hw5soln - ECE 130C HW5 Solution May 26, 2008 1. (a) The...

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ECE 130C HW5 Solution May 26, 2008 1. (a) The incidence matrix is: A = - 1 1 0 0 - 1 0 1 0 0 - 1 1 0 0 - 1 0 1 - 1 0 0 1 0 0 - 1 1 Entry ( i,j ) in the incidence matrix relates edge i with node j of the graph. ( 1 0 0 1 - 1 0 ) , ( 0 0 1 - 1 0 1 ) , ( 0 1 0 0 - 1 1 ) are the vectors that span N ( A T ).These come from sending the currents around each of the loops. (b) C ( A ) : r = 3, the first 3 columns are a basis. N ( A ) : (1 , 1 , 1 , 1) is a basis. C ( A T ) : r = 3, rows 1,2,4 are a basis. N ( A T ) : the three vectors in part(a) are a basis. 2. z = ( x + y ) / 2 Az = ( Ax + Ay ) / 2 3. (2 + 3 t )1 = (2 + 3 t ), (2 + 3 t ) t = 2 t + 3 t 2 ,(2 + 3 t ) t 2 = 2 t 2 + 3 t 3 , (2 + 3 t ) t 3 = 2 t 3 + 3 t 4 T = 2 0 0 0 3 2 0 0 0 3 2 0 0 0 3 2 0 0 0 3 4. Let p ( x ) ,q ( x ) ± S Z 1 0 ( cp ( x ) + dq ( x )) dx = c Z 1 0 p ( x ) dx + d Z 1 0 q ( x ) dx = 0.
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hw5soln - ECE 130C HW5 Solution May 26, 2008 1. (a) The...

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