AB is nonsingular, so are A and B.
8. If A and B are n ×n matrices, and either one is singular, then their products AB and BA
are singular.
9. Every elementary matrix is nonsingular, and its inverse is an elementary matrix of the
same type.
10. An n × n matrix A is nonsingular if and only if AX = B has a solution for every
n × 1 B.
Proof These statements use the uniqueness of the inverse of a matrix. This allows us to show
that a matrix is the inverse of another matrix by showing that it behaves like the inverse (the
product of the two matrices is the identity matrix).
Conclusion (2) of the theorem is true because
−1
−1
−1
−1
−1
(B A )(AB) = B (A A)B = B B = I n .
Similarly
−1
−1
(AB)(B A ) = I n .
−1
This proves that B A −1 behaves like the inverse of AB, hence this must be inverse.
For conclusion (3) observe that the equation
−1 −1
AA = A A = I n
−1
is symmetric in the sense that A −1 is the inverse of A, but also A is the inverse of A . The latter
phrasing means that
−1 −1
A = (A ) .
For conclusion (4), first write
t
t
−1 t
−1 t
I n = (I n ) = (AA ) = (A ) A .
Similarly,
−1 t
t
A (A ) = I n .
−1 t
t −1
These two equations show that (A ) = (A ) .
The key to conclusion (5) lies in recalling (Section 7.1.1) that the columns of AB are A times
the columns of B. Using this, we can attempt to build an inverse for A a column at a time. To
find B so that AB = I n , we must be able to choose the columns of B so that
0
⎛ ⎞
⎜ 0 ⎟
⎛ ⎞
⎜ ⎟
b 1 j
⎜.⎟
.
⎜ . ⎟
b 2 j
⎜ ⎟
⎜ ⎟
column j of AB = A⎜ . ⎟ = column j of I n = 1 ,
⎜
⎟
⎜ ⎟
. ⎜ ⎟
0
⎝ . ⎠ ⎜ ⎟
⎜ ⎟
b nj ⎜ . ⎟
.
⎝ . ⎠
0
having 1 in the jth place and zeros elsewhere.
If now A R = I n , then the system just written for column j of B has a unique solution for
−1
j = 1,··· ,n. These solutions form the columns of B such that AB = I n , yielding A .
(Actually we must show that BA = I n also, but we will not go through these details.)
Conversely, if A is nonsingular, then this system has a unique solution for j = 1,··· ,n
−1
because these solutions are the columns of A . Then A R = I n . This proves conclusion (5).
Conclusion (6) follows directly from (5).
For conclusion (7), suppose AB is nonsingular. Then for some matrix K, (AB)K = I n . Then
A(BK) = I n ,so A is nonsingular. Similarly, B is nonsingular.
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October 14, 2010 14:23 THM/NEIL Page-228 27410_07_ch07_p187-246