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Matrix Algebra

Solution to Simultaneous Algebraic Equations Gauss Elimination

Method:

Consider n simultaneous equations ,

a 11 x 1 + a 12 x 2 +, a 13 x 3 + a 1n x n = b1a 21 x 1 + a 22 x 2 + a 23 x 3 + a 2n x n = b2

a 31 x 1 + a 32 x2 + a 33 x 3 + a 3n x n = b3

an1 x 1 + a n2 x 2 + a n3 x 3 + a nn x n = bn

write the given set of equations in matrix form,

=

In Gauss elimination method the variables x2, x n-1, will besuccessively eliminated using Row Operations. This step is called forward elimination

. The given matrix will be converted to into an upper triangular matrix, Lower triangular

elements become zeros.

After forward elimination the nth

equation (last equation) become simple , it as anequation with one variable xn , determine xn. Now using (n-1)

th equation xn-1 can be

determined. Similarly using (n-2)nd

equation xn-2 can be determined. Using (n-3)rd

equation xn-3 can be determined. Continue up to first equation until all the unknowns aredetermined. This is called backward substitution.

Forward EliminationStep 1 :a11 becomes pivot, eliminate x1 from row2 , row3, row4, row n etc

Row2 a 21 = a 21 - ( a 21 / a 11) a 11 a 21 becomes 0a 22 = a 22 - ( a 21 / a 11) a 12 a 22 changes

a 23 = a 23 - ( a 21 / a 11) a 13 a 23 changes

x 1

x 2

x 3

x n

a 11 a 12 a 13 . a 1n

a 21 a 22 a 23 . a 2n

a 31 a 32 a 33 . a 3n

.

.

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.

.

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a n1 a n2 a n3 . a nn

b 1b 2

b 3

b n

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etc up to a 1n = a 2n - ( a 21 / a 11) a 1n a 2n changes

b2 = b2 - ( a 21 / a 11) b1 b2 changeswhatever we did to make a 21 = 0 applied the same to other elements

of that row

Row3 a 31 = a 31 - ( a 31 / a 11) a 11 a 31 becomes 0a 32 = a 32 - ( a 31 / a 11) a 12 a 32 changes

a 33 = a 33 - ( a 31 / a 11) a 13 a 33 changes

etc up to a 3n = a 3n - ( a 31 / a 11) a 1n a 3n changes

b3 = b3 - ( a 31 / a 11) b1 b3 changes

whatever we did to make a 31 = 0 applied the same to other elements

of that row

Row n a n1 = a n1 - ( a n1 / a 11) a 11 a n1 becomes 0a n2 = a n2 - ( a n1 / a 11) a 12 a n2 changes

a n3 = a n3 - ( a n1 / a 11) a 13 a n3 changes

etc up to a nn = a nn - ( a n1 / a 11) a 1n a nn changesb3 = b3 - ( a 31 / a 11) b1 b3 changeswhatever we did to make a n1 = 0 applied the same to other elements of that row.

Now, re-write the whole matrix equation. First row remains same, elements of other

rows will be different.

Step2 : :a22 becomes pivot, eliminate x2 from row3 , row4, row5, etc., row n

following the same methodNow, re-write the whole matrix equation. First row , Second row remains same,

elements of other rows will be different

Step3 : :a33 becomes pivot, eliminate x3 from row4 , row5, row6, etc., row n

following the same methodNow, re-write the whole matrix equation. First row , Second row , Third row remains

same, elements of other rows will be different

Continue until the variables x2, x3, x4 . x n-1 will be

successively eliminated and all the lower triangular elements becomes zero.

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Backward substitution.After forward elimination the nth equation (last equation) become simple , it as an

equation with one variable xn , determine xn. Now , using (n-1)th

equation xn-1 can be

determined. Similarly using (n-2)nd

equation xn-2 can be determined. Using (n-3)rd

equation xn-3 can be determined. Continue up to first equation until all the unknowns aredetermined. The method is best understood by solving problems.

Different Methods used to Solve Set of Simultaneous Equations in FEM.

Method of Matrix Inversion

Gauss elimination method

Cholesky Decomposition TechniqueGauss-Seidal Iteration Technique

Relaxation Method

Numerical examples illustratingGauss elimination method :

Problem 1. Solve the following set of equation by Gaussian elimination

technique.

