# 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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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
1n
steps of forward elimination. At the
end of
1n
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
1n
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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