ME 210 Applied Mathematics for Mechanical Engineerscourses.me.metu.edu.tr/courses/me210/... · ME...

ME 210 Applied Mathematics for Mechanical Enginee rs

Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 1/26

Gauss(ian) Elimination

This is a systematic method in which

• First variable is eliminated from the (m–1) equations starting from the 2nd

equation up until the last equation using 1st equation,

• Second variable is eliminated from the (m–2) equations starting from the 3rd equation up until the last equation using 2nd equation,

• This process is continued until no more variable and/or equation is left.

• Then, the solution is obtained in the reverse order starting from the last one by

a simple back substitution .

This is a powerful method which also yields all the information on the consistency of equations as well as the uniqueness of the solution, if exists.

The method has several variants and some advanced versions. However, its very

basic version will be covered here.



The procedure is applied to the augmented matrix [A¦b] by performing a set of basic (elementary) row operations such that the matrix [A¦b] will be transformed into what is so-called as the row echelon form in which all elements of [C] = [A¦b] below its elements cii {i = 1, 2, …, min(m,n+1)} become zero.

Some basic row echelon forms of [A ¦ b]

xx0000

xxx000

xxxx00

xxxxx0

xxxxxx

L

L

MMMOMMM

L

L

L

xxx000

xxxx00

xxxxx0

xxxxxx

L

MMMOMMM

L

L

L

x00000

xx0000

xxx000

xxxx00

xxxxx0

xxxxxx

L

L

L

MMMOMMM

L

L

L

000000 L

000000 L

000000 L

000000 LSign of redundancy

Sign of inconsistencySign of

unique solutionSign of

multiple solutions

IMPORTANT WARNING: NEVER PERFORM ANY COLUMN OPERATIONS! WHY?



Row operations used in Gauss Elimination Method are as follows.

• Interchange of two rows.

• Multiplication of any row by a nonzero scalar.

• Replacement of a row by the sum of multiple of itself and another multiple of

the pivot row .

The pivot row of a [A¦b] matrix is its row located at the same position as the location of the column containing the coefficients of the variable to be eliminated.

Examples:

72233

01124

65108 ← The pivot row

1st variable is to be eliminated in rows below it.

3300

1100

0740

8642

← The pivot row

3rd variable is to be eliminated in rows below it.



→←←

1100

0740

3310

8642

R2R4R4R2

3310

0740

1100

8642

3300

1100

0740

8642

72233

01124

65108

pivot elementpivot element

pivot element

pivot element

The pivot or pivot element of a [A¦b] matrix is the element of the pivot row located

at the same position as the location of the pivot row.

Examples:

The pivot element of a pivot row should be a nonzero entry.

If there is zero in its location, then its row should be interchanged by one of the

rows below that row to have a nonzero entry at its position in order declare it as

pivot element and its row as pivot row .

Example:



−−−

−=→

3111

53101

2212

b][AM

3xxx

53x10xx

22xx2x

321

321

321

−=++−=−+=+−

Example: (m=n and a unique solution exists)

444 3444 21

R1R3*2:R32element Pivot1Row row Pivot

neliminatio 1 Column

4410

88210

2212

+=

=

−−

−→

444 3444 21

M

R1R2*2:R22element Pivot1Row row Pivotneliminatio 1 Column

3111

88210

2212

b][A

−=

=

−−−

−→

44444 344444 21

R2R3*21:R321 element Pivot2Row row Pivot


929200

88210

2212

−=

=

−−

−→

Back substitution:

92 x3 = –92 → x3 = –1

21 x2 – 8 x3 = 8 → 21 x2 – 8 (–1) = 8 → x2 = 0

2 x1 – x2 + 2 x3 = 2 → 2 x1 – (0) + 2 (–1) = 2 → x1 = 2

−=

1

0

2

[x]

Solution:

