SYSTEM OF EQUATIONS

System of linear algebraic equations

Index:Basics Gauss elimination methodGauss iterative method:Gauss Jacobi methodGauss sidle method

Basics: A system of linear equation in n unknowns can be rewritten as:

This system consists of linear equations, each with coefficients, and has unknowns which have to fulfill the set of equations simultaneously. To simplify notation, it is possible to rewrite the above equations in matrix notation:

Simply ,we can say

Where A is called matrix of coefficient of order m*n

X is a simple vector of order B is a simple vector of order

We can write a system of linear equations as a matrix by writing only the coefficients and constants that appear in the equations.

This is called the augmented matrix of the system.

Elementary Row OperationsTo solve linear systems correspond to operations on the rows of the augmented matrix of the system.

For example, adding a multiple of one equation to another corresponds to adding a multiple of one row to another.

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For example,Add a multiple of one row to another.Multiply a row by a nonzero constant.Interchange two rows.

Note that performing any of these operations on the augmented matrix of a system does not change its solution.

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We use the following notation to describe the elementary row operations:SymbolDescriptionRi + kRj RiChange the ith row by adding k times row j to it.Then, put the result back in row i.kRiMultiply the ith row by k.Ri RjInterchange the ith and jth rows.

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Gauss eliminationGauss iterative methodGauss Jacobigauss seidel

Gaussian EliminationIn general, to solve a system of linear equations using its augmented matrix, we use elementary row operations to arrive at a matrix in a certain form.This form is described as follows.

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Row-Echelon Form

A matrix is in row-echelon form if it satisfies the following conditions:The first nonzero number in each row (reading from left to right) is 1. This is called the leading entry.The leading entry in each row is to the right of the leading entry in the row immediately above it.All rows consisting entirely of zeros are at the bottom of the matrix.

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Reduced Row-Echelon FormA matrix is in reduced row-echelon form if it is in row-echelon form and also satisfies the following condition. Every number above and below each leading entry is a 0.

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Not in row-echelon form:

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Row-Echelon FormReduced Row-Echelon Form

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Start by obtaining 1 in the top left corner.Then, obtain zeros below that 1 by adding appropriate multiples of the first row to the rows below it.

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We now discuss a systematic way to put a matrix in row-echelon form using elementary row operations.

We see how the process might work for a 3 x 4 matrix.

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Next, obtain a leading 1 in the next row.Then, obtain zeros below that 1.

At each stage, make sure every leading entry is to the right of the leading entry in the row above it.Rearrange the rows if necessary.

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Solve the system of linear equations using Gaussian elimination.

We first write the augmented matrix of the system.Then, we use elementary row operations to put it in row-echelon form.

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We now have an equivalent matrix in row-echelon form.The corresponding system of equations is:

We use back-substitution to solve the system.

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y + 4(2) = 7 y = 1 x + 2(1) (2) = 1 x = 3

The solution of the system is:(3, 1, 2)

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Gauss JacobiLet the set of equations given be:

Where the element of the diagonal are the largest in each column.We can get the values of unknowns as follows:

Rewrite the equations as:

Then make initial approximation of the solution x1,x2,..xn.

27Convergence of Jacobi iteration

Evaluate the infinity(maximum row sum) norm of EDiagonally dominant matrixIf A is a diagonally dominant matrix, then Jacobi iteration converges for any initial vector

28Example 1 (Jacobi Iteration)

Diagonally dominant matrix

29Example 1 continued...

Matrix is diagonally dominant, Jacobi iterations are converging

30Example 2

The matrix is not diagonally dominant

31Example 2 continued...

The residual term is increasing at each iteration, so the iterations are diverging.Note that the matrix is not diagonally dominant

Gauss SeidelLike for Gauss Jacobi well first ensure that the elements of the diagonal are the maximum valuesThen solve the 1s equation for the 1st unknown, 2nd for the 2nd unknown, etc.First iteration: Find the value of x1 by substituting the initial values of other unknownsFind the value of x2 by substituting the current value of x1 and initial values of other unknowns. and continue......Second iteration: Substitute all the values obtained in 1st iteration and find x1Use the current values of other unknowns, find x2 and continue....

33Convergence of Gauss-Seidel iterationGS iteration converges for any initial vector if A is a diagonally dominant matrixGS iteration converges for any initial vector if A is a symmetric and positive definite matrixMatrix A is positive definite ifxTAx>0 for every nonzero x vector

Gauss-Seidel Method: Example Given the system of equations

With an initial guess ofThe coefficient matrix is:

Will the solution converge using the Gauss-Siedel method?

Checking if the coefficient matrix is diagonally dominant

The inequalities are all true and at least one row is strictly greater than, Therefore: The solution should converge using the Gauss-Siedel Method

Rewriting each equation

With an initial guess of

After Iteration #1

Substituting the x values into the equationsAfter Iteration #2

Iterationa1a2a31234560.500000.146790.742750.946750.991770.999194.90003.71533.16443.02813.00343.00013.09233.81183.97083.99714.00014.0001

Repeating more iterations, the following values are obtained

The solution obtained is close to the exact solution of .

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