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Assignment 1 solutions

# Assignment 1 solutions - MIT OpenCourseWare...

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MIT OpenCourseWare http://ocw.mit.edu 18.085 Computational Science and Engineering I Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms .

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18.085 - Mathematical Methods for Engineers I Solutions - Problem Set 1 Prof. Gilbert Strang Section 1.1 2) T 3 = 1 1 0 1 2 1 0 1 2 U = 1 1 0 0 1 1 0 0 1 U T = 1 0 1 1 0 1 0 0 1 U T U = 1 0 1 1 0 1 0 0 1 1 1 0 0 1 1 0 0 1 = 1 1 0 1 2 1 0 1 2 = T 3 UU 1 = 1 1 0 0 1 1 0 0 1 1 1 1 0 1 1 0 0 1 = 1 0 0 0 1 0 0 0 1 = I T 3 = U T U T 1 3 = ( U T U ) 1 = ( U 1 )( U 1 ) T = 1 1 1 0 1 1 0 0 1 1 0 0 1 1 0 1 1 1 = 3 2 1 2 2 1 1 1 1 # 5) K 2 = 2 1 1 2 K 1 2 = 1 4 1 2 1 1 2 = 1 3 2 1 1 2 # 1
Given that 3 2 1 1 K 1 3 = 2 4 2 4 1 2 3 4 3 2 1 1 3 6 4 2 2 4 6 3 K 1 = 4 5 1 2 3 4 Since det ( K 2 ) = 3, det ( K 3 ) = 4, det ( K 4 ) = 5 determinant of K 5 = 6 # From MATLAB, det ( K 5 ) = 6 # 5 4 3 2 1 4 8 6 4 2 3 6 9 6 3 2 4 6 8 4 1 2 3 4 5 1 inv ( K 5 ) = 6 # 5 4 3 2 1 4 8 6 4 2 det ( K ) inv ( K ) = 3 6 9 6 3 2 4 6 8 4 1 2 3 4 5 # 2 3 1 2 2 [ 1 2 ] 3 [ 2 4 ] 20) = + 4 5 2 4 4 5 2 4 = + 4 8 6 12 10 20 16 8 = 14 28 # For columns times row, n by n would require n 3 multiplications 22) MATLAB’s code n = 1000 ; e = ones ( n, 1) ; K = spdiags ([ e, 2 e, e ] , 1:1 , n, n ); u = K \ e ; plot ( u ) ; 2

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4 x 10 Section 1.2 0 200 400 600 800 1000 0 2 4 6 8 10 12 14 Ax if x 0 1) u ( x ) = Bx if x 0 0 if x = 0 u �� ( x ) = = δ ( x ) # if x = 0 2 A A 0 B 2 B A n if n 0 U n = = B n if n 0 . . . .
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