C3092214 - AB is nonsingu-lar When you become good enough at writing proofs you can use the following way though I do not recommend you to do it

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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 nonsingu-lar. / 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.

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