Unique solution



Solve the following set of equations by Gauss elimination

1 2 3 4

1 2 3 4

1 2 3 4

1 2 3 4

x + 2 x + 3 x + 4 x = 5

2 x + x + 4 x + x = 2

3 x + 4 x + x + 5 x = 14

2 x + 3 x + 5 x + 2 x = 4

1 2 3 4 5 1 2 3 4 5

2 1 4 1 2 0 -3 -2 -7 -8 =>

3 4 1 5 14 0 0 -6 5

2 3 5 2 4

11

0 0 0 -71 -71

Solution:

x4 = 1 x3 = -1 x2 = 1 x1 = 2

Example: (m=n and a unique solution exists)

Unique solution



Example: (m>n, but a unique solution exists)

43421

M

R2*3R1*2:R23element Pivot1Row row Pivot


1331

3570

423

b][A

−=

=

→

44 344 21

R3*3-R1:R33element Pivot1Row row Pivotneliminatio 1 Column

3570

3570

423

==

−−→

43421

R3R2:R37 element Pivot2Row row Pivot


000

3570

423

+=

=

→

Back substitution:

7 x2 = 35 → x2 = 5

3 x1 + 2 x2 = 4 → 3 x1 + 2 (5) = 4 → x1 = –2

−=

5

2[x]

Unique solution

Solution:133xx

9x2x

42x3x

21

21

21

=+−=−

=+

−−=→

1331

912

423

b][AM

redundancy

Twoindependentequations



Note that m = 3, n = 2, rank [A] = 2, rank [A ¦ b] = 2; hence a solution exists and it is

unique.

Notice that the last line of the augmented matrix [A ¦ b] is transformed into all zeros

indicating the redundancy existing in the set of equations.

Therefore, m – n = 3 – 2 = 1 of 3 equations must be redundant.

As a matter of fact, for large scale systems, number of such all–zero rows gives the number of redundant equations, which is when subtracted from number of

equations m, gives the number of independent equations.

However, it is not too clear to spot the structure of this redundancy by a simple inspection, which becomes even impossible when m and/or becomes large.



=→

=++=++=++

10233

9122

6111

b][A

102x3x3x

9x2x2x

6xxx

321

321

321

M

Note that m = 3, n = 3, rank[A] = 2, rank[A¦b] = 3.

Hence the equations are inconsistent, therefore no solution exists.

For large scale systems, number of such rows gives the number of inconsistentequations.

No solution

44 344 2144 344 21444 3444 21

M

R2R3:R3R33R1:R3R2R1*2:R2

5000

3100

6111

8100

3100

6111

10233

3100

6111

b][A

−−−

→

→

→indication of

inconsistency

Example: (m=n , no solution)

row echelon form is obtained

additional row operation



=→

=++=+=+=+

4233

3110

2101

1011

b][A

42x3x3x

3xx

2xx

1xx

321

32

31

21

M

44 344 2144 344 2144 344 2144 344 21

M

R4R3:R4R2R3:R3R1*3R4:R4R1R2:R2

3000

4200

1110

1011

1200

4200

1110

1011

1200

3110

1110

1011

4233

3110

1110

1011

b][A

−+−−

−→

−→

−→

−→

No solutionNote that m = 4, n = 3, rank [A] = 3, rank [A¦b] = 4


Example: (m>n , no solution)

indication of inconsistency



444 3444 21444 3444 21

M

R2*2R3:R3R1R3:R3

10000

21100

10011

52200

21100

10011

b][A

−−

→

→

Example: (m<n , no solution)

No solution

Note that m = 3, n = 4, rank[A] = 2, rank[A¦b] = 3


=→

=+++=+=+

62211

21100

10011

b][A

62x2xxx

2xx

1xx

4321

43

21

M

indication of inconsistency





=→

=+=+=+

844

422

211

b][A

84x4x

42x2x

2xx

21

21

21

M

Example: (m>n , multiple solutions)

Family of solutions

Set x2 = c. Then, the solution is expressed as: 1

2

x 2 - c 2 -1 = = + c

x c 0 1

4342143421

M

R1*4R3:R3R1*2R2:R2

000

000

211

844

000

211

b][A

−−

⇒

⇒ indication of 2 redundant equations

only 1 independent equation

Note that m = 3, n = 2, rank[A] = 1, rank[A¦b] = 1Hence a solution exists and it is multiple with {n–rank[A] = 2–1 = 1} parameter.



