《结构工程中的数学方法》课程教学课件（讲稿）intro_gauss

2AlgebraicproblemsinmatrixformSystemsofLinearEquationsGaussianelimination-introductoryexample(fromslide33a)equationsofequilibriumwith:a=2.50m,b=1.25m,q=32.0kN/mX225.0(E.)@ZM。=0:2.5.x+1.25.X2三25.0(E2)@ZM=0:1.25.x1+1.25.x2-X3 =X1+I ZF, = 0 :=-40.0(E:)+0.6·X3-X2D2 (E)-(E) →(E):qro52.5.X +1.25.X225.0(E)0+1.25.X225.0(E2)-2.0.X3=b=-40.0(E:)+0.6·X3-X2X31.25 (E) +(E2) → (E3) :25.0(E.)2.5.X+1.25.X2X,=-16.0kN三025.0(E2)+1.25.X2-2.0.X3X, = 52.0 kN=0-25.0(E:)X, = 20.0 kN-1.25.X3=36aMichael Beer,Engineering Mathematics

Michael Beer, Engineering Mathematics equations of equilibrium with: Systems of Linear Equations Gaussian elimination − introductory example (from slide 33a) 2 Algebraic problems in matrix form 36a + ∑M 0: C = + ∑M 0: D = + ∑F 0: y = ● ( ) ( ) ( ) 1 2 1 1 23 2 2 3 3 2.5 x 1.25 x 25.0 E 1.25 x 1.25 x x 25.0 E x 0.6 x 40.0 E ⋅+ ⋅ = ⋅ + ⋅ −= − + ⋅ =− a 2.50 m, b 1.25 m, q 32.0 kN/m = = = 2E E E: ⋅−→ ( 21 2 ) ( ) ( ) ( ) ( ) ( ) 1 2 1 2 3 2 2 3 3 2.5 x 1.25 x 25.0 E 0 1.25 x 2.0 x 25.0 E x 0.6 x 40.0 E ⋅+ ⋅ = + ⋅ −⋅ = − + ⋅ =− 1.25 E E E : ⋅+→ ( 32 3 ) ( ) ( ) ( ) ( ) ( ) 1 2 1 23 2 3 3 2.5 x 1.25 x 25.0 E 0 1.25 x 2.0 x 25.0 E 0 1.25 x 25.0 E ⋅+ ⋅ = + ⋅ −⋅ = − ⋅ =− x1 x2 x3 0.5 a D q b a C x 20.0 kN 3 = x 52.0 kN 2 = x 16.0 kN 1 = −

2AlgebraicproblemsinmatrixformSystemsofLinearEquationsGaussianelimination-introductoryexample(cont'd)effectofinappropriateequationsofequilibriumX2= 25.0(E.)①ZM。=0:2.5.X+1.25.X2X1.E-X, = 25.0(E2)@ZM,=0:1.25.X +1.25X2+1.25.X2-2.0-X,=25.0(E:)QZM=0:2 (E2)-(E) → (E2):Dqro2.5X+1.25.X2=25.0(E.)5.0 +1.25.x2 -2.0 X, = 25.0(E2)1.25.x2-2.0.X,= 25.0(E:)工bX3 (-1)·(E)+(E,)→(E):25.02.Xi+X2=20kN2.5.x+1.25X2(E)=0(E2)5.X,-8.X,=100kN+1.25.X2-2.0.X,= 25.000three unknowns,but only two equations+0 =(E:)infinitely many solutions as (E3) can be expressed by (Ei) and (E2),(E3)isalinearcombinationof(Ei)and(E2)36bMichaelBeer,EngineeringMathematics