5x1 + 3x2 + 2x3 + x4 = 4

4x1 + 3x2 - 3x3 - 2x4 = 5

x1 + 2x2 - 2x3 + 3x4 = 6

-4x1 + 3x2 5x3 + 2x4 = 7

Solution : Write the given equations in Matrix Form

[ CO ] [ X ] = [ CONS ]

[ a ] [ x ] = [ b ]

5 3 2 1

*

x1

=

4

4 3 -3 -2 x2 5

1 2 -2 3 x3 6

-4 3 -5 2 x4 7

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Step 1 : a i j = a i j (a i1/ a 11) a 1j b i = b i (a i1/ a 11) b 1

i = 2, j = 1,2,3,4

Row 2 i = 2 j = 1,2,3,4

4 (4/5) 5 = 4 (4) = 0.0 3 (4/5) 3 = 3 2.4 = 0.6

-3 (4/5)2 = -3 1.6 = -4.6 -2 (4/5) 1 = -2 0.8 = -2.8

Row 3 i = 3 j = 1,2,3,4

1 (1/5)5 = 1 1 = 0.0 2 (1/5)3 = 2 -0 .6 = 1.4

-2 (1/5) 2 = -2 0.4= -2.4 3 (1/5) 1 = 3 0.2 = 2.8

Row 4 : i = 4 j = 1,2,3,4

-4 ( -4 / 5)5 = 4 - 4 = 0 3 ( -4 / 5)3 = 3+2.4 = 5.4-5 ( -4 /5) 2 = -5 + 1.6 = -3 .4 2 ( -4 /5) 1 = 3 0.2 = 2.8

b i = b i (a i1/ a11) b 1

i = 2 b2 = 5 (4/5)4 = 1.8

i = 3 b3 = 6 (1/5)4 = 5.2

i = 4 b4 = 7 (-4/5)4 = 10.2

The modified matrix equation , after eliminating x1

from 2nd , 3rd and 4th equations.

5 3 2 1

*

x1

=

4

0 0.6 -4.6 -2.8 x2 1.8

0 1.4 -2.4 2.8 x3 5.2

0 5.4 -3.4 2.8 x4 10.2

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Step 2 : To eliminate x2 from Row 3 and Row 4

a i j = a i j (a i2/ a 22) a 2j b i = b i (a i2/ a 22 ) b 2 i = 3, j = 2,3,4

Row 3 i =3 j = 2,3,4

1.4 (1.4/0.6) 0.6 =1.4 1.4 = 0.0 j = 2-2.4 (1.4/0.6) (-4.6) = 2.4+10.73 =8.33 j = 3

2.8 (1.4/0.6)(-2.8) = -2.8 6.53 = 9.33 j = 4

Row 4 i = 4 j = 2,3,4

5.4 (5.4/0.6)0.6 = 5.4-5.4 = 0.0 j = 2

-3.4 (5.4/0.6)(-4.6) = -3.4+41.4 = 38 j = 3

2.8 (5.4/0.6)2.8 = 2.8 +25.2 = 28 j = 4

b i = b i (a i2/ a22) b 2

i = 3 b3 = b3 (a32/a22)b2

5.2 - (1.4/0.6) 1.8 = 5.2 4.2 = 1

i = 4 b4 = b4 (a42/a22)b2

10.2 - (5.4/0.6)1.8 = 10.2 -16.2 = -6

The modified matrix after step 2 eliminating x2 from 3rd and 4th equations.

Step 3 : To eliminate x3 from Row 4

a i j = a i j (a i3/ a 33) a 3j b i = b i (a i3/ a 33) b 3 i = 4, j = 3,4

a43 = a43 (a43/ a33) a33 = 38 (38 / 8.33) 8.33 = 38 38 = 0.0

5 3 2 1

*

x1

=

4

0 0.6 -4.6 -2.8 x2 1.8

0 0 8.33 9.32 x3 1

0 0 38 28 x4 -6

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a44 = a44- (a43/ a33) a34 = 28 - (38 / 8.33) 9.32

= 28 42.52 = -1 4.52

b i = b i (a i3/ a33) b 3 i = 4,

b4 = b4 - (a43/ a33) b3 b4 = -6 - (38 / 8.33) 1 = - 6 - 4.56 = - 10.56

The modified matrix, after step 3 , eliminating x3 from 4th equation.

Back Substitution:

The modified equations are

5 x1 + 3 x2 + 2 x3 + x4 = 4

0.6 x2 4.6 x3 - 2.8 x4 = 1.88.33 x3 + 9.32 x4 = 1

-14.52 x4 = -10.56

x4 = ( -1 0.56/ -1 4.52) = 0.727

8.33 x3 + 9.32 (0.727 ) = 1

x3 = (1 - 6.776) / 8.33 = - 0.693

0.6 x2 4.6( - 0.693) -2.8 (0.727) = 1.8

0.6 x2 = (1.8 - 3.1878 + 2.0356)x2 = (1.8 - 3.1878 + 2.0356) / 0.6 = 1.079

5x1 + 3 (1.079) + 2 ( - 0.693) + 0.727 = 4

x1 = ( 4 - 3 (1.079) + 2 (0.693) - 0.727 ) / 5 = 0.155

x1= 0.155 x2=1.079 x3= - 0.693 x4= 0.727

5 3 2 1

*

x1

=

4

0 0.6 -4.6 -2.8 x2 1.8

0 0 8.33 9.32 x3 1

0 0 0 -

14.52

x4 -

10.56

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Prob2 : Solve using gauss elimination method

