2AlgebraicproblemsinmatrixformGaussianEliminationAlgorithmExample (cont'd).augmented matrix after operations Pii50(E.)1.07010-1.6051-1.6051(E.)00006.8099(E))005.6179-0.8024874.99800(E.)5.617974.998-0.80248forwardelimination,operationsPi2toeliminateX,inrowsi=3and4a2(E,) →(E,), not necessary as a32 = 0, (E3) is not modified》 P32 : (E3)on22(E,) → (E.), not necessary as a42 = 0, (E) is not modified》P42 :(E4)a,2augmented matrix afteroperations Pi2》seeabove(notmodified)46Michael Beer, Engineering Mathematics
46 Gaussian Elimination Algorithm ( ) ( ) ( ) ( ) 1 2 3 4 1.0701 0 1.6051 1.6051 50 E 0 6.8099 0 0 0 E 0 0 5.6179 0.80248 74.998 E 0 0 0.80248 5.6179 74.998 E − − − − Example (cont'd) ● augmented matrix after operations Pi1 ● forward elimination, operations Pi2 to eliminate x2 in rows i = 3 and 4 , not necessary as a32 = 0, (E3 » ) is not modified » ( ) ( ) ( ) 32 32 3 2 3 22 a P : E E E a − → , not necessary as a42 = 0, (E4 ( ) ( ) ( ) ) is not modified 42 42 4 2 4 22 a P : E E E a − → ● augmented matrix after operations Pi2 » see above (not modified) 2 Algebraic problems in matrix form Michael Beer, Engineering Mathematics
2AlgebraicproblemsinmatrixformGaussian EliminationAlgorithmExample (cont'd).augmented matrix after operations Pi250(E.)1.07010-1.6051-1.6051(E.)00006.8099(E))005.6179-0.8024874.9980L0(E.)74.998-0.802485.6179forward elimination, operations Pi3 to eliminate X3 in row i= 4-0.80248a (E) = (E.)》 P43 : (E)(E) → (E4)5.6179033newcoefficientsinrow4:-0.802485.6179=0a41=0,a42=0,a43=-0.802485.6179-0.80248a44=5.6179(-0.80248)=5.50335.6179-0.80248a4s=74.99874.998=85.7115.617947Michael Beer,EngineeringMathematics
47 Gaussian Elimination Algorithm ( ) ( ) ( ) ( ) 1 2 3 4 1.0701 0 1.6051 1.6051 50 E 0 6.8099 0 0 0 E 0 0 5.6179 0.80248 74.998 E 0 0 0.80248 5.6179 74.998 E − − − − Example (cont'd) ● augmented matrix after operations Pi2 ● forward elimination, operations Pi3 to eliminate x3 in row i = 4 » ( ) ( ) ( ) ( ) ( ) 43 43 4 3 4 3 4 33 a 0.80248 P : E E E E E a 5.6179 − − =− → ( ) 41 42 43 44 45 0.80248 a 0, a 0, a 0.80248 5.6179 0 5.6179 0.80248 a 5.6179 0.80248 5.5033 5.6179 0.80248 a 74.998 74.998 85.711 5.6179 − = = =− − ⋅ = − = − ⋅− = − =− ⋅= new coefficients in row 4: 2 Algebraic problems in matrix form Michael Beer, Engineering Mathematics
2AlgebraicproblemsinmatrixformGaussian Elimination AlgorithmExample(cont'd)augmented matrix after operations Pi3 (result of forward elimination)(E.)1.0701050-1.6051-1.6051(E,)00006.8099(E)0074.9985.6179-0.802480(E.)005.503385.711backwardsubstitution》determinationoftheunknownX4fromrow 485.711a45文15.574=X=×。.10-4=0.00155745.5033244》substitution of ×.in row 3 and determination of the unknown X374.998-(-0.80248)·15.574-axa35X=15.5745.6179a33=X=x,·10-4=0.001557448Michael Beer,EngineeringMathematics
