Solution of Simultaneous Linear Algebraic Equations

IntroductionApplications and Problem Setup: Scaffolding

Basic Concepts of SolutionHomework

Solution of Simultaneous Linear AlgebraicEquations

Mike Renfro

February 13, 2008

Mike Renfro Solution of Simultaneous Linear Algebraic Equations



Basic Form of the EquationsDetour: Multiplication in Matrix AlgebraSolution Types

Introduction

Systems of linear algebraic equations are common representationsof many engineering problems, especially problems involvingequilibrium of some sort:

Statics:∑

Fy = 0

Dynamics:∑

F = ma

Thermodynamics: min = mout

etc.





Basic Form of the Equations

a11x1 + a12x2 + · · · + a1nxn = b1

a21x1 + a22x2 + · · · + a2nxn = b2

...

an1x1 + an2x2 + · · · + annxn = bn

Typical characteristics:

n equations, with nunknown variables.

Only linear terms involvingunknowns, no x2, sin(x), orother nonlinear termsallowed.

Particular unknowns may beomitted from someequations entirely, in whichcase the corresponding aij

coefficient is assumed to be0.





Reformatted Form of the Equations

The system of equations

a11x1 + a12x2 + · · · + a1nxn = b1

a21x1 + a22x2 + · · · + a2nxn = b2

...

an1x1 + an2x2 + · · · + annxn = bn

can be rewritten asa11 a12 · · · a1n

a21 a22 · · · a2n...

an1 an2 · · · ann

x1

x2...xn

=

b1

b2...

bn





Detour: Multiplication in Matrix Algebra

Two matrices [A] and [B] can be multiplied if and only if theyare conformable, that is, if the number of columns in [A] isequal to the number of rows in [B].

If [A] has m rows and n columns, and [B] has n rows and pcolumns, then their product [C ] will have m rows and pcolumns. For example, a 3 × 3 matrix can be multiplied by a3 × 1 vector, yielding a 3 × 1 vector.

The values in [C ] are found with the formulacij =

∑nk=1 aikbkj — multiply each element of the i ’th row of

[A] by the corresponding element of the j ’th column of [B],then sum the products.





Matrix Multiplication Example

1 2 34 5 67 8 9

101112

Are these two matrices conformable? How many columns are in[A]? 3. How many rows are in {b}? 3. How large will the productmatrix be? [A] is 3 × 3, {b} is 3 × 1, so [A]{b} will be 3 × 1.





Matrix Multiplication Example

1 2 34 5 67 8 9

101112

=

1 × 10 + 2 × 11 + 3 × 124 × 10 + 5 × 11 + 6 × 127 × 10 + 8 × 11 + 9 × 12

=

68167266

Compare to: a11 a12 a13

a21 a22 a23

a31 a32 a33

x1

x2

x3

=

a11x1 + a12x2 + a13x3

a21x1 + a22x2 + a23x3

a31x1 + a32x2 + a33x3

=

b1

b2

b3





Solution Methods: Direct and Indirect

Direct solution methods can provide exact solutions to asystem of equations, assuming we can avoid roundoff andother numerical errors. However, these methods can be proneto substantial unavoidable errors under certain conditions.

Direct methods include Cramer’s rule, Gauss elimination, andLU decomposition.

Indirect solution methods are much simpler to program fromscratch, and arrive at a final solution by starting with aninitial guess and continually refining the guess until it satisfiesall the equations in the system. They are less prone toroundoff errors, but might not find a solution at all.

Indirect methods include the Jacobi and Gauss-Seidelmethods.




Solution of Simultaneous Linear Algebraic Equations:Static Analysis of a Scaffolding

3 bars supported by 6 cablesform a simple scaffolding. Giventhe positions and magnitudes for3 loads applied to the bars, findthe tension in each cable.




Governing Equations for Bar 1

Force equilibrium ∑Fy = 0

TA + TB − TC − TD − TF − P1 = 0

Moment equilibrium ∑M = 0

−9TB + TC + 4TD + 7TF + 5P1 = 0





Force equilibrium ∑Fy = 0

TC + TD − TE − P2 = 0

Moment equilibrium∑M = 0

−3TD + 2TE + P2 = 0





Force equilibrium∑Fy = 0

TE + TF − P3 = 0

Moment equilibrium∑M = 0

−4TF + P3 = 0




Assembling Equations

At this point, we have six independent equations (two for eachbar), and six unknowns (cable tensions). Reformat the sixequilibrium equations to isolate the unknown tensions on theleft-hand side of the equations. Make sure the tension variables arein the same order in each equation:

TA +TB −TC −TD −TF = P1

−9TB +TC +4TD +7TF = −5P1

TC +TD −TE = P2

−3TD +2TE = −P2

TE +TF = P3

−4TF = −P3




Matrix form setup

Make sure that the left-hand side of each equation containsnothing but unknowns and multipliers.