x1- 2x2 + 6 x3 = 0 2x1 +2x2 + 3x3 = 3 -x 1 + 3x2 = 2

=

Step 1 : a

i j = a i j (a i1 / a11) a1j I = 2,3 j = 2,3

2- (2/1)* 1 = 0 2 (2/1) (-2) = 6 3 (2/1) (6) = -9

-1- (-1/1) 1 = 0 3 (-1/1) (-2) = 1 0 (-1/1) 6 = 6

bi = bi - (a i1 / a11) b1

3 (2/1) *0 = 3 2 (-1/1) *0 = 2

Modified Matrix After step1

=

Step 2 : a i j = a i j (a i2 / a22) a2j i = 3 j = 2,31 (1/6)* (6) = 0 6 (1/6) (-9) = 7.5

bi = bi - (a i2 / a22) b22 (1/6) 3 = 1.5

x1

x2

x3

1 -2 6

2 2 3

-1 3 0

0

3

2

x1

x2

x3

1 -2 6

0 6 -9

0 1 6

0

3

2

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Modified Matrix After step2

=

Back Substitution :

7.5 x3 = 1.5 x3 = 1.5 / 7.5 = 0.2 x3 = 0.2

6 x2 9x3 = 3 x2=(3 + 9(0.2))/6 =0.8 x2 = 0.8

x1 2x2 +6x3 = 0 x1 = 2(0.8)6(0.2) = 0.4 x1 = 0.4

Prob 3 :Solve using gauss elimination method

4x1 +6x2 +8x3 = 2 8x1 +4x2 + 6x3 = 4 6x1 + 2x2+4x3= 6

Solution : writing the equations in matrix form

=

Step 1 : a i j = a i j (a i1 / a11) a1j i = 2,3 j = 2,3

8 - (8/4)* 4 = 0 4 (8/4) (6) = -8 6 -( 8/4)8 = -10

6 -( 6/4) 4 = 0 2 (6/4 (6) = -7 4 (6 /4) 8 = -8

bi = bi - (a i1 / a11) b1

4 (8/4) 2 = 0 6 (6/4) 2 = 3

x1

x2

x3

1 -2 6

0 6 -9

0 0 7.5

0

3

1.5

4 6 8

8 4 6

6 2 4

x1

x2

x3

2

4

6

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Modified matrix equation after step 1

=

Step 2 : a i j = a i j (a i2 / a22) a2j i = 3 j = 2,3

-7 (-7/-8) (-8) = 0

-8 -(-7 / -8 )( -10) = -8 +70/8 = -8 + 8.75 = 0.75

bi = bi - (a i2 / a22) b2

3 (-7/-8) 0 = 3

Modified matrix equation after step 2 :

0.75 x3 = 3 x3 = 3/0.75 = 4 x3 = 4

-8 x2 -10 x3 = 0 -8x2 -10 (4) = 0 -8x2 = 40 x2 = -5

4x1+6x2 + 8x3 = 2 4x1+ 6(-5) +8(4) = 2

x1 = (2+30 -32 ) / 4 =0

x1 = 0 x2 = -5 x3= 4

4 6 8

0 -8 -10

0 -7 -8

X1

X2

X3

2

0

3

4 6 8

0 -8 -10

0 0 0.75

X1

X2

x3

2

0

3

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Prob 4 : Solve using gauss elimination method

3x1 -3x2 -2x3 = 5 2x1+2x2+3x3= 6 3x1 - 5x2+2x3= 7

Step 1 :

a i j = a i j (a i1 / a11) a1j i = 2,3 j = 2,3

2 - (2/3)3 = 0 2 - (2/3)(-3) = 4 3 -( 2/3)(-2) = 4.33

3 -( 3/3) 3 = 0 -5 (3/3 (-3) = -2 2 (3/3)(-2) = 4

bi = bi - (a i1 / a11) b1

6 (2/3) 5 = 6-10/3 = 2.667 7 (3/3) 5 = 2

Modified matrix is

Step 2 :

a i j = a i j (a i2 / a22) a2j i = 3 j = 2,3

-2 - (-2/4)4 = 0 4 - (-2/4)(4.333) = 4+ 2.166 = 6.166

bi = bi - (a i2 / a22) b2

2 (-2/4) 2.667 = 2+(2.667/2) = 2+1.333 = 3.333

6.166 x3 = 3.333 x3 = ( 3.333/6.166) = 0.504

4x2 + 4.333x3 = 2.667 4x2 = 2.667 - 4.333 (0.504)

x2 = 0.483 /4 = 0.120

3 -3 -2 5

2 2 3 6

3 -5 2 7

3 -3 -2 5

0 4 4.33 2.667

0 -2 4 2

3 -3 -2 5

0 4 4.33 2.667

0 0 6.166 3.333

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3x1 -3x2 2x3 = 5 3x1 -3(0.120) -2(0.504) = 5

x1 = (5 + 0.360 +1.008) / 3 = 2.122

x1 = 2.122 x2 = 0.120 x3 = 0.504

THUS COMPLETE GAUSSIAN ELIMINATION

TECHNIQUE

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