Example: (m<n , multiple solutions)

−−−−

−−−−

=→

=+−+−=−++−=+−+−=+−+−

112321

245942

359863

46111184

b][A

1x2x3x2xx

24x5x9x4x2x

35x9x8x6x3x

46x11x11x8x4x

54321

54321

54321

54321

M

44444 344444 2144444 344444 21

M

R1*4R4:R4

R1*2R3:R3

R1*3R2:R2R4&R1eInterchang

023100

069300

023100

112321

46111184

245942

359863

112321

b][A

−−−

−−−

−−−−

→

−−−−

−−−−

→

Note that m = 4, n = 5, and it can further be shown that rank[A] = 2, rank[A¦b] = 2Hence a solution exists and it is multiple with {n–rank[A] = 5–2 = 3} parameters.



44444 344444 21

M

R33R4:R4

000000

069300

023100

112321

023100

069300

023100

112321

b][A

+

−−−−−

→

−−−

−−−−

→

only 2 independent

equations

4444 34444 21

M

R2*3R3:R3

000000

000000

023100

112321

000000

069300

023100

112321

b][A

+

−−−−

→

−−−−−

→

3-parameter solutions


indication of1 redundant

equation


indication of2 redundant equations



−−−−

→

000000

000000

023100

112321

b][AM

The 2nd row (corresponding to one the only two independent equations), which is normally reserved for finding the second unknown x2 reads as

– x3 – 3x4 + 2x5 = 0indicating that not all x3, x4, x5 can be chosen arbitrarily.

But, we still have one more unknown that can be chosen arbitrarily. Since the 1st row is actually reserved to determine the 1st unknown x1.

So, take x5 = c1 and x4 = c2 then x3 = 2c1 –3c2

Take x2 = c3 then x1 = 1 – c1 + 2c2 – 3(2c1 –3c2) +2c3 = 1 – 7c1 + 11c2 + 2c3

+

−+

−

+

=

−

++−

=

0

0

0

1

2

c

0

1

3

0

11

c

1

0

2

0

7

c

0

0

0

0

1

c

c

3c2c

c

2c11c7c1

x

x

x

x

x

321

1

2

21

3

321

5

4

3

2

1

Therefore, the solution is expressed as



Example: (homogeneous equations, m=n, only the trivial solution exists)

Note that equations are all homogeneous and m = 3, n = 3, rank[A] = 3Hence only the trivial solution exists.

=→

=+=+=+

0110

0101

0011

b][A

0xx

0xx

0xx

32

31

21

M

444 3444 21444 3444 21R2R3:R3R1R2:R2

0200

0110

0011

0110

0110

0011

+−

−→

−→

The 3rd row indicates that

2 x3 = 0 thus x3 = 0

Back substitution of this in the 2nd equation gives

– x2 + x3 = 0 thus x2 = 0Back substitution of this in the 1st equation gives

x1 + x2 = 0 thus x1 = 0

Hence, the trivial solution x1 = x2 = x3 = 0 is the only solution.



