Page 1
Edurev123
2. Solutions of System of Linear Equations
2.1 The equation ?? ?? +???? +?? =?? has two real roots ?? and ?? . Show that the
iterative method given by: ?? ?? +?? =-
(?? ?? ?? +?? )
?? ?? ,?? =?? ,?? ,?? ….. is convergent near ?? =?? ,
if |?? |>|?? |.
(2009 : 6 Marks)
Solution:
Let the iterations are given by
?? ?? +1
=-
(?? ?? ?? +?? )
?? ?? =?? (?? ?? ) (say) (??)
Now by the known theorem, if ?? (?? ) and ?? '
(?? ) are continuous in an interval about a root a
of the equation ?? =?? (?? ) and if |?? '
(?? )|<1 for all ?? in the internal, then the successive
approximations ?? 1
,?? 2
,…. are given by ?? ?? =?? (?? ?? -1
)(?? =1,2,…) converges to the root ??
provided that the initial approximation ?? 0
is chosen in the interval.
? The iterations converges to ?? if
|?? (?? )|=|
-?? ?? ?? 2
|<1 ???????? (??)
Note that ?? '
(?? ) is continuous near ' ?? ' if the iteration converges to ?? =?? . So, we require
|?? '
(?? )|?=|
?? ?? 2
|<1
|?? |<|?? |
2
?????????????????????????????????????????????????????????????????????????(???? )
Now, given that ?? and ?? are roots of the equation
?????????????????????????????????? 2
+???? +?? =0
So, ??? +?? =-?? ????????????????????????????????????????????????????? =?? ????????????????????????????????????????????????????????????????????????????????????????????(?????? )
From (ii) and (iii)
?????????????????????????????????????????????|???? |<|?? |
2
??????????????????????????|?? |(|?? |-|?? |)>0
???????????????????????????????????????????????|?? |>0
2.2 Given the system of equations
Page 2
Edurev123
2. Solutions of System of Linear Equations
2.1 The equation ?? ?? +???? +?? =?? has two real roots ?? and ?? . Show that the
iterative method given by: ?? ?? +?? =-
(?? ?? ?? +?? )
?? ?? ,?? =?? ,?? ,?? ….. is convergent near ?? =?? ,
if |?? |>|?? |.
(2009 : 6 Marks)
Solution:
Let the iterations are given by
?? ?? +1
=-
(?? ?? ?? +?? )
?? ?? =?? (?? ?? ) (say) (??)
Now by the known theorem, if ?? (?? ) and ?? '
(?? ) are continuous in an interval about a root a
of the equation ?? =?? (?? ) and if |?? '
(?? )|<1 for all ?? in the internal, then the successive
approximations ?? 1
,?? 2
,…. are given by ?? ?? =?? (?? ?? -1
)(?? =1,2,…) converges to the root ??
provided that the initial approximation ?? 0
is chosen in the interval.
? The iterations converges to ?? if
|?? (?? )|=|
-?? ?? ?? 2
|<1 ???????? (??)
Note that ?? '
(?? ) is continuous near ' ?? ' if the iteration converges to ?? =?? . So, we require
|?? '
(?? )|?=|
?? ?? 2
|<1
|?? |<|?? |
2
?????????????????????????????????????????????????????????????????????????(???? )
Now, given that ?? and ?? are roots of the equation
?????????????????????????????????? 2
+???? +?? =0
So, ??? +?? =-?? ????????????????????????????????????????????????????? =?? ????????????????????????????????????????????????????????????????????????????????????????????(?????? )
From (ii) and (iii)
?????????????????????????????????????????????|???? |<|?? |
2
??????????????????????????|?? |(|?? |-|?? |)>0
???????????????????????????????????????????????|?? |>0
2.2 Given the system of equations
?? ?? +?? ?? =?? ?? ?? +?? ?? +?? =?? ?? ?? +?? ?? +???? =?? ???? +???? =??
State the solubility ?? uniqueness conditions for the system. Give the solution
when it exists.
(2010 : 20 Marks)
Solution:
From numerical analysis and computer programming.
