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1SYSTEMS OF EQUATIONSDAVID I. SCHWARTZAdapted from Introduction to Maple, by David I. Schwartz, Prentice Hall, 1999.Excerpt from Introduction to Maple 2EXAMPLE PROBLEMFigure 0-1: Example DeviceA device composed of two elastic bars capped with rigidplates is shown in Fig. 0-1. Loads are statically applied onboth plate ① and plate ②. For now, call the connections at ①and ② as nodes.Figure 0-2: Model of DeviceYou can model elastic bars as springs as shown in Fig. 0-2.Assume that Hooke’s law governs spring behavioras discussed in Chapter 6.Figure 0-3: Free Body DiagramsAfter slowly applying loads, the bars deform and reach anew resting or equilibrium position. The applied loads andinternal bar forces must balance according to equilibrium.Assume no twisting or rotation of the plates occur.Figure 0-4: Hooke’s LawFrom Hooke’s Law, relate each spring’s internal force withto relative displacement, the measure of how much eachspring stretches. Bars and have relative displacements and , respectively.Figure 0-5: Relative DisplacementsDisplacement at ② stretches spring and compresses spring. Displacement at ① stretches spring . Therefore,determine ’s “relative stretch” by subtracting node ②’sdisplacement from that of node ① as shown in Fig. 0-5.Figure 0-6: Combine EquationsEase your computation by expressing the equations in termsof nodal values as shown in Fig. 0-6. Combine equilibrium,Hooke’s Law, and displacement relations into two equationsin terms of , , and .p2p1ab②①p2p1ab②①u2u1p ku=①p1papap2pb②pap1=pbp2pa+=p1p2+=pakaua=pbkbub=puk1abuaubBeforeAfteru2u1② ①bau1u2ua+=uau1u2–=∴ubu2=ubuabau1( )aau2( )p1pa=kauakau1u2–( )==p2pbp1–=kbu2kau1u2–( )–=k–au1kakb+( )u2+=pkuExcerpt from Introduction to Maple 3SYSTEMS OF EQUATIONSRearrange the equations of Fig. 0-6 into the system of equations,(1)(2)A system of equations collects simultaneous equations with common unknown variables alsocalled unknowns or indeterminates. The system in Eq. 1 and 2 has two unknowns, and .Assume all other variables have predetermined values.Linear systems contain all first-order equations in the form . BothEq. 1 and 2 contain terms with powers no greater than one, and thus, constitute a linear system.On the other hand, non-linear systems of equations contain terms with powers higher than one.GAUSSIAN ELIMINATIONSolve for the unknowns and as shown in Fig. . First, assume values , ,, and in Step ①. Apply gaussian elimination as demonstrated by Steps ②, ③,and ④. By adjusting coefficients, you can eliminate common terms. After dividing out leadingcoefficients, solve for unknowns by backsubstitution.DEPENDENCYThe following conditions characterize a linearly independent system of equations:• The number of equations match the number of unknowns.• No equation is a multiple of another equation in the system.A linearly independent system produces only one, or unique, solution for each unknown asdemonstrated in Figure 1. You might encounter systems with duplicates of equations, such as thesystem and . Such a system produces an infinite number of andsolutions and is called linearly dependent.Denote the resulting “matrix form” of the equations as(3)where matrix multiplication and equality are implied. Each term is described below:Substitute valuesinto Eqs. 1 and 2.Reduce thesecond equation byadding the first.Divide equationsby the first orleading coefficient.Backsubstituteresults into the firstequation.FIGURE 1: GAUSSIAN ELIMINATIONkau1kau2– p1=k–au1kakb+( )u2+ p2=u1u2a0a1x1… anxn+ +=u1u2ka2=kb3=p110=p220=①②③④2u12u2– 10=2u1– 5u2+ 20=2u12u2– 10=3u230=u1u2– 5=u210=u115=u210=x y+ 1=2x 2y+ 2=xyKu p=Excerpt from Introduction to Maple 4MANUAL SOLUTIONAs demonstrated in this document, you can perform Gaussian elimination to solve the system ofequations that Eq. 3 represents. But, why should you use the matrix formulation? Vectors andmatrices store equation data in a compact form that computer programs can readily manipulate.Though more complex techniques exist, you can solve the matrix formulation with row reduction,which is a method that mimics Gaussian elimination as demonstrated in Table 2.The following steps illustrate row reduction:• Step ①: Cast the equations into a matrix formulation.• Step ②: Rewrite the system into a matrix that includes the source vector written to the right.You may draw a vertical bar to serve as a reminder to separate the coefficient matrix.• Step ③: Row reduction dictates that you may add a row to any other row. So, add the top rowto the bottom row. This process is equivalent to adding an equation to another equation withinthe given system.• Step ④: Row reduction also dictates you can multiply any row by any constant. So, divide thetop row by 2, and divide the bottom row by 3. Note that you can perform this action inconjunction with adding rows to each other.• Step ⑤: You keep performing row reduction until the coefficient matrix becomes the identitymatrix, a matrix with values of 1 on the diagonal and 0 elsewhere. The final values in theright-hand column of the matrix represent the solution vector.Note that a linearly dependent system of equations will yield at least one row that contains onlyvalues of zero.SystemThe system of equations, , is a set of simultaneous linear equations.CoefficientMatrixThe coefficient matrix, , collects the constants in front of unknowns:• Square matrices, , have same the number of unknowns and equations.• Elements of these matrices typically reflect models’ physical parameters.SourceVectorThe source vector, , applies modeled inputs or “sources” to the system:• In the spring example, these source terms are loads.• Source values are typically known or assumed.SolutionVectorThe solution vector, , collects the unknown variables you wish to find:• The matrix formulation separates known and unknown variables.• Manipulating the coefficient matrix and source vector with linear algebratechniques like Gaussian elimination (Fig. ) finds the unknowns.Ku p=Ku p=K2 2–2– 5=KKp1020=puu1u2=uExcerpt from Introduction to Maple 5TABLE 2: ROW REDUCTIONStep Matrix Formulation Operations Results①②③④⑤2 2–2– 5u1u21020=2u12u2– 10=2u1– 5u2+ 20=2u12u2– 10=2u1– 5u2+ 20=22–2–51020202–310302u12u2– 10=2u1– 5u2+ 20=+0u13u2+ 30=2u12u2– 10=3u230=101–151012---2u12u2– 10=( ) u1u2–→ 5=13---0u13u2+ 30=( ) u2→ 10=u1u2–


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CORNELL CS 100 - SYSTEMS OF EQUATIONS

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