Example: (m>n, unique solution with a redundant row arising at the mid-elimination)

−−−−

=→

=++=++−=−++=−+−=+

105230

02064

75131

35031

40102

b][A

105x2x3x

02x6x4x

75xx3xx

35x3xx

4x2x

432

421

4321

421

31

M

−−−−

→

−−−−

+−+

105230

82260

1010160

1010160

40102

105230

02064

75131

35031

40102

operation no : R5R4R1*2 : R4R3*2R1 : R3R2*2R1 : R2




indication of redundancy

−

− →

−−−−

+−+−

+

1020300

212100

00000

1010160

40102

105230

82260

1010160

1010160

40102

R5*2R2 : R5R4R2 : R4

R3R2 : R3

−−

→

−

−

00000

1020300

212100

1010160

40102

1020300

212100

00000

1010160

40102

bottom the torow redundant the Take



4444 34444 21Form EchelonRow

R4R3*3 : R4

00000

1616000

212100

1010160

40102

00000

1020300

212100

1010160

40102

−−

− →

−−

−

• There are no inconsistencies in the row-echelon form.

• Therefore, the system has at least one solution.

• In the row-echelon form, one row is redundant leaving 4 linearly

independent equations.

• Since there is the same number of linearly independent equations as the

number unknowns (4), this system has a unique solution .




Back-substitution:

Solution:


1x1616x 44 −=→−=

10x212x212xx 3343 =→−=−→−=+

35

x1010106x1010xx6x 22432 −=→=++→=−+

3x4102x4x2x 1131 −=→=+→=+

[ ]

−

−−

=

1

10

5/3

3

x



−−−−

→

−−−−

+−+

105230

82260

1210160

1010160

40102

105230

02064

85131

35031

40102

operation no : R5R4R1*2 : R4R3*2R1 : R3R2*2R1 : R2

−−−−

=→

=++=++−=−++=−+−=+

105230

02064

85131

35031

40102

b][A

105x2x3x

02x6x4x

85xx3xx

35x3xx

4x2x

432

421

4321

421

31

M

The only change in the previous example

Example: (m>n, inconsistency arising at the mid-elimination)

(same as before)



−−

− →

−−−−

+−+−

+

1020300

212100

20000

1010160

40102

105230

82260

1210160

1010160

40102

R5*2R2 : R5R4R2 : R4

R3R2 : R3

Example: (m>n, inconsistency arising at the mid-elimination)

The 3rd row indicates an inconsistency, i.e.

which is not possible.

Therefore, there is no need to continue the elimination; this system has no solution .

(same as before)

20x0x0x0x 4421 −=+++



Inverse of a Square Matrix

Adjoint matrix of [A]

Determinant of [A]

Adj[A] = [Cij]T

Transpose of the Cofactor Matrix, [Cij]T

Cofactor Matrix (Recall)

Matrix formed from cofactors of elements (minors with sign)

Properties of matrix inversion

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

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

det{[A]-1} = 1 / det[A]

{[A]T}-1 = {[A]-1}T

{[B][C]}-1 = [C]-1[B]-1

det[A]Adj[A]

[A] 1 =−

Cij = (–1)i+j Mij



Unique Solution of a Set of Linearly-independent, A lgebraic Equationsby Matrix Inverse

[A] [x] = [b] →→→→ [A]-1 [A] [x] = [A]-1 [b]

Example

Find the solution of the following system of linear equations by using

matrix inverse.

{ 32144 344 21[b][x]

3

2

1

[A]

321

321

321

3

5

2

x

x

x

111

3101

212

3xxx

53x10xx

22xx2x

−=

−−

−⇒

−=++−=−+=+−

→→→→ [x] = [A]-1 [b]



46](11)(4)2)(1)(1)[(

111

2110

410

111

2110

212

111

3101

212

det[A]442

R3*2R1:R1R3R2:R2

=−−−=−

−=−

−−

=−

−−

=−

++

321321

443442144 344 21

−−=

−−

−−

−−

−−

−−

−

−−−

−−

=21817

143

11213

101

12

31

22

310

2111

12

11

22

11

2111

101

11

31

11

310

][Cij

−

−=−

21111

842

17313

461

[A] 1

−=

−×+×−+×−×+×+×

−×−+×+×=

−

−

−== −

1

0

2

3)(2151)(211

3)(85422

3)(17)(53213

461

3

5

2

21111

842

17313

461

[b][A][x] 1

The solution is:



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