# Gaussian Elimination - Numerical Analysis - Solved Exam, Exams for Numerical Analysis. Central University of Kerala

## Numerical Analysis

Description: Main Points are:Gaussian Elimination, Set of Equations, Simultaneous Linear Equations, Back Substitution, Velocity Data, Naïve Gauss Elimination Method, Solution Vector, Matrix Multiplication, Values of Velocity
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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
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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