Gauss Jordan

E. T. S. I. Caminos, Canales y Puertos 1

Gauss Jordan

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1. Motivate Study of Systems of Equations and particularly Systems of Linear Equations

2. Review steps of Gaussian Elimination

3. Examine how roundoff error can enter andbe magnified in Gaussian Elimination

4. Introduce Pivoting and Scaling as defenses against roundoff.

5. Consider what an engineer can do to generate well formulated problems.

Learning Objectives for Lecture

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Systems of Equations

• In Part 2 we have tried to determine the value x, satisfying f(x)=0. In this part we try to obtain the values x1,x2, xn, satisfying the system of equations:

• These systems can be linear or nonlinear, but in this part we deal with linear systems:

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• where a and b are constant coefficients, and n is the number of equations.

• Many of the engineering fundamental equations are based on conservation laws. In mathematical terms, these principles lead to balance or continuity equations relating the system behavior with respect to the amount of the magnitude being modelled and the extrenal stimuli acting on the system.

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• Matrices are rectangular sets of elements represented by a single symbol. If the set if horizontal it is called row, and if it is vertical, it is called column.

Row 2

Column 3

Row vector

Column

vector

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• There are some special types of matrices:

Symmetric matrix Identity matrix

Diagonal matrix Upper triangular matrix

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Banded matrix

All elements are null with the exception of thoise in a band centered around the main diagonal. This matrix has a band width of 3 and has the name of tridiagonal.

Half band width

Lower triangular matrix

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Linear Algebraic Equationsa11x1 + a12x2 + a13x3 + … + a1nxn = b1

a21x1 + a22x2 + a23x3 + … + a2nxn = b2

…..an1x1 + an2x2 + an3x3 + … + anxn = bn

where all aij's and bi's are constants.

In matrix form:

11 12 13 1n 1 121 22 23 2n 2 2

3 331 32 33 3n

n nn1 n2 n3 nn

a a a a x ba a a a x b

= x ba a a a

x ba a a a

n x n n x 1 n x 1or simply [A]{x} = {b}


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• Matrix representation of a system

Matrix product:

Resulting dimensions

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• Graphic Solution: Systems of equations are hyperplanes (straight lines, planes, etc.). The solution of a system is the intersection of these hyperplanes.

Compatible and determined system. Vectors are linearly independent. Unique solution. Determinant of A is non-null.

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Incompatible system, Linearly dependent vectors. Null determinant of A. There is no solution.

Compatible but undetermined system. Linearly dependent vectors. Null determinant of A. There exists an infinite number of solutions.

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Compatible and determined system. Linearly independent vectors. Nonnull determinant of A, but close to zero. There exists a solution but it is difficult to find precisely. It is an ill conditioned system leading to numerical errors.

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Gauss elimination• Naive Gauss elimination method: The Gauss’ method has two phases:

Forward elimination and backsustitution. In the first, the system is reduced to an upper triangular system:

• First, the unknown x1 is eliminated. To this end, the first row is multiplied by -a21/a11 and added to the second row. The same is done with all other succesive rows (n-1 times) until only the first equation contains the first unknown x1.

Pivotequation

substract

pivot

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Gauss elimination

• This operation is repeated with all variables xi, until an upper triangular matrix is obtained.

• Next, the system is solved by backsustitution.

• The number of operations (FLOPS) used in the Gauss method is:

Pass 1 Pass 2

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b. By elementary row manipulations, reduce [A|b] to [U|b'] where U is an upper triangular matrix:

DO i = 1 to n-1DO k = i+1 to nRow(k) = Row(k) - (aki/aii)*Row(i)ENDDO

ENDDO

11 12 1n 1 1 11 12 1n 121 22 2n 2 2 21 22 2n 2

n1 n2 nn n n n1 n2 nn n

a a a x b a a a ba a a x b a a a b

a a a x b a a a b

1. Forward Elimination (Row Manipulation):

a. Form augmented matrix [A|b]:

