# ch3 - Solutions of Selected Problems in Chapter Test B 3....

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Solutions of Selected Problems in Chapter Test B 3. (a) Note that N ( A ) = { x R 5 | A x = 0 } . We can derive RREF of the corresponding augmented matrix ( A | 0 ) as 1 3 0 2 3 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 . For x = ( x 1 , x 2 , x 3 , x 4 , x 5 ) T , we have a general solution as x 2 = α, x 4 = β, x 5 = γ, x 1 = - 3 α - 2 β - 3 γ, x 3 = - β - γ. Now observe x 1 x 2 x 3 x 4 x 5 = α - 3 1 0 0 0 + β - 2 0 - 1 1 0 + γ - 3 0 - 1 0 1 . This means N ( A ) is spanned by { ( - 3 , 1 , 0 , 0 , 0) T , ( - 2 , 0 , - 1 , 1 , 0) T , ( - 3 , 0 , - 1 , 0 , 1) T } . Since these vectors are also linearly independent, they form a basis and nullity( A )=dim( N ( A )) = 3 . (b)

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## This note was uploaded on 07/17/2008 for the course MATH 571 taught by Professor Kim during the Spring '08 term at Ohio State.

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ch3 - Solutions of Selected Problems in Chapter Test B 3....

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