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
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Topics for Today • LU (Fnish) • 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
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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
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Pivoting • Let's solve the matrix problem at right by Gauss elimination. • First, we'll subtract the frst row ±rom 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
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Pivoting • Fortunately, the
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This note was uploaded on 09/16/2008 for the course BMEN 3001 taught by Professor Victorbarocas during the Spring '08 term at Minnesota.

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Biomechanics_3 - Lecture 4 BMEn 3001 Biomechanics 10...

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