Gauss elimination

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Gauss elimination

2. Back Substitution

Solve the upper triangular system [U]{x} = {b´}

xn = b'n / unn

DO i = n-1 to 1 by (-1)

END

n

i ij jj i 1

iii

b u x

xu

11 12 13 1n 1 1

22 23 2n 2 2

33 3n 3 3

nn n n

u u u u x b

0 u u u x b

0 0 u u x b

0 0 0 u x b

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Consider the system of equations

123

x50 1 2 11 40 4 x = 22 6 30 3x

To 2 significant figures, the exact solution is:

true

0.016

x 0.041

0.091

We will use 2 decimal digit arithmetic with rounding.

Gauss elimination (example)

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Start with the augmented matrix:

Multiply the first row by –1/50

and add to second row.

Multiply the first row by –2/50 and add to third row:

Multiply the second row by –6/40

and add to third row:

50 1 2 11 40 4 22 6 30 3

50 1 2 10 40 4 20 6 30 3

50 1 2 10 40 4 20 0 29 2.7


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Now backsolve:

50 1 2 10 40 4 20 0 29 2.7

32.7

x 0.09329

(vs. 0.091, t = 2.2%)

(vs. 0.041, t = 2.5%)

(vs. 0.016, t = 0%)

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2 4xx 0.040

40

3 21

1 2x xx 0.016

50


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Consider an alternative solution interchanging rows:

After forward elimination, we obtain:

Now backsolve:

x3 = 0.095 (vs. 0.091, et = 4.4%)x2 = 0.020 (vs. 0.041, et = 50%)x1 = 0.000 (vs. 0.016, et = 100%)

Apparently, the order of the equations matters!

2 6 30 350 1 2 11 40 4 2

2 6 30 3

0 150 750 74

0 0 200 19


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WHAT HAPPENED?

• When we used 50 x1 + 1 x2 + 2 x3 = 1 to solve for x1, there was little change in other equations.

• When we used 2 x1 + 6 x2 + 30 x3 = 3 to solve for x1 it made BIG changes in the other equations. Some coefficients for other equations were lost!

The second equation has little to do with x1.

It has mainly to do with x3.

As a result we obtained LARGE numbers in the table, significant roundoff error occurred and information was lost.

Things didn't go well!

• If scaling factors | aji / aii | are 1 then the effect of roundoff

errors is diminished.


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Effect of diagonal dominance:

As a first approximation roots are:

xi bi / aii

Consider the previous examples: true

0.016x = 0.041

0.091

123

50 1 2 1 x 1/50 =0.021 40 4 2 x 2/40 =0.05

x 3/30 =0.102 6 30 3

123

2 6 30 3 x 3/2 =1.550 1 2 1 x 1/1 =1.0

x 2/4 = 0.501 40 4 2


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Goals: 1. Best accuracy (i.e. minimize error)

2. Parsimony (i.e. minimize effort)

Possible Problems:A. Zero on diagonal term ÷ by zero.B. Many floating point operations (flops) cause numerical

precision problems and propagation of errors.C. System may be ill-conditioned: det[A] 0.D. No solution or an infinite # of solutions: det[A] = 0.

Possible Remedies:A. Carry more significant figures (double precision).B. Pivot when the diagonal is close to zero.C. Scale to reduce round-off error.


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PIVOTING

A. Row pivoting (Partial Pivoting) - In any good routine, at each step i, find

maxk | aki | for k = i, i+1, i+2, ..., nMove corresponding row to pivot position.

(i) Avoids zero aii

(ii) Keeps numbers small & minimizes round-off,

(iii) Uses an equation with large | aki | to find xi

Maintains diagonal dominance. Row pivoting does not affect the order of the variables. Included in any good Gaussian Elimination routine.

Gauss elimination (pivoting)

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B. Column pivoting - Reorder remaining variables xj

for j = i, . . . ,n so get largest | aji |Column pivoting changes the order of the unknowns, xi, and thus leads to complexity in the algorithm. Not usually done.

