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Chapter 5 Simultaneous Linear Equations. Many engineering and scientific problems can be formulated in terms of systems of simultaneous linear equations.

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Chapter 5 Simultaneous Linear Equations

  • Many engineering and scientific problems can be formulated in terms of systems of simultaneous linear equations.

  • In a system consisting of only a few equations, a solution can be found analytically using the standard methods from algebra, such as substitution.


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Kirchhoff’s law

I: current R: resistance

V: voltage


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Assume R1=2, R2=4, R3=5, V1=6, and V2=2. We get the following system of three linear simultaneous equations.

The solution to these three equations produces the current flows in the network.


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General Form for a System of Equations

aij : known coefficient

Xj : unknown variable

Ci : known contant

  • Assume # of unknowns = # of equations

  • Assume the equations are linearly independent; that is, any one equation is not a linear combination of any of the other equations.


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The linear system can be written in a matrix-vector form:

Combining A and C, it can be expressed as:



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Solution of Two Equations

It can be solved by substitution.

With substitution, we get


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Classification of Systems of Equations

  • Systems that have unique solutions




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Permissible Operations

  • Rule 1: The solution is not changed if the order of the equations is changed.


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Gaussian Elimination divided by a nonzero constant without changing the solution.

  • Gaussian elimination procedure:

    Phase 1: forward pass

    Phase 2: back substitution

  • The the forward pass is to apply the three permissible operations to transform the original matrix to an upper-triangular matrix:


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Written in terms of individual equations: divided by a nonzero constant without changing the solution.

  • Back substitution:


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Example: Gaussian Elimination Procedure divided by a nonzero constant without changing the solution.

Represented in the matrix form:


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  • As the divided by a nonzero constant without changing the solution.step 1 in the forward pass, we will convert the element a11 (a11 is called the pivot for row 1) to 1 and eliminate, that is set to zero, all the other elements in the first column.


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  • Step 2: divided by a nonzero constant without changing the solution.

  • Step 3:


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  • Step 4: divided by a nonzero constant without changing the solution.

    It represents:


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  • Step 3: divided by a nonzero constant without changing the solution.

  • The solution is:

X1 = 1, X2 = 2, X3 = 3, X4 = 4


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Gauss-Jordan Elimination divided by a nonzero constant without changing the solution.

  • The Gaussian elimination procedure requires a forward pass to transform the coefficient matrix into an upper-triangle form.

  • In Gauss-Jordan elimination, all coefficients in a column except for the pivot element are eliminated.

  • In Gauss-Jordan elimination, the solution is obtained directly after the forward pass; there is no back substitution phase.

  • The Gauss-Jordan method needs more computational effort than Gaussian elimination.


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Example: Gauss-Jordan Elimination divided by a nonzero constant without changing the solution.

  • Step1:

  • Step 2:


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  • Step 3: divided by a nonzero constant without changing the solution.

  • Step 4:


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Accumulated Round-off Errors divided by a nonzero constant without changing the solution.

  • Problems with round-off and truncation are most likely to occur when the coefficients in the equations differ by several orders of magnitude.

  • Round-off problems can be reduced by rearranging the equations such that the largest coefficient in each equation is placed on the principal diagonal of the matrix.

  • The equations should be ordered such that the equation having the largest pivot is reduced first, followed by the equation having the next largest, and so on.


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Programming Gaussian Elimination divided by a nonzero constant without changing the solution.

  • Forward pass:

  • Loop over each row i, making each row i in turn the pivot row.

  • Normalize the elements of the pivot row (row i) by dividing each element in the row by aii as follows:


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3. divided by a nonzero constant without changing the solution.Loop over rows( i + 1) to n below the pivot row and reduce the elements in each row as follows:

  • Back substitution

1. For the last row n:

2. For rows (n-1) through 1,


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LU Decomposition divided by a nonzero constant without changing the solution.

  • A matrix A can be decomposed into L and U, where L is a lower-triangular matrix and U is a upper-triangular matrix.

    LU=A


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  • L divided by a nonzero constant without changing the solution.and U can be determined as follows:


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AX=C divided by a nonzero constant without changing the solution.

LUX=C, we have LE=C and UX=E

  • To calculate LE=C (forward substitution)

  • To calculate UX=E (back substitution)


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Relation between LU Decomposition and Gaussian Elimination divided by a nonzero constant without changing the solution.

