This preview shows page 1. Sign up to view the full content.
10. (Statement of the problem.)
Proof.
Since
AB
=
I
, det(
AB
) = det(
I
).
Since det(
I
) = 1, by Theorem 2.2.3, det(
A
)det(
B
) = 1.
It follows that det(
A
)
6
= 0. Hence,
A
is nonsingular, and
A

1
exists.
By
AB
=
I
, we get
A

1
(
AB
)
A
=
A

1
(
I
)
A
. That is, (
A

1
A
)(
BA
) =
A

1
A
.
Now we get
BA
=
I
.
In the deﬁnition of a nonsigular matrix, we say that a matrix
This is the end of the preview. Sign up
to
access the rest of the document.
Unformatted text preview: A is nonsingular if there is a matrix B such that AB = I and BA = I. Now since AB = I implies BA = I , we only need one of the equations in the deﬁnition, and the other follows automotically. / 1...
View
Full
Document
This note was uploaded on 04/01/2008 for the course MATH 309 taught by Professor Samiabdul during the Spring '08 term at Michigan State University.
 Spring '08
 samiabdul
 Linear Algebra, Algebra

Click to edit the document details