L04_Solving_Cubic_Polynomials

L04_Solving_Cubic_Polynomials - CE507 Lecture 4 I The Cubic...

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CE507 Lecture 4 Numerical Solution of the Cubic Polynomial I. The Cubic Polynomial From [ ] A v v λ = ± ± Using 0 A I λ = to solve for λ 11 12 13 21 22 23 31 32 33 0 a a a a a a a a a λ λ λ = ( ) 22 23 21 23 21 22 11 12 13 32 33 31 33 31 32 0 a a a a a a a a a a a a a a a λ λ λ λ λ + = ( ) ( )( ) ( ) ( ) 11 22 33 23 32 12 21 33 13 32 21 13 31 22 0 a a a a a a a a a a a a a a λ λ λ λ λ = Simplifying ( ) ( ) 3 2 1 2 3 1 11 22 33 2 2 1 3 0 1 2 det ii ij ji ij I I I I a a a a I I a a I a λ λ λ + = = = + + = = I.1 Example 2 1 1 2 3 4 1 1 2 A = ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 2 3 2 3 2 3 1 9 2 2 1 2 1 1 2 1 3 3 4 1 1 1 1 4 2 2 1 2 3 4 2 4 2 3 2 3 1 2 1 2 1 1 I I I = + = = × + × + ×− + × + × + ×− + − × + − × + − ×− = − = + = − so 3 2 3 3 0 λ λ λ + = with factors ( )( )( ) 1 1 3 0 λ λ λ + =
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