2AlgebraicproblemsinmatrixformSystemsof Equationsin EngineeringExamplesstructural analysis:evaluation of eguilibrium conditionsreliabilityanalysis:responsesurfacemethodsstructural design:determination of optimum dimensions.material modelling:determination of parameters from dataForm of eguations.linear/nonlinearexplicit/implicitalgebraic/transcendental.homogeneous/non-homogeneousbasic cases and common applicationsexplicit linear algebraicequationsSolution methodsdirectmethodsiterativemethods31MichaelBeer,EngineeringMathematics
Michael Beer, Engineering Mathematics 31 2 Algebraic problems in matrix form Systems of Equations in Engineering Examples ● structural analysis: evaluation of equilibrium conditions ● reliability analysis: response surface methods ● structural design: determination of optimum dimensions ● material modelling: determination of parameters from data basic cases and common applications explicit linear algebraic equations Form of equations ● linear / nonlinear ● explicit / implicit ● algebraic / transcendental ● homogeneous / non-homogeneous Solution methods ● direct methods ● iterative methods
2AlgebraicproblemsinmatrixformSystemsofLinearEquationsSystem structure.n equations for the determination of n unknowns x,(E.)b,x+x十ainxn+aix++(E,)b,=ax+a22x2+..+a2,X,+.+ax主:::：S(E)= b,a,x,十a.x,a,x,a,X2十十十..：..:：:b.(E.)anx,annxnan2x,anx++aj: constant coefficients, e.g., stiffness elementsb,:constant values (right-hand side),e.g.,loads(atleastoneb,+Otoformanon-homogeneoussystem)X,:unknowns,e.g.,displacementsthreecasesforsolution(graphicalrepresentationforn=2):2)nosolution1)singlesolution3)infinitelymanysolutionsE1EiEE2E2E232Michael Beer,Engineering Mathematics
32 Systems of Linear Equations System structure ● n equations for the determination of n unknowns xn aij: constant coefficients, e.g., stiffness elements bi : constant values (right-hand side), e.g., loads (at least one bi ≠ 0 to form a non-homogeneous system) xn: unknowns, e.g., displacements ( ) ( ) ( ) ( ) 11 1 12 2 1j j 1n n 1 1 21 1 22 2 2 j j 2n n 2 2 n i1 1 i2 2 ij j in n i i n1 1 n2 2 nj j nn n n n a x a x . a x . a x b E a x a x . a x . a x b E S a x a x . a x . a x b E a x a x . a x . a x b E + ++ ++ = + ++ ++ = = + ++ ++ = + ++ ++ = ● three cases for solution (graphical representation for n=2): 1) single solution 2) no solution 3) infinitely many solutions E1 E2 E1 E2 E1 E2 2 Algebraic problems in matrix form Michael Beer, Engineering Mathematics
2AlgebraicproblemsinmatrixformSystems of Linear EquationsSystem structure (cont'd)matrixformanaina12a,[b,](E.)X1(E,)b2a21a2na22aziX2........：：：.(E.)b,X,a.iai2aainintroductory:目...：：example:bn(E.)Xnanan2annaAb真a1la12an=b,a.(E,)a(E,)a21=b,a22aa2na2n+representationwith：..augmented matrix(E.)a.n+1 =baiaina,2a.iA=[A /b]目：::::(E.)=b.a.an2annaniann+1n133Michael Beer,EngineeringMathematics
33 Systems of Linear Equations System structure (cont'd) ● matrix form ( ) ( ) ( ) ( ) 11 12 1j 1n 1 1 1 21 22 2 j 2n 2 2 2 i1 i2 ij in i i i n n n n1 n2 nj nn a a . a . a x b E a a . a . a x b E a a . a . a x b E a a . a . a x b E ⋅ = A xb ⋅ = ( ) ( ) ( ) ( ) 11 12 1j 1n 1 n 1 1 1 21 22 2 j 2n 2 n 1 2 2 i1 i2 ij in i n 1 i i n n1 n2 nj nn n n 1 n a a . a . a a b E a a . a . a a b E a a . a . a a b E a a . a . a a b E + + + + = = = = representation with augmented matrix ˆ = A Ab 2 Algebraic problems in matrix form introductory example Michael Beer, Engineering Mathematics
