Macroeconomics Exam Review 139

Macroeconomics Exam Review 139 - Solutions for Foundations...

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That is, every element of 𝐵 is nonnegative. Consequently since x + 𝛼 z 0 𝐵 ( x + 𝛼 z ) 0 (3.45) On the other hand, substituting 𝐴 = 𝐼 𝐵 in (3.45) ( 𝐼 𝐵 )( x + 𝛼 z ) > 0 x + 𝛼 z > 𝐵 ( x + 𝛼 z ) which implies that the first component of 𝐵 ( x + 𝛼 z ) is negative, contradicting (3.45). This contradiction establishes that z 0 . Suppose 𝐴 x = 0 . A fortiori 𝐴 x 0 . By the previous part this implies x 0 . On the other hand, it also implies that 𝐴 x = 𝐴 ( x ) = 0 so that x 0 . We conclude that x = 0 is the only solution to 𝐴 x = 0 . 𝐴 is nonsingular. Since 𝐴 is nonsingular, the system 𝐴 x = y has a unique solution x for any y 0 . By the first part, x 0 . 3.114 Suppose 𝐴 is productive. By the previous exercise, 𝐴 is nonsingular with inverse 𝐴 1 . Let e 𝑖 be the 𝑖 th unit vector. Since e 𝑖 0 , there exists x 𝑖 0 such that 𝐴 x 𝑖 = e 𝑖 Multiplying by 𝐴 1 x 𝑖 = 𝐴 1 𝐴 x 𝑖 = 𝐴 1 e 𝑖 = 𝐴 1 𝑖 where 𝐴 1 𝑖 is the 𝑖 column of 𝐴 1 . Since x 𝑖 0 for every 𝑖 , we conclude that 𝐴 1 0 . Conversely, assume that 𝐴 1 0 . Let 1 = (1 , 1 , . . . , 1) denote a net output of 1 for
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