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MIT18_024S11_soln-pset3

MIT18_024S11_soln-pset3 - Solutions for PSet 3 1 First we...

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Solutions for PSet 3 1. First we note that T ( P ) is again a parallelepiped. To see this notice that if x T ( P ) then x = T ( i n =1 c i v i ) where 0 c i 1. By linearity x = i c i T ( v i ) for 0 c i 1. That is, T ( P ) = { x R n | x = c i T ( v i ) , 0 c i 1 } and this i is precisely the definition of a parallelepiped. Now, we determine the volume. If V denotes the matrix formed such that the i th row is v i , then: V ol ( P ) = | det ( V ) | Let W denote matrix with w i = m ( T ) v i as the i th row, then by the definition of T ( P ) above: V ol ( T ( P )) = | det ( W ) | = | det ( m ( T ) V ) | = | det ( m ( T )) det ( V ) | = | det ( m ( T )) | vol ( P ) . 2. (14.4:23) Since F is continuous, we can apply the first fundamental theorem of calculus to determine a derivative for the expression on the right hand side of the equation. Thus F ( x ) = e x A + xe x A + x 1 F ( x ) x 1 2 1 x F ( t ) dt = 2 e x A + xe x A x > 0 . Now we apply the second fundamental theorem of calculus and the above result to get: x x F ( x ) F (1) = F ( t ) dt = A (2 e t + te t ) dt 1 1 Integrating by parts we have: x x F ( x ) F (1) = A (2 e t + te t e t ) = A ( e t + te t ) = A ( e x + xe x ) 2 Ae 1 1 Substituting F (1) = eA we observe F ( x ) = Ae x + Axe x Ae 3. First the vectors (3 , 2 , 4) (1 , 0 , 0) = (2 , 2 , 4) and (1 , 1 , 1) (1 , 0 , 0) = (0 , 1 , 1) are parallel to the plane P . Thus, the vector (2 , 2 , 4) × (0 , 1
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