Macroeconomics Exam Review 181

# Macroeconomics Exam Review 181 - z = ( ?, 1 )( x , ) = ? x...

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3.240 We have already shown (Exercise 3.239) that the alternatives I and II are mutu- ally incompatible. If Gordan’s system II ? ? y = 0 has a semipositive solution y 0 , then we can normalize y such that 1y =1andthe system ? ? y = 0 1y =1 has a nonnegative solution. Conversely, if Gordan’s system II has no solution , the system ± y = c where ± = ( ? ? 1 ) and c =( 0 , 1) = (0 , 0 ,..., 0 , 1), 0 ∈ℜ ± ,i sthe( ² + 1)st unit vector has no solution y 0. By the Farkas lemma, there exists z ∈ℜ ² +1 such that ± z 0 cz < 0 Decompose z into z =( x ) with x ∈ℜ ² . The second inequality implies that ³< 0 since cz =( 0 , 1) ( x )= ³< 0 Since ± =( ?, 1
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Unformatted text preview: z = ( ?, 1 )( x , ) = ? x + 1 or ? x 1 &gt; x solves Gordans system I. 3.241 Let a 1 , a 2 ,..., a be a basis for . Let ? = ( a 1 , a 2 ,..., a ) be the matrix whose columns are a . To say that contains no positive vector means that the system ? x &gt; has no solution. By Gordans theorem, there exists some y such that ? ? y = that is a y = ya = 0 , = 1 , 2 ,..., so that y . 190...
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## This note was uploaded on 01/16/2012 for the course ECO 2024 taught by Professor Dr.dumond during the Fall '10 term at FSU.

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