2.4 Gauss and Gauss-Jordan Methods

2.4 Gauss and Gauss-Jordan Methods

Ayman Hashem Sakka

Islamic University of GazaFaculty of Science

Department of Mathematics

First Semester 2013-2014

Ayman Hashem Sakka 2.4 Gauss and Gauss-Jordan Methods


In Gaussian elimination, to eliminate the elementsbelow a diagonal element aii , we use the rowoperations

Rk −→ Rk −akiaii

Ri , k = i + 1, i + 2, · · · , n.

Thus problems occur when a diagonal element aii iszero or close to zero.

To avoid such problems we may interchange rowsand columns so that the elements aii has largestmagnitude. This process is called complete pivoting.If we interchange rows only, then the process is calledpartial pivoting.





Ri , k = i + 1, i + 2, · · · , n.







Ri , k = i + 1, i + 2, · · · , n.




Definition

( Partial pivoting)Partial pivoting is interchanging rows so that |aii | ≥ |aji | forj = i + 1, i + 2, · · · , n, before eliminating the elements below adiagonal element aii .

Example

Use Gaussian elimination and back-substitution to solve the linearsystem

−0.002x1 + 4x2 + 4x3 = 7.998,−2x1 + 2.906x2 − 5.387x3 = −4.481,3x1 − 4.031x2 − 3.112x3 = −4.143.

Use four-digit arithmetic.


Definition

( Partial pivoting)Partial pivoting is interchanging rows so that |aii | ≥ |aji | forj = i + 1, i + 2, · · · , n, before eliminating the elements below adiagonal element aii .

Example

Use Gaussian elimination and back-substitution to solve the linearsystem

−0.002x1 + 4x2 + 4x3 = 7.998,−2x1 + 2.906x2 − 5.387x3 = −4.481,3x1 − 4.031x2 − 3.112x3 = −4.143.



Solution:

Elimination: R2 −→ R2 − 1000R1, R3 −→ R3 + 1500R1

[A : b] −→

−0.002 4 4 | 7.9980 −3997 −4005 | −80020 5996 5997 | 11990

Elimination: R3 −→ R3 + 1.5R2

[A : b] −→

−0.002 4 4 | 7.9980 −3997 −4005 | −80020 0 −10.5 | −10

Back-substitution: x3 = 0.9524, x2 = 1.048, x1 = 1.8


Solution:


[A : b] −→

−0.002 4 4 | 7.9980 −3997 −4005 | −80020 5996 5997 | 11990


[A : b] −→

−0.002 4 4 | 7.9980 −3997 −4005 | −80020 0 −10.5 | −10



Solution:


[A : b] −→

−0.002 4 4 | 7.9980 −3997 −4005 | −80020 5996 5997 | 11990


[A : b] −→

−0.002 4 4 | 7.9980 −3997 −4005 | −80020 0 −10.5 | −10



Remark

Exact solution: x1 = x2 = x3 = 1.

Example

Use Gaussian elimination with partial pivoting andback-substitution to solve the linear system

−0.002x1 + 4x2 + 4x3 = 7.998,−2x1 + 2.906x2 − 5.387x3 = −4.481,3x1 − 4.031x2 − 3.112x3 = −4.143.



Remark

Exact solution: x1 = x2 = x3 = 1.

Example

Use Gaussian elimination with partial pivoting andback-substitution to solve the linear system

−0.002x1 + 4x2 + 4x3 = 7.998,−2x1 + 2.906x2 − 5.387x3 = −4.481,3x1 − 4.031x2 − 3.112x3 = −4.143.



Solution:

Pivoting: Since |a31| = max {|a11|, |a21|, |a31|} ,, we interchangethe first and third rows. Thus

[A : b] −→

3 −4.031 −3.112 | −4.143−2 2.906 −5.387 | −4.481−0.002 4 4 | 7.998

Elimination: R2 −→ R2 + 0.6667R1, R3 −→ R3 + 0.0007R1

[A : b] −→

3 −4.031 −3.112 | −4.1430 −0.2185 −7.462 | −7.2130 3.997 3.998 | 7.995


Solution:

Pivoting: Since |a31| = max {|a11|, |a21|, |a31|} ,, we interchangethe first and third rows. Thus

[A : b] −→

3 −4.031 −3.112 | −4.143−2 2.906 −5.387 | −4.481−0.002 4 4 | 7.998

Elimination: R2 −→ R2 + 0.6667R1, R3 −→ R3 + 0.0007R1

[A : b] −→

3 −4.031 −3.112 | −4.1430 −0.2185 −7.462 | −7.2130 3.997 3.998 | 7.995


Pivoting: Since |a32| = max {|a22|, |a32|} ,, we interchange thesecond and third rows. Thus

[A : b] −→

3 −4.031 −3.112 | −4.1430 3.997 3.998 | 7.9950 −0.2185 −7.462 | −7.213

Elimination: R3 −→ R3 − 0.0547R2

[A : b] −→

3 −4.031 −3.112 | −4.1430 3.997 3.998 | 7.9950 0 −7.681 | −7.680


Remark

We use pivoting to minimize round-off error and to avoid divisionby zero .



