04.06.1

Chapter 04.06

Gaussian Elimination

After reading this chapter, you should be able to:

1. solve a set of simultaneous linear equations using Naïve Gauss elimination,

2. learn the pitfalls of the Naïve Gauss elimination method,

3. understand the effect of round-off error when solving a set of linear equations with

the Naïve Gauss elimination method,

4. learn how to modify the Naïve Gauss elimination method to the Gaussian elimination

with partial pivoting method to avoid pitfalls of the former method,

5. find the determinant of a square matrix using Gaussian elimination, and

6. understand the relationship between the determinant of a coefficient matrix and the

solution of simultaneous linear equations.

How is a set of equations solved numerically?

One of the most popular techniques for solving simultaneous linear equations is the Gaussian

elimination method. The approach is designed to solve a general set of

n

equations and

n

unknowns

11313212111 ... bxaxaxaxa nn =++++

22323222121 ... bxaxaxaxa nn =++++

. .

. .

. .

nnnnnnn

bxaxaxaxa

=++++ ...

332211

Gaussian elimination consists of two steps

1. Forward Elimination of Unknowns: In this step, the unknown is eliminated in each

equation starting with the first equation. This way, the equations are reduced to one

equation and one unknown in each equation.

2. Back Substitution: In this step, starting from the last equation, each of the unknowns

is found.

Forward Elimination of Unknowns:

In the first step of forward elimination, the first unknown,

1

x

is eliminated from all rows

below the first row. The first equation is selected as the pivot equation to eliminate 1

x

. So,

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04.06.2 Chapter 04.06

to eliminate

1

x

in the second equation, one divides the first equation by

11

a

(hence called the

pivot element) and then multiplies it by 21

a. This is the same as multiplying the first

equation by

1121

/aa

to give

1

11

21

1

11

21

212

11

21

121 ... b

a

a

xa

a

a

xa

a

a

xa nn =+++

Now, this equation can be subtracted from the second equation to give

1

11

21

21

11

21

2212

11

21

22 ... b

a

a

bxa

a

a

axa

a

a

annn −=

−++

−

or

22222 ... bxaxa nn

′

=

′

++

′

where

nnn a

a

a

aa

a

a

a

aa

1

11

21

22

12

11

21

2222

−=

′

−=

′

This procedure of eliminating

1

x

, is now repeated for the third equation to the

th

n

equation

to reduce the set of equations as

11313212111 ...

bxaxaxaxa

nn =++++

22323222 ... bxaxaxa nn ′

=

′

++

′

+

′

33333232 ... bxaxaxa nn ′

=

′

++

′

+

′

. . .

. . .

. . .

nnnnnn bxaxaxa

′

=

′

++

′

+

′

...

3322

This is the end of the first step of forward elimination. Now for the second step of forward

elimination, we start with the second equation as the pivot equation and 22

a

′

as the pivot

element. So, to eliminate

2

x

in the third equation, one divides the second equation by 22

a

′

(the pivot element) and then multiply it by

32

a′

. This is the same as multiplying the second

equation by

2232 /aa ′′

and subtracting it from the third equation. This makes the coefficient of

2

x

zero in the third equation. The same procedure is now repeated for the fourth equation till

the

th

n

equation to give

11313212111 ... bxaxaxaxa nn =++++

22323222 ... bxaxax

ann ′

=

′

++

′

+

′

33333 ... bxaxa nn

′′

=

′′

++

′′

. .

. .

. .

nnnnn bxaxa ′′

=

′′

++

′′ ...

33

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Gaussian Elimination 04.06.3

The next steps of forward elimination are conducted by using the third equation as a pivot

equation and so on. That is, there will be a total of

1−n

steps of forward elimination. At the

end of

1−n

steps of forward elimination, we get a set of equations that look like

++ 212111 xaxa

11313 ... bxaxa nn =++

22323222 ... bxaxaxa nn ′

=

′

++

′

+

′

33333 ... bxaxa nn ′′

=

′′

++

′′

. .

. .

. .

( ) ( )

11 −−

=

n

nn

n

nn

bxa

Back Substitution:

Now the equations are solved starting from the last equation as it has only one unknown.

)1(

)1(

−

−

=

n

nn

n

n

n

a

b

x

Then the second last equation, that is the

th

)1( −n

equation, has two unknowns:

n

x

and

1−n

x

,

but

n

x

is already known. This reduces the

th

)1( −n

equation also to one unknown. Back

substitution hence can be represented for all equations by the formula

( ) ( )

( )

1

1

11

−

+=

−− ∑

−

=i

ii

n

ij j

i

ij

i

i

ia

xab

x for

1,,2,1 −−= n

ni

and

)1(

)1(

−

−

=

n

nn

n

n

n

a

b

x

Example 1

The upward velocity of a rocket is given at three different times in Table 1.

Table 1 Velocity vs. time data.

Time,

t

(s) Velocity,

v

(m/s)

5

106.8

8 177.2

12

279.2

The velocity data is approximated by a polynomial as

( )

125 ,

32

2

1

≤≤++= tatatatv

The coefficients

321 and a, , aa

for the above expression are given by

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04.06.4 Chapter 04.06

=

2.279

2.177

8.106

112144

1864

1525

3

2

1

a

a

a

Find the values of

321

and a,, aa

using the Naïve Gauss elimination method. Find the velocity

at

11 ,9 ,5.7 ,6=t

seconds.

Solution

Forward Elimination of Unknowns

Since there are three equations, there will be two steps of forward elimination of unknowns.

First step

Divide Row 1 by 25 and then multiply it by 64, that is, multiply Row 1 by

2.5664/25 =

.

[ ] [ ]

( )

56.28.106 1525 ×

gives Row 1 as

[ ] [ ]

408.27356.28.1264

Subtract the result from Row 2

[ ] [ ]

[ ] [ ]

208.9656.18.40

408.27356.28.1264

2.1771 864

−−−

−

to get the resulting equations as

−=

−−

2.279

208.96

8.106

112144

56.18.40

1525

3

2

1

a

a

a

Divide Row 1 by 25 and then multiply it by 144, that is, multiply Row 1 by

5.76144/25 =

.

[ ] [ ]

( )

76.58.106 1525 ×

gives Row 1 as

[ ] [ ]

168.61576.

58.28144

Subtract the result from Row 3

[ ] [ ]

[ ] [ ]

968.33576.48.16 0

168

.61576.58.28144

2.2791 12144

−−−

−

to get the resulting equations as

−

−=

−−

−−

968.335

208.96

8.106

76.48.160

56.18.40

1525

3

2

1

a

a

a

Second step

We now divide Row 2 by –4.8 and then multiply by –16.8, that is, multiply Row 2 by

3.54.816.8/ =−−

.

[ ] [ ]

( )

5.3208.96 56.18.40 ×−−−

gives Row 2 as

[ ] [ ]

728.33646.58.160 −−−

Subtract the result from Row 3

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