Make sure that each equation has the unknowns in the sameorder.

Make sure that the right-hand side of each equation containsnothing but known constants.




Matrix form

Assuming n as the number of equations or unknowns in theproblem, make some space below for a matrix with n rows and ncolumns, a vector with n rows and 1 column, and another vectorwith n rows and 1 column as shown:

· · · · · ·· · · · · ·· · · · · ·· · · · · ·· · · · · ·· · · · · ·

······

=

······




Filling out the matrix equation

1 Populate the unknowns vector with the list of unknownvariables.

2 Populate the right-hand side vector with the right-hand sidesof the equations.

3 Populate each row of the coefficient matrix with coefficientsfrom the left-hand side of the equations.

1 1 −1 −1 0 −10 −9 1 4 0 70 0 1 1 −1 00 0 0 −3 2 00 0 0 0 1 10 0 0 0 0 −4

TA

TB

TC

TD

TE

TF

=

P1

−5P1

P2

−P2

P3

−P3




Detour: Determinants and RankRank and Determinant Example: Scaffolding

Basic Concepts of Solution

A homogeneous system of equations is one where {b} = {0}. Ithas a trivial solution of {x} = {0}, but only has a non-trivialsolution if the determinant of [A], |[A]| = 0.A non-homogeneous system of equations is one where {b} 6= {0}.It has a proper solution if and only if rank(A) = rank([A b]) = n,where n is the number of equations in the system.





Detour: Determinants and Rank

The determinant of a square matrix is found by multiplying alongits diagonals, then adding and subtracting the products.∣∣∣∣[ 1 2

3 4

]∣∣∣∣ = (1 × 4) − (2 × 3) = −2

∣∣∣∣∣∣ 1 2 3

4 5 67 8 9

∣∣∣∣∣∣ =(1 × 5 × 9) +(2 × 6 × 7) +(3 × 4 × 8)−(1 × 6 × 8) −(2 × 4 × 9) −(3 × 5 × 7)

= 0





Detour: Determinants and Rank

The rank of a matrix is the size of the largest square submatrix itcontains with a non-zero determinant.

rank

([1 23 4

])= 2

The 2× 2 matrix has a non-zero determinant itself, so its rank is 2.

rank

1 2 34 5 67 8 9

= 2

The 3 × 3 matrix has a zero determinant, but has several 2 × 2submatrices with non-zero determinants, including∣∣∣∣[ 1 2

4 5

]∣∣∣∣ = −3





Rank and Determinant Example: Scaffolding

∣∣∣∣∣∣∣∣∣∣∣∣

1 1 −1 −1 0 −10 −9 1 4 0 70 0 1 1 −1 00 0 0 −3 2 00 0 0 0 1 10 0 0 0 0 −4

∣∣∣∣∣∣∣∣∣∣∣∣= −108

therefore, its rank is 6. The rank of [A b] is also 6, since the Amatrix itself is an easily-found submatrix with a non-zerodeterminant. Assuming that the {b} vector is non-zero, thescaffolding equation only has a solution ifrank([A b]) = rank(A) = n, where n is the number of equationsin the system (6).




Homework

Consider the three-mass, four-spring system shown below.Expanding the equations of motion from

∑Fx = max for each

mass using its free-body diagram results in the followingdifferential equations:

x1 +

(k1 + k2

m1

)x1 −

(k2

m1

)x2 = 0

x2 −(

k2

m2

)x1 +

(k2 + k3

m2

)x2 −

(k3

m2

)x3 = 0

x3 −(

k3

m3

)x2 +

(k3 + k4

m3

)x3 = 0

where k1 = 10 N/m, k2 = k3 = 40 N/m, andm1 = m2 = m3 = m4 = 1 kg. Assuming values for x1 · · · x3 arealso known, reformat the system of equations in matrix form.




Homework Figure


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Solution of Simultaneous Linear Algebraic Equations