effect of inappropriate equations of equilibrium Systems of Linear Equations Gaussian elimination − introductory example (cont'd) 2 Algebraic problems in matrix form 36b + ∑M 0: C = + ∑M 0: D = ● ( ) ( ) ( ) 1 2 1 1 23 2 23 3 2.5 x 1.25 x 25.0 E 1.25 x 1.25 x x 25.0 E 1.25 x 2.0 x 25.0 E ⋅+ ⋅ = ⋅ + ⋅ −= + ⋅ −⋅ = (−⋅ + → 1 E E E: ) ( 32 3 ) ( ) ( ) x1 x2 x3 0.5 a D q b a C 5 x 8 x 100 kN 2 3 ⋅ −⋅ = 2 x x 20 kN 1 2 ⋅+ = E + ∑M 0: E = 2E E E: ⋅−→ ( 21 2 ) ( ) ( ) ( ) ( ) ( ) 1 2 1 23 2 23 3 2.5 x 1.25 x 25.0 E 0 1.25 x 2.0 x 25.0 E 1.25 x 2.0 x 25.0 E ⋅+ ⋅ = + ⋅ −⋅ = ⋅ −⋅ = ( ) ( ) ( ) 1 2 1 23 2 3 2.5 x 1.25 x 25.0 E 0 1.25 x 2.0 x 25.0 E 0 0 0E ⋅+ ⋅ = + ⋅ −⋅ = + = infinitely many solutions as (E3) can be expressed by (E1) and (E2), (E3) is a linear combination of (E1) and (E2) three unknowns, but only two equations ! Michael Beer, Engineering Mathematics

2AlgebraicproblemsinmatrixformSystemsofLinearEquationsGaussianelimination-introductory example (cont'd)inappropriateequationsof eguilibrium-numericalproblemidentification2.5 1.250/25.0A=1.251.25-1.025.0n=301.25-2.025.0》determinantof matrixAdet(A)=2.5.1.25.(-2.0)-1.25.1.25.(-2.0)-2.5.1.25.(-1.0)=0no uniquesolutionexists(nosolutionORinfinitelymanysolutions)》rank of matrixA>, (2.5,1.25,0) + 22 (1.25,1.25,-1) + 2 (0,1.25,-2) = 0= 21 = 1, ≥2=-2, 入3= 1 = one linear dependency = rk(A)= 2》rank of augmented matrixA2 (2.5,1.25,0,25)+ 22 (1.25,1.25,-1,25) +2 -(0,1.25,-2,25) = 0= 21 = 1, 22=-2, ≥3 = 1 = one linear dependency = rk(A)=2rk(A) = rk(A) → solvable; rk(A) = 2 < 3 = n infinitely many solutions36cMichael Beer,Engineering Mathematics

inappropriate equations of equilibrium − numerical problem identification Systems of Linear Equations Gaussian elimination − introductory example (cont'd) 2 Algebraic problems in matrix form 36c ● 2.5 1.25 0 25.0 ˆ 1.25 1.25 1.0 25.0 0 1.25 2.0 25.0     = − −   A det (A) = 2.5·1.25·(−2.0) − 1.25·1.25·(−2.0) − 2.5·1.25·(−1.0) = 0 no unique solution exists (no solution OR infinitely many solutions) λ ⋅ 12 3 (2.5,1.25,0 1.25,1.25, 1 0,1.25, 2 ) +λ ⋅ ( − +λ ⋅ − = ) ( ) 0 ⇒ λ1 = 1, λ2 = −2, λ3 = 1 ⇒ one linear dependency ⇒ rk(A) = 2 » rank of matrix A λ ⋅ 12 3 (2.5,1.25,0,25 1.25,1.25, 1,25 0,1.25, 2,25 ) +λ ⋅ ( − +λ ⋅ − = ) ( ) 0 » rank of augmented matrix  rk(A) = rk(Â) solvable » determinant of matrix A ⇒ λ1 = 1, λ2 = −2, λ3 = 1 ⇒ one linear dependency ⇒ rk(Â) = 2 n = 3 ; rk(Â) = 2 < 3 = n infinitely many solutions Michael Beer, Engineering Mathematics