mase201sp09_Feb10_PRINTABLE

mase201sp09_Feb10_PRINTABLE - Solving Systems of Linear...

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Methods that avoid calculating a matrix inverse explicitly: • Gauss elimination method with pivoting – most efficient for solving [A][x]=[b] for a single [b] – MATLAB built-in function >> x= A\b • Gauss-Jordan method – no back substitution step but takes more operations an Gauss elimination 2/10/2009 1 than Gauss elimination – MATLAB built-in function >> result=rref([A b]) – Implementation of rref makes it slower than \ Calculating a matrix inverse – For N x N matrix [A], equivalent to solving A\b for N [b]’s – MATLAB built-in functions >> invA= inv(A) >> invA= A^(-1) MASE201 Spring 2009 The Eigenvalue Problem In systems of linear equations, we may be interested in solving problems that take the form Find a scalar λ such that [ a ] [u] – λ [u] = [0] where [ a ] is a square matrix This can also be written 2/10/2009 2 ([ a ] – λ [ I ]) [u] = [0] This is called the “eigenvalue problem” When [a] is very small, e.g. 2 x 2 or 3 x 3, then problem is solved using the characteristic equation: det([ a ] – λ [ I ]) = 0 MASE201 Spring 2009 Eigenvalues of a 2 x 2 matrix The solution to det([ a ] – λ [ I ]) = 0 yields two eigenvalues: λ 1 , λ 2 which are the roots of the characteristic equation: ( )( ) ( ) ( ) ( ) 0 ]) det([ 0 0 det 22 11 2 12 21 22 11 22 11 2 12 21 22 11 22 21 12 11 a a a a a a a a a a a a a a a a a = + + = + + = = λ 2/10/2009 3 ( ) ( ) ]) det([ 2 ]) det([ 4 , 2 22 11 22 11 2 1 a a a a a a + ± + = In MATLAB, we could use the roots built-in function to find the roots of the 2 nd order polynomial a = [2 3; 3 8]; roots ([ 1 -(a(1,1)+a(2,2)) det(a)]) %find roots ans = 9.2426 0.7574 MASE201 Spring 2009 Eigenvectors of a 2 x 2 matrix • Once we have the eigenvalues, we substitute them into: ([ a ] – λ i [ I ]) [u] = [0] where i = 1 or 2 ( ) ( ) ( ) ) 1 ( 2 1 11 12 ) 1 ( 1 ) 1 ( 2 1 22 ) 1 ( 1 21 ) 1 ( 2 12 ) 1 ( 1 1 11 ) 1 ( 2 ) 1 ( 1 1 22 21 12 1 11 : row first the From 0 0 or 0 0 u a a u u a u a u a u a u u a a a a = = + + = 2/10/2009 4 ( ) ( ) ( ) ( )( ) ) 1 ( 1 1 22 1 11 21 12 ) 1 ( 1 1 22 21 1 11 12 ) 1 ( 1 ) 1 ( 2 ) 1 ( 1 1 22 21 : ng Substituti : row second the From u a a a a u a a a a u u u a a = = = Since det([ a ] – λ i [ I ]) = ( a 11 - λ 1 )( a 22 - λ 1 ) – a 12 a 21 = 0, this equation is satisfied, but it also shows us that we can only find the ratio u 1 (1) / u 2 (1) . A usual, but not mandatory, procedure, is to choose values so that ( u 1 (1) ) 2 + ( u 2 (1) ) 2 = 1 MASE201 Spring 2009
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mase201sp09_Feb10_PRINTABLE - Solving Systems of Linear...

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