48 Gaussian Elimination Algorithm ( ) ( ) ( ) ( ) 1 2 3 4 1.0701 0 1.6051 1.6051 50 E 0 6.8099 0 0 0 E 0 0 5.6179 0.80248 74.998 E 0 0 0 5.5033 85.711 E − − − Example (cont'd) ● augmented matrix after operations Pi3 (result of forward elimination) ● backward substitution 45 4 4 4 4 44 a 85.711 ˆ ˆ x 15.574 x x 10 0.0015574 a 5.5033 − = = = ⇒ =⋅ = » determination of the unknown x4 from row 4 ( ) 35 34 4 3 33 4 3 3 a axˆ 74.998 0.80248 15.574 ˆx 15.574 a 5.6179 x x 10 0.0015574 ˆ − − ⋅ −− ⋅ = = = ⇒ =⋅ = » substitution of in row 3 and determination of the unknown x ˆx4 3 2 Algebraic problems in matrix form Michael Beer, Engineering Mathematics
2AlgebraicproblemsinmatrixformGaussian Elimination AlgorithmExample(cont'd)backward substitutionrecall:[1.0701050-1.6051-1.6051》substitutionof ×,andinrow200006.8099and determination of the unknown X2005.6179-0.8024874.998a2s -2a2, - x,00Lo85.7115.5033i3七a220-[0.15.574+0.15.574：0=×,=×,.104=06.8099》substitution of ×4,x,and ×, in row 1 and determination of the unknown Xias-Zay-x-2ai50-[0.0+(-1.6051)·15.574+(-1.6051)·15.57493.4451.0701=X=×,·10-4=0.009344549MichaelBeer,EngineeringMathematics
49 Gaussian Elimination Algorithm 1.0701 0 1.6051 1.6051 50 0 6.8099 0 0 0 0 0 5.6179 0.80248 74.998 0 0 0 5.5033 85.711 − − − Example (cont'd) ● backward substitution 4 25 2 j j j 3 2 22 4 2 2 a axˆ ˆx a 0 0 15.574 0 15.574 0 x x 10 0 ˆ 6.8099 = − − ⋅ ∑ = − ⋅ +⋅ = =⇒ = ⋅ = » substitution of and in row 2 and determination of the unknown x2 ˆx4 recall: ˆx3 ( ) ( ) 4 15 1j j j 2 1 11 4 1 1 a axˆ ˆx a 50 0 0 1.6051 15.574 1.6051 15.574 93.445 1.0701 x x 10 0.0093445 ˆ = − − ⋅ ∑ = − ⋅ +− ⋅ +− ⋅ = = ⇒ =⋅ = » substitution of , and in row 1 and determination of the unknown x ˆx4 ˆx3 ˆx2 1 2 Algebraic problems in matrix form Michael Beer, Engineering Mathematics
2AlgebraicproblemsinmatrixformGaussian Elimination AlgorithmExample(cont'd).solutionsummaryrecall:50kNE93.452[x]93.445300X2·104.10-m23X315.57415.575E315.57415.575X454increased3m3mnumberofdigits1.07010-1.6051-1.60515000006.8099eremarks0-1.605108.02551.6051》another order of equations may have008.0255-1.60511.6051requiredrowswitchingtoensurepivot elements aii different from zero》pre-eliminationofx,possible(decoupledsystem)=reductionof numericaleffort by splittingthesystem into sub-systems》X3=x4canbepre-identified from structural properties(reduced system size)=reductionof numerical effortexploiting symmetry conditions》alwayscheckplausibilityofresults50MichaelBeer,Engineering Mathematics
50 Gaussian Elimination Algorithm Example (cont'd) ● solution summary 3 m 3 m 3 m 3 m 50 kN 1 2 3 4 5 x y 1 2 4 3 4 x 93.445 x 0 = 10 m x 15.574 x 15.574 − ⋅ ● remarks » another order of equations may have required row switching to ensure pivot elements ajj different from zero » pre-elimination of x2 possible (decoupled system) ⇒ reduction of numerical effort by splitting the system into sub-systems » x3 = x4 can be pre-identified from structural properties (reduced system size) ⇒ reduction of numerical effort exploiting symmetry conditions » always check plausibility of results 1.0701 0 1.6051 1.6051 50 0 6.8099 0 0 0 1.6051 0 8.0255 1.6051 0 1.6051 0 1.6051 8.0255 0 − − − − recall: 4 93.452 0 10 15.575 15.575 − ⋅ increased number of digits 2 Algebraic problems in matrix form Michael Beer, Engineering Mathematics