C. Complete or Full pivotingPerforming both row pivoting and column pivoting.(If [A] is symmetric, needed to preserve symmetry.)


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How to fool pivoting:

Multiply the third equation by 100 and then performing pivoting will yield:

Forward elimination then yields (2-digit arithmetic):

Backsolution yields:

x3 = 0.095 (vs. 0.091, et = 4.4%)x2 = 0.020 (vs. 0.041, et = 50.0%)x1 = 0.000 (vs. 0.016, et = 100%)

The order of the rows is still poor!!

200 600 3000 300

50 1 2 1

1 40 4 2

200 600 3000 300

0 150 750 74

0 0 200 19


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SCALINGA. Express all equations (and variables) in comparable units so all

elements of [A] are about the same size.

B. If that fails, and maxj |aij| varies widely across the rows, replace each row i by:

aij

This makes the largest coefficient |aij| of each equation equal to 1 and the largest element of [A] equal to 1 or -1

NOTE: Routines generally do not scale automatically; scaling can cause round-off error too!

SOLUTIONS • Don't actually scale, but use hypothetical scaling factors to determine

what pivoting is necessary.• Scale only by powers of 2: no roundoff or division required.

ij

j ij

a

max | a |

Gauss elimination (scaling)

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If the units of x1 were expressed in µg instead of mg the matrix might read:

50 1 2 1

1 40 4 2

2 6 30 3

50000 1 2 1

1000 40 4 2

2000 6 3 3

1 0.00002 0.00001 0.00001

1 0.04 0.004 0.002

1 0.003 0.015 0.0015

How to fool scaling:A poor choice of units can undermine the value of scaling.

Begin with our original example:

Scaling then yields:

Which equation is used to determine x1 ? Why bother to scale ?

Gauss elimination (scaling)

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OPERATION COUNTING

In numerical scientific calculations, the number of multiplies & divides often determines CPU time. (This represents the numerical effort!)

One floating point multiply or divide (plus any associated adds or subtracts) is called a FLOP. (The adds/subtracts use little time compared to the multiplies/divides.) FLOP = FLoating point OPeration.

Examples: a * x + b a / x – b

Gauss elimination (operation counting)

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Useful identities in counting FLOPS:m m

i 1 i 1

1) c f (i) c f (i)

m m m

i 1 i 1 i 1

2) [f (i) g(i)] f (i) g(i)

m

i 1

3) 1 1 1 1 m

m

i k

4) 1 m k 1

m 2

i 1

m(m 1) m5) i 1 2 3 4 m (m)

2 2

O

m 32 2 2 2 2

i 1

m(m 1)(2m 1) m6) i 1 2 m (m )

6 3

O

O(mn) means that there are terms of order mn and lower.


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Simple Example of Operation Counting:

DO i = 1 to nY(i) = X(i)/i – 1

ENDDO

X(i) and Y(i) are arrays whose values change when i changes. In each iteration

X(i)/i – 1

represents one FLOP because it requires one division (& one subtraction).

The DO loop extends over i from 1 to n iterations:n

i 1

1 n FLOPS


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Another Example of Operation Counting:

DO i = 1 to nY(i) = X(i) X(i) + 1DO j = i to n

Z(j) = [ Y(j) / X(i) ] Y(j) + X(i)ENDDO

ENDDO

With nested loops, always start from the innermost loop.