[A : b] −→

3 −4.031 −3.112 | −4.1430 3.997 3.998 | 7.9950 −0.2185 −7.462 | −7.213


[A : b] −→

3 −4.031 −3.112 | −4.1430 3.997 3.998 | 7.9950 0 −7.681 | −7.680


Remark




[A : b] −→

3 −4.031 −3.112 | −4.1430 3.997 3.998 | 7.9950 −0.2185 −7.462 | −7.213


[A : b] −→

3 −4.031 −3.112 | −4.1430 3.997 3.998 | 7.9950 0 −7.681 | −7.680


Remark




[A : b] −→

3 −4.031 −3.112 | −4.1430 3.997 3.998 | 7.9950 −0.2185 −7.462 | −7.213


[A : b] −→

3 −4.031 −3.112 | −4.1430 3.997 3.998 | 7.9950 0 −7.681 | −7.680


Remark



Gauss-Jordan Method

Gauss-Jordan method is an elimination method totransform the coefficient matrix A of a linear systemAx = b to the identity matrix I . As a result of thisprocess, the right-hand side b will be transformed tothe solution of the system.

Example

Use Gauss-Jordan method to solve the linear system

2x1 + 2x2 − 2x3 = 2,3x1 + 5x2 + x3 = 5,x1 + 2x2 = 6.


Gauss-Jordan Method

Gauss-Jordan method is an elimination method totransform the coefficient matrix A of a linear systemAx = b to the identity matrix I . As a result of thisprocess, the right-hand side b will be transformed tothe solution of the system.

Example

Use Gauss-Jordan method to solve the linear system

2x1 + 2x2 − 2x3 = 2,3x1 + 5x2 + x3 = 5,x1 + 2x2 = 6.


Remarks

(1) Pivoting is normally used with Gauss-Jordan methodto preserve arithmetic accuracy.

(2) To solve linear systems with the same coefficientmatrix A, but with different right-hand sidesb(1), b(2), · · · , b(k), we apply the row operations to

[A : b(1), b(2), · · · , b(k)]

and reduce it to

[A : b(1), b(2), · · · , b(k)].

Then the solutions of the systems are obtained byapplying back-substitution to the equivalent systems

Ax = b(j), j = 1, 2, · · · , k .


Remarks

(1) Pivoting is normally used with Gauss-Jordan methodto preserve arithmetic accuracy.

(2) To solve linear systems with the same coefficientmatrix A, but with different right-hand sidesb(1), b(2), · · · , b(k), we apply the row operations to

[A : b(1), b(2), · · · , b(k)]

and reduce it to

[A : b(1), b(2), · · · , b(k)].

Then the solutions of the systems are obtained byapplying back-substitution to the equivalent systems

Ax = b(j), j = 1, 2, · · · , k .


Example

Solve the systems Ax = b(i) by Gaussian elimination if

A =

4 −1 32 3 73 6 −1

, b(1) =

001

, b(2) =

−213

.

Solution: We will apply the elimination to the matrix

[A|b(1), b(2)] =

4 −1 3 | 0 −22 3 7 | 0 13 6 −1 | 1 3

.


Example

Solve the systems Ax = b(i) by Gaussian elimination if

A =

4 −1 32 3 73 6 −1

, b(1) =

001

, b(2) =

−213

.

Solution: We will apply the elimination to the matrix

[A|b(1), b(2)] =

4 −1 3 | 0 −22 3 7 | 0 13 6 −1 | 1 3

.


R2 −→ R2 − 0.5R1, R3 −→ R3 − 0.75R1

[A|b(1), b(2)] −→

4 −1 3 | 0 −20 3.5 5.5 | 0 20 6.75 −3.25 | 1 4.5

.

R3 −→ R3 − 1.92857R2

[A|b(1), b(2)] −→

4 −1 3 | 0 −20 3.5 5.5 | 0 20 0 −13.8571 | 1 0.6428

.Therefore the solutions of the systems are

x (1) =

0.08240.1134−0.0721

, x (2) =

−0.30410.6443−0.0463

.


R2 −→ R2 − 0.5R1, R3 −→ R3 − 0.75R1

[A|b(1), b(2)] −→

4 −1 3 | 0 −20 3.5 5.5 | 0 20 6.75 −3.25 | 1 4.5

.R3 −→ R3 − 1.92857R2

[A|b(1), b(2)] −→

4 −1 3 | 0 −20 3.5 5.5 | 0 20 0 −13.8571 | 1 0.6428

.

Therefore the solutions of the systems are

x (1) =

0.08240.1134−0.0721

, x (2) =

−0.30410.6443−0.0463

.