  • In the LU decomposition, matrix U is equivalent to the upper triangular matrix obtained in the forward pass in Gaussian elimination.

  • The calculation of UX=E is equivalent to the back substitution in Gaussian elimination.


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Example: LU Decompostion divided by a nonzero constant without changing the solution.

Applying LU decomposition:


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Thus, the L and U matrices are divided by a nonzero constant without changing the solution.

Forward substitution:


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Back substitution: divided by a nonzero constant without changing the solution.


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Cholesky Decomposition for Symmetric Matrices divided by a nonzero constant without changing the solution.

  • A symmetric matrix A:

  • Cholesky decomposition for a symmetric matrix A


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Matrix divided by a nonzero constant without changing the solution.L can be computed as follows:


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Example: Cholesky Decomposition divided by a nonzero constant without changing the solution.

We can obtain the following:


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Therefore, the divided by a nonzero constant without changing the solution.L matrix is

The validity can be verified as


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Iterative Methods divided by a nonzero constant without changing the solution.

  • Elimination methods like the Gaussian elimination procedure are often called direct equation-solving methods. An iterative method is a trial-and error procedure.

  • In iterative methods, we can assume a solution, that is, a set of estimates for the unknowns, and successively refine our estimate of the solution through some set of rules.

  • A major advantage of iterative methods is that they can be used to solve nonlinear simultaneous equations, a task that is not possible using direct elimination methods.


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Jacobi Iteration divided by a nonzero constant without changing the solution.

  • Each equation is rearranged to produce an expression for a single unknown.


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Example: Jacobi Iteration divided by a nonzero constant without changing the solution.

Rearrange each equation as follows:


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Assume an initial estimate for the solution: divided by a nonzero constant without changing the solution.X1=X2=X3=1. First iteration:

Second iteration:

The solution is shown in the next table.


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Table: divided by a nonzero constant without changing the solution.Example of Jacobi Iteration


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Gauss-Seidel Iteration divided by a nonzero constant without changing the solution.

  • In the Jacobi iteration procedure, we always complete a full iteration cycle over all the equations before updating our solution estimates. In the Gauss-Seidel iteration procedure, we update each unknown as soon as a new estimate of that unknown is computed.

  • Example: Gauss-Seidel Iteration


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Assume an initial solution estimate of divided by a nonzero constant without changing the solution.X1=X2=X3=1.

First iteration:

Second iteration:


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Table: divided by a nonzero constant without changing the solution.Example of Gauss-Seidel Iteration

Jocobi iteration method requires 13 iterations to reach the accuracy of 3 decimal places. Gauss-Seidel iteration method needs 7 iterations.


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Convergence Considerations of the Iterative Methods divided by a nonzero constant without changing the solution.

  • Both the Jacobi and Gauss-Seidel iterative methods may diverge.

  • Interchange the order of equations in the above example, and solve it by the Gauss-Seidel method:


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Table: divided by a nonzero constant without changing the solution. Divergence of Gauss-Seidel Iteration

  • The solution in the above table would not converge. That is, we can not get a solution.

  • The divergence of the iterative calculation does not imply there is no solution, since it is a permissible operation to interchange the order in a set of equations.


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Convergence and Divergence of Gauss-Seidel Iteration divided by a nonzero constant without changing the solution.

If we solve both equations individually for X1 we get


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  • It will divided by a nonzero constant without changing the solution.converge when the absolute value of the slope of f1 is less than the absolute value the slope of f2. Thus the equations should be arranged such that X1 is expressed in terms of X2 with the following conditions:

  • For a system with more than 2 equations, we should select the equation with the largest coefficient as the first equation, the equation with the largest coefficient in the remaining equations as the second equation, andso on.


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Cramer’s rule divided by a nonzero constant without changing the solution.

Cramer’s rule for obtaining Xi:

An example of |Ai| :


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Example: Cramer’s Rule divided by a nonzero constant without changing the solution.


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The solution is divided by a nonzero constant without changing the solution.


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Matrix Inversion divided by a nonzero constant without changing the solution.

  • The inverse of a matrix P is defined by the following equation

    in which I is the identity or unit matrix and both P and I are square matrices.

  • The values of P-1can be computed by solving a set of n2 simultaneous equations.

.


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Let the elements of divided by a nonzero constant without changing the solution.P-1 and P be denoted asqij and pij.


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