2AlgebraicproblemsinmatrixformSystems of Linear EquationsSolvabilitysolvable if and only if rk(A) = rk(A)exactly one(unique)solution if,further,rk(A)=n》rk(A):rankofthematrixAofcoefficientsrk(A):rankoftheaugmentedmatrixAofcoefficients》n:number of unknownsX,rankof a matrix A = maximal number of linearly independent rows of Arow vectors (aii, ai2, ..., aij, .., ain) are linearly independentif the solution to Za. (a,aiz...,ay..,a..) = o is , = O, i = 1, .., n, only."quadratic"system of equations:uniquesolutionifandonly ifdet(A)+0》 det(A):determinant of matrix Adet(A)=Zsgn(βi,β2..,β.).aieap,.angm(n!summands)B1.P2....,Bn(βi,β2,...,βn):permutationsof indicessgn(βi,β2,..,βn):signatureof permutation(alternating+1and-1)34Michael Beer, Engineering Mathematics
34 Solvability ● solvable if and only if rk(A) = rk(Â), exactly one (unique) solution if, further, rk(Â) = n » rk(A): rank of the matrix A of coefficients » rk(Â): rank of the augmented matrix  of coefficients » n: number of unknowns xi ● "quadratic" system of equations: unique solution if and only if det(A) ≠ 0 » det(A): determinant of matrix A rank of a matrix A = maximal number of linearly independent rows of A row vectors (ai1, ai2, ., aij, ., ain) are linearly independent if the solution to ( ) is λi = 0, i = 1, ., n, only. n i i1 i2 ij in i 1 a ,a ,.,a ,.,a = ∑ λ = 0 ( ) ( ) 12 n 12 n 12 n 1 2 n , ,., det sgn , ,., a a . a ββ β ββ β A = ββ β ⋅ ⋅ ⋅ ⋅ ∑ (β1, β2, ., βn): permutations of indices sgn(β1, β2, ., βn): signature of permutation (alternating +1 and −1) (n! summands) Systems of Linear Equations 2 Algebraic problems in matrix form Michael Beer, Engineering Mathematics
2AlgebraicproblemsinmatrixformSystems of LinearEquationsSolvability.exampleproblem interpretation:(E) stipulates X + 2·X2 + X3 = 31213(E1)consequently,2.(x+2.x2+x3)shouldyield2.3=62421(E2)but (E2) demands 2X1 + 4·X2 + 2·X3 = 1 + 6-212-3(E3)whichis a contradictionnosolutionexists》rankofmatrixA21 (1,2,1) + 22 :(2,4,2) + 23 (-3, -2,1)= 0 for 21 = 2, 22 = -1, 23 = 0lineardependencybetween vectorsin rows1 and2ofmatrixA=rk(A)=2》rank of augmented matrixA21 (1,2,1,3) +22 (2,4,2,1)+ 23 (-3, -2,1,2) = 0 for 2, = 0, i = 1,2,3, onlythe three row vectors in matrix A are linearly independent =rk(A)=3rk(A) + rk(A) no solution exists》determinantofmatrixAdet (A) = 1·4·1 + 2·2·(-3) + 1·2·(-2) - 1·4·(-3) - 1·2·(-2) - 2·2·1 = 0no unique solution exists (no solution OR infinitelymany solutions)also:det(A)= O if a row ora column in A contains zeros only35MichaelBeer,EngineeringMathematics
35 Solvability ● example 1 213 2 421 3 212 − − (E1) (E2) (E3) λ ⋅ +λ ⋅ +λ ⋅ − − = 12 3 (1,2,1 2,4,2 3, 2,1 ) ( ) ( ) 0 for λ1 = 2, λ2 = −1, λ3 = 0 linear dependency between vectors in rows 1 and 2 of matrix A ⇒ rk(A) = 2 » rank of matrix A problem interpretation: (E1) stipulates x1 + 2·x2 + x3 = 3 consequently, 2·(x1 + 2·x2 + x3) should yield 2·3 = 6 but (E2) demands 2·x1 + 4·x2 + 2·x3 = 1 ≠ 6 which is a contradiction no solution exists λ ⋅ +λ ⋅ +λ ⋅ − − = 12 3 (1,2,1,3 2,4,2,1 3, 2,1,2 ) ( ) ( ) 0 the three row vectors in matrix  are linearly independent ⇒ rk(Â) = 3 » rank of augmented matrix  for λi = 0, i = 1,2,3, only rk(A) ≠ rk(Â) no solution exists » determinant of matrix A det (A) = 1·4·1 + 2·2·(−3) + 1·2·(−2) − 1·4·(−3) − 1·2·(−2) − 2·2·1 = 0 no unique solution exists (no solution OR infinitely many solutions) !also: det(A) = 0 if a row or a column in A contains zeros only Systems of Linear Equations 2 Algebraic problems in matrix form Michael Beer, Engineering Mathematics