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10. (Statement of the problem.)
Proof.
“=
⇒
” Suppose that
AB
is nonsingular.
By Theorem 2.2.2, det(
AB
)
6
= 0.
By Theorem 2.2.3, det(
A
)det(
B
) = det(
AB
), hence, det(
A
)det(
B
)
6
= 0.
So we have both det(
A
)
6
= 0 and det(
B
)
6
= 0.
By Theorem 2.2.2, both
A
and
B
are nonsingular.
“
⇐
=” Suppose that both
A
and
B
are nonsigular.
By Theorem 2.2.2, both det(
A
)
6
= 0 and det(
B
)
6
= 0.
By Theorem 2.2.3, det(
A
)det(
B
) = det(
AB
), hence, det(
AB
)
6
= 0.
By Theorem 2.2.2,
AB
is nonsingular.
/
(Or you can say:
“
⇐
=” If both
A
and
B
are nonsigular. then by Theorem 1.3.3,
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Unformatted text preview: AB is nonsingular. / When you become good enough at writing proofs, you can use the following way, though I do not recommend you to do it now . Proof. AB is nonsingular. ⇐⇒ det( AB ) 6 = 0. (By Theorem 2.2.2) ⇐⇒ det( A )det( B ) 6 = 0. (By Theorem 2.2.3) ⇐⇒ Both det( A ) 6 = 0 and det( B ) 6 = 0. ⇐⇒ Both A and B are are nonsingular. (By Theorem 2.2.2) / 1...
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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

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