《结构工程中的数学方法》课程教学课件（讲稿）intro_LU_decomp

2AlgebraicproblemsinmatrixformSystemsofLinearEquationsLUdecomposition-introductoryexample(fromslide33a)systemofequationswith:a=2.50m,b=1.25m,=32.0kN/m25.0751.2502.5.LUdecomposition:A=L.UA=25.01.25 1.25-1.0Ui2U13福1.2500-1.00.6-40.0OU22U23212.50o00U33]Gaussianforward elimination1.25022.5 1.250(E,)(E.) = (E2)(E)→(E2)2.50al1.251.25-1.0=A[2.5025.01.25X,-1.00.6O012.50.625-1.0X20-1.00.6-40.0X3used inforward231-1.032(E.)(E,) = (E)(E2)→(E)eliminationifa31+0a.0.625+o27A (elim.)2.501.2525.0X100100.625-1.012.5X200.51L=00-1.0-20.0X30152bMichael Beer,EngineeringMathematics

Michael Beer, Engineering Mathematics system of equations with: Systems of Linear Equations LU decomposition − introductory example (from slide 33a) 2 Algebraic problems in matrix form 52b ● a 2.50 m, b 1.25 m, q 32.0 kN/m = = = 1 2 3 2.5 1.25 0 x 25.0 0 0.625 1.0 x 12.5 0 0 1.0 x 20.0           −⋅ =       − −          2.5 1.25 0 25.0 ˆ 1.25 1.25 1.0 25.0 0 1.0 0.6 40.0     = − − −   A Gaussian forward elimination ( ) ( ) ( ) ( ) ( ) 21 2 12 1 2 11 a 1.25 E EE E E a 2.50 − =− → ( ) ( ) ( ) ( ) ( ) 32 3 23 2 3 22 a 1.0 E EE E E a 0.625 − − =− → ● 1 2 3 2.5 1.25 0 x 25.0 0 0.625 1.0 x 12.5 0 1.0 0.6 x 40.0           −⋅ =       − −          ● LU decomposition: A LU = ⋅ 1 00 0.5 1 0 0 1     =       L  11 12 13 22 23 33 u u u 0 u u 0 0 u       21 31 32 100 l 10 ll1           2.5 1.25 0 1.25 1.25 1.0 0 1.0 0.6     − = −   A 21 21 11 1.25 a l 2.50 a = = 31 31 11 a l 0 a = = used in forward elimination if a31 ≠ 0 (elim.) A

2AlgebraicproblemsinmatrixformSystemsofLinearEquationsLudecomposition-introductoryexample(cont'd)systemofequationswith:a=2.50m,b=1.25m,q=32.0kN/m1.25025.02.5.LUdecomposition:A=L.UA=25.01.251.25-1.0U31.2520-1.00.6-40.00U23U22212.50a1l00U3Gaussianforward elimination1.25O2002.51.25(E,)福(E.) = (E2)E)→(E2)2.500-1.01.25.250[2.51.2525.0七0<0.6)00.625-1.012.5X20-1.00.6-40.0X3-1.0232[51 = 0 = a31-1.00.6250(E.)(E2) → (E.)(E2) = (E)a10.625dA(elim.)=1.02.501.2525.0= 0.625, U23U22X100710-1.00.62512.5X200.51L=1.000-1.0-20.0X30U= A(elim.)-1.6152cMichael Beer, Engineering Mathematics

Michael Beer, Engineering Mathematics system of equations with: Systems of Linear Equations LU decomposition − introductory example (cont'd) 2 Algebraic problems in matrix form 52c ● a 2.50 m, b 1.25 m, q 32.0 kN/m = = = 1 2 3 2.5 1.25 0 x 25.0 0 0.625 1.0 x 12.5 0 0 1.0 x 20.0           −⋅ =       − −          2.5 1.25 0 25.0 ˆ 1.25 1.25 1.0 25.0 0 1.0 0.6 40.0     = − − −   A Gaussian forward elimination ( ) ( ) ( ) ( ) ( ) 21 2 12 1 2 11 a 1.25 E EE E E a 2.50 − =− → ( ) ( ) ( ) ( ) ( ) 32 3 23 2 3 22 a 1.0 E EE E E a 0.625 − − =− → ● 1 2 3 2.5 1.25 0 x 25.0 0 0.625 1.0 x 12.5 0 1.0 0.6 x 40.0           −⋅ =       − −          ● LU decomposition: A LU = ⋅ 1 00 0.5 1 0 0 1.6 1     =     −   L 11 12 13 22 23 33 u u u 0 u u 0 0 u       21 31 32 100 l 10 ll1           2.5 1.25 0 1.25 1.25 1.0 0 1.0 0.6     − −   21 21 11 1.25 a l 2.50 a = = u 0.625, u 1.0 22 = 23 = − 31 31 11 a l 0 a = = 32 32 22 1.0 a l 0.625 a − = = u 1.0 33 = − (elim.) U A= (elim.) A