[Y(j)/X(i)] * Y(j) + X(i) represents 2 FLOPSn n

j i j i

2 2 1 2(n i 1) FLOPS


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For the outer i-loop: X(i) • X(i) + 1 represents 1 FLOP

n n n

i 1 i 1 i 1

1 2 n i 1) 3 2n 1 2 i

n(n 1)3 2n n 2

2

= 3n +2n2 - n2 - n

= n2 + 2n = n2 + O(n)


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Forward Elimination:DO k = 1 to n–1

DO i = k+1 to nr = A(i,k)/A(k,k)

DO j = k+1 to n A(i,j)=A(i,j) – r*A(k,j)ENDDO

B(i) = B(i) – r*B(k)ENDDO

ENDDO


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Operation Counting for Gaussian Elimination

Back Substitution:X(n) = B(n)/A(n,n)

DO i = n–1 to 1 by –1SUM = 0

DO j = i+1 to nSUM = SUM + A(i,j)*X(j)

ENDDO

X(i) = [B(i) – SUM]/A(i,i)

ENDDO


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Forward Elimination

Inner loop:

Second loop:

n

j=k+1

1 = n - (k +1) +1 = n - k

n

i=k+1

2 + (n - k) (2 n) k (n k) = (n2 + 2n) – 2(n + 1)k + k2


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Forward Elimination (cont'd)

Outer loop =1

2 2

1

[( 2 ) 2( 1) ]n

k

n n n k k

n 1 n 1 n 1

2 2

k 1 k 1 k 1

(n 2n) 1 2(n 1) k k

2 (n 1))(n)(n 2n)(n 1) 2(n 1)

2

(n 1)(n)(2n 1)

6

3n 2= + (n )3

O


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Back Substitution

Inner Loop:

Outer Loop:

n

j i 1

1 n (i 1) 1

n - i

n 1 n 1 n 1

i 1 i 1 i 1

1 (n i) (1 n) 1 i

(n 1)n

(1 n)(n 1)2

2n= + (n)

2O


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Total flops = Forward Elimination + Back Substitution= n3/3 + O (n2) + n2/2 + O (n) n3/3 + O (n2)

To convert (A,b) to (U,b') requires n3/3, plus terms of order n2 and smaller, flops.

To back solve requires:

1 + 2 + 3 + 4 + . . . + n = n (n+1) / 2 flops;

Grand Total: the entire effort requires n3/3 + O(n2) flops altogether.


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Diagonalization by both forward and backward elimination in each column.

Perform elimination both backwards and forwards until:

Operation count for Gauss-Jordan is: (slower than Gauss elimination)

11 12 13 1n 1 1

21 22 23 2n 2 2

31 32 33 3n 3 3

n1 n2 n3 nn nn n

a a a ... a x b

a a a ... a x b

=a a a ... a x b

...

a a a ... a x b

1 1

2 2

3 3

n n

x x1 0 0 0

x x0 1 0 0

= x x0 0 1 0

x x0 0 0 1

32n

O(n )2

Gauss-Jordan Elimination

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Example (two-digit arithmetic):

50 1 2 11 40 4 22 6 30 3

1 0.02 0.04 0.020 40 4 20 6 30 3

1 0 0.038 0.0190 1 0.1 0.050 0 29 2.7

1 0 0 0.0150 1 0 0.0410 0 1 0.093

x1 = 0.015 (vs. 0.016, t = 6.3%)x2 = 0.041 (vs. 0.041, t = 0%)x3 = 0.093 (vs. 0.091, t = 2.2%)

Gauss-Jordan Elimination

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The solution of: [A]{x} = {b} is: {x} = [A]-1{b}

where [A]-1 is the inverse matrix of [A]

Consider: [A] [A]-1 = [ I ]

1) Create the augmented matrix: [ A | I ]

2) Apply Gauss-Jordan elimination:

==> [ I | A-1 ]

Gauss-Jordan Matrix Inversion

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Gauss-Jordan Matrix Inversion (with 2 digit arithmetic):

50 1 2 1 0 0A I = 1 40 4 0 1 0

2 6 30 0 0 1

1 0.02 0.04 0.02 0 00 40 4 -0.02 1 00 6 30 -0.04 0 1

MATRIX INVERSE [A-1]