R2 −→ R2 − 0.5R1, R3 −→ R3 − 0.75R1

[A|b(1), b(2)] −→

4 −1 3 | 0 −20 3.5 5.5 | 0 20 6.75 −3.25 | 1 4.5

.R3 −→ R3 − 1.92857R2

[A|b(1), b(2)] −→

4 −1 3 | 0 −20 3.5 5.5 | 0 20 0 −13.8571 | 1 0.6428

.Therefore the solutions of the systems are

x (1) =

0.08240.1134−0.0721

, x (2) =

−0.30410.6443−0.0463

.Ayman Hashem Sakka 2.4 Gauss and Gauss-Jordan Methods

Scaling

Scaling is the operation of adjusting the coefficientsof a set of equations so that they are all of the sameorder of magnitude. To scale a linear system, wedivide each equation by the magnitude of the largestcoefficient in each equation.

That is, if A =

a11 a12 · · · a1na21 a22 · · · a2n

... · · ·...

...am1 am2 · · · amn

and

αi = max {|ai1|, · · · , |ain|}, then

[A |b ]−scaling− −→

a11α1

a12α1

· · · a1nα1

| b1α1

a21α2

a22α2

· · · a2nα2

| b2α2

...... · · ·

... |...

am1αm

am2αm

· · · amnαm

| bmαm


Scaling

Scaling is the operation of adjusting the coefficientsof a set of equations so that they are all of the sameorder of magnitude. To scale a linear system, wedivide each equation by the magnitude of the largestcoefficient in each equation.

That is, if A =

a11 a12 · · · a1na21 a22 · · · a2n

... · · ·...

...am1 am2 · · · amn

and

αi = max {|ai1|, · · · , |ain|}, then

[A |b ]−scaling− −→

a11α1

a12α1

· · · a1nα1

| b1α1

a21α2

a22α2

· · · a2nα2

| b2α2

...... · · ·

... |...

am1αm

am2αm

· · · amnαm

| bmαm


Example

Consider the system Ax = b, where

A =

200 1 1−1 100 2−50 3 −3

, b =

202101−50

.(a) Solve the system by Gaussian elimination with

pivoting.

(b) Solve the system by Gaussian elimination withpivoting and scaling.

Use three digit arithmetic rounding.


Solution:

(a)

[A : b ]R2+0.005R1−→

200 1 1 | 2020 100 2.01 | 1020 3.25 −3.75 | 0.5

R3−0.033R2−→

200 1 1 | 2020 100 2.01 | 1020 0 −2.82 | −2.87

Back-substitution givesx3 = 1.02, x2 = 0.999, x1 = 1


Solution:

(a)

[A : b ]R2+0.005R1−→

200 1 1 | 2020 100 2.01 | 1020 3.25 −3.75 | 0.5

R3−0.033R2−→

200 1 1 | 2020 100 2.01 | 1020 0 −2.82 | −2.87



Solution:

(a)

[A : b ]R2+0.005R1−→

200 1 1 | 2020 100 2.01 | 1020 3.25 −3.75 | 0.5

R3−0.033R2−→

200 1 1 | 2020 100 2.01 | 1020 0 −2.82 | −2.87



(b) Scaling implies

[A : b ] −→

1 0.005 0.005 | 1.01−0.01 1 0.02 | 1.01−1 0.06 −0.06 | −1

Elimination: R2 −→ R2 + 0.01R1, R3 −→ R3 + R1

[A : b ] −→

1 0.005 0.005 | 1.010 1 0.02 | 1.020 0.065 −0.055 | 0.01

R3 −→ R3 − 0.065R2

[A : b ] −→

1 0.005 0.005 | 1.010 1 0.02 | 1.020 0 −0.0563 | −0.0563

Back-substitution gives x3 = 1, x2 = 1, x1 = 1


(b) Scaling implies

[A : b ] −→

1 0.005 0.005 | 1.01−0.01 1 0.02 | 1.01−1 0.06 −0.06 | −1


[A : b ] −→

1 0.005 0.005 | 1.010 1 0.02 | 1.020 0.065 −0.055 | 0.01

R3 −→ R3 − 0.065R2

[A : b ] −→

1 0.005 0.005 | 1.010 1 0.02 | 1.020 0 −0.0563 | −0.0563



(b) Scaling implies

[A : b ] −→

1 0.005 0.005 | 1.01−0.01 1 0.02 | 1.01−1 0.06 −0.06 | −1


[A : b ] −→

1 0.005 0.005 | 1.010 1 0.02 | 1.020 0.065 −0.055 | 0.01

R3 −→ R3 − 0.065R2

[A : b ] −→

1 0.005 0.005 | 1.010 1 0.02 | 1.020 0 −0.0563 | −0.0563



(b) Scaling implies

[A : b ] −→

1 0.005 0.005 | 1.01−0.01 1 0.02 | 1.01−1 0.06 −0.06 | −1


[A : b ] −→

1 0.005 0.005 | 1.010 1 0.02 | 1.020 0.065 −0.055 | 0.01

R3 −→ R3 − 0.065R2

[A : b ] −→

1 0.005 0.005 | 1.010 1 0.02 | 1.020 0 −0.0563 | −0.0563



Remark

The exact solution of the above example is x1 = x2 = x3 = 1.Thus we see that scaling minimizes round-off error.


Remark

The exact solution of the above example is x1 = x2 = x3 = 1.Thus we see that scaling minimizes round-off error.


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2.4 Gauss and Gauss-Jordan Methods