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Biomechanics_3

# Biomechanics_3 - Lecture 4 BMEn 3001 Biomechanics 10...

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Lecture 4 BMEn 3001 Biomechanics 10 September 2008

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Last Lecture Linear Algebraic Systems LU Decomposition • Pivoting
Topics for Today LU (finish) • Pivoting Matrix Inversion The Eigenproblem Condition Number

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An LU Decomposition Note that we can get back the original matrix by multiplying the lower and upper parts. Note also that A = 1 0 0 2 1 0 4 0 1 1 1 1 0 2 4 0 0 2 = 1 1 1 2 0 2 4 4 6 A L + U
Use LU to solve... 1 1 1 2 0 2 4 4 6 x 1 x 2 x 3 = 12 2 62

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Storage The L and U matrices only overlap on the diagonal, and L has one's on the diagonal, so we really don't need to store them. In fact, if we don't need to get the original matrix back, we can store L and U over it. U 11 U 12 U 13 L U 1 N L 21 U 22 U 23 L U 2 N L 31 L 32 U 33 L U 3 N M M M O M L N 1 L N 2 L N 3 L U NN
Pivoting Let's solve the matrix problem at right by Gauss elimination. First, we'll subtract the first row from the second row, right? 1 2 0 1 2 3 0 1 1 x y z = 7 10 3

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Pivoting All looks good, but our next step should be to divide the second row by its diagonal element, which is zero! 1 2 0 0 0 3 0 1 1 x y z = 7 3 3 The situation looks dire, but the determinant of the original matrix is not zero, so there is hope. (1•2•1+2•3•0+0•1•1-0•2•0-2•1•1-1•3•1 = -3) 1 2 0 1 2 3 0 1 1 x y z = 7 10 3
Pivoting Fortunately, the equations don't care

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