1 0 0 0.02 0.00029 0.0014

0 1 0 0.00037 0.026 0.0036

0 0 1 0.0013 0.0054 0.036

1 0 0.038 0.02 0.005 0

0 1 0.1 0.0005 0.025 0

0 0 28 0.037 0.15 1


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50 1 2 0.020 -0.00029 -0.0014 0.997 0.13 0.0022 40 4 -0.00037 0.026 -0.0036 0.000 1.016 0.0012 6 30 -0.0013 -0.0054 0.036 0.001 0.012 1.056

CHECK:[ A ] [ A ]-1 = [ I ]

[ A ]-1 { b } = { x }

0.020 -0.0029 -0.0014 1 0.015-0.00037 0.026 -0.0036 2 0.033-0.0013 -0.0054 0.036 3 0.099

true

0.016x 0.041

0.091

0.016x 0.040

0.093

GaussianElimination


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LU decomposition

• LU decomposition - The LU decomposition is a method that uses the elimination techniques to transform the matrix A in a product of triangular matrices. This is specially useful to solve systems with different vectors b, because the same decomposition of matrix A can be used to evaluate in an efficient form, by forward and backward sustitution, all cases.

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LU decomposition

Decomposition Initial system

Transformed system 1Substitution

Transformed system 2

Forward sustitutionBackward sustitution

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LU decomposition

• LU decomposition is very much related to Gauss method, because the upper triangular matrix is also looked for in the LU decomposition. Thus, only the lower triangular matrix is needed.

• Surprisingly, during the Gauss elimination procedure, this matrix L is obtained, but one is not aware of this fact. The factors we use to get zeroes below the main diagonal are the elements of this matrix L.

Substract

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LU decomposition

resto

resto

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Basic Approach

Consider [A]{x} = {b}

a) Gauss-type "decomposition" of [A] into [L][U] n3/3 flops [A]{x} = {b} becomes [L] [U]{x} = {b}; let [U]{x} {d}

b) First solve [L] {d} = {b} for {d} by forward subst. n2/2 flops

c) Then solve [U]{x} = {d} for {x} by back substitution n2/2 flops

LU decomposition (Complexity)

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1ij

1 0L = a 1

0ij

0 0L = a 0

ij0

0 aU =

0 0

ij1

1 aU =

0 1

1 0D = 0 1

[A] = [L] + [U0]

[A] = [L0] + [U]

[A] = [L0] + [U0] + [D]

11ij nn

a 0L = a a

11 ij

nn

a aU =

0 a

[A] = [L1] [U]

[A] = [L] [U1]

LU Decompostion: notation

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LU Decomposition VariationsDoolittle [L1][U] General [A]Crout [L][U1] General [A]Cholesky[L][L] T Pos. Def. Symmetric [A]

Cholesky works only for Positive Definite Symmetric matrices

Doolittle versus Crout:• Doolittle just stores Gaussian elimination factors where Crout

uses a different series of calculations (see C&C 10.1.4).• Both decompose [A] into [L] and [U] in n3/3 FLOPS

• Different location of diagonal of 1's• Crout uses each element of [A] only once so the same array

can be used for [A] and [L\U] saving computer memory!

LU decomposition

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Matrix Inversion

Definition of a matrix inverse:

[A] [A]-1 = [ I ]

==> [A] {x} = {b}

[A]-1 {b} = {x}

First Rule: Don’t do it.

(numerically unstable calculation)

LU decomposition

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Matrix InversionIf you really must --

1) Gaussian elimination: [A | I ] –> [U | B'] ==> A-1

2) Gauss-Jordan: [A | I ] ==> [I | A-1 ] Inversion will take

n3 + O(n2)flops if one is careful about where zeros are (taking advantage of the sparseness of the matrix)

Naive applications (without optimization) take 4n3/3 + O(n2) flops. For example, LU decomposition requires n3/3 + O(n2) flops. Back solving twice with n unit vectors ei:

2 n (n2/2) = n3 flops.

Altogether: n3/3 + n3 = 4n3/3 + O(n2) flops

LU decomposition

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