The given system of equation can be written as
[
2 3 0 0 1
2 4 1 0 2
0 2 6 ?? 4
0 0 4 ?? ?? ]
Applying ?? 2
??? 2
-?? 1
, we get
[
2 3 0 0 1
0 1 1 0 1
0 2 6 ?? 4
0 0 4 ?? ?? ]
[
2 3 0 0 1
0 1 1 0 1
0 0 4 ?? 2
0 0 4 ?? ?? ]
[
2 3 0 0 1
0 1 1 0 1
0 0 4 ?? 2
0 0 0 ?? -?? ?? -2
]
Unique solution if
?? -?? ?=0 and ?? -2?0
?? ???? and ?? ?2
Infinite solution if
?? -?? ?=0 or ?? =?? ?? -2?=0 or ?? =2
?? ?=?? and ?? ?2 or ?? ??? and ?? =2
Unique solution:
Page 3
Edurev123
2. Solutions of System of Linear Equations
2.1 The equation ?? ?? +???? +?? =?? has two real roots ?? and ?? . Show that the
iterative method given by: ?? ?? +?? =-
(?? ?? ?? +?? )
?? ?? ,?? =?? ,?? ,?? ….. is convergent near ?? =?? ,
if |?? |>|?? |.
(2009 : 6 Marks)
Solution:
Let the iterations are given by
?? ?? +1
=-
(?? ?? ?? +?? )
?? ?? =?? (?? ?? ) (say) (??)
Now by the known theorem, if ?? (?? ) and ?? '
(?? ) are continuous in an interval about a root a
of the equation ?? =?? (?? ) and if |?? '
(?? )|<1 for all ?? in the internal, then the successive
approximations ?? 1
,?? 2
,…. are given by ?? ?? =?? (?? ?? -1
)(?? =1,2,…) converges to the root ??
provided that the initial approximation ?? 0
is chosen in the interval.
? The iterations converges to ?? if
|?? (?? )|=|
-?? ?? ?? 2
|<1 ???????? (??)
Note that ?? '
(?? ) is continuous near ' ?? ' if the iteration converges to ?? =?? . So, we require
|?? '
(?? )|?=|
?? ?? 2
|<1
|?? |<|?? |
2
?????????????????????????????????????????????????????????????????????????(???? )
Now, given that ?? and ?? are roots of the equation
?????????????????????????????????? 2
+???? +?? =0
So, ??? +?? =-?? ????????????????????????????????????????????????????? =?? ????????????????????????????????????????????????????????????????????????????????????????????(?????? )
From (ii) and (iii)
?????????????????????????????????????????????|???? |<|?? |
2
??????????????????????????|?? |(|?? |-|?? |)>0
???????????????????????????????????????????????|?? |>0
2.2 Given the system of equations
?? ?? +?? ?? =?? ?? ?? +?? ?? +?? =?? ?? ?? +?? ?? +???? =?? ???? +???? =??
State the solubility ?? uniqueness conditions for the system. Give the solution
when it exists.
(2010 : 20 Marks)
Solution:
From numerical analysis and computer programming.
The given system of equation can be written as
[
2 3 0 0 1
2 4 1 0 2
0 2 6 ?? 4
0 0 4 ?? ?? ]
Applying ?? 2
??? 2
-?? 1
, we get
[
2 3 0 0 1
0 1 1 0 1
0 2 6 ?? 4
0 0 4 ?? ?? ]
[
2 3 0 0 1
0 1 1 0 1
0 0 4 ?? 2
0 0 4 ?? ?? ]
[
2 3 0 0 1
0 1 1 0 1
0 0 4 ?? 2
0 0 0 ?? -?? ?? -2
]
Unique solution if
?? -?? ?=0 and ?? -2?0
?? ???? and ?? ?2
Infinite solution if
?? -?? ?=0 or ?? =?? ?? -2?=0 or ?? =2
?? ?=?? and ?? ?2 or ?? ??? and ?? =2
Unique solution:
(?? -?? )?? ?=?? -2
?? ??=
?? -2
?? -?? 4?? +???? ?=2
???? ?=
2-
?? (?? -2)
?? -?? 4
=
2?? -????
4(?? -?? )
?? +?? ?=1??? =1-?? =1-
2?? -????
4(?? -?? )
=
2?? -4?? +????
4(?? -?? )
2?? +3?? ?=1??? =
1
2
(1-3?? )
?? ?=
1
2
{1-3×
(2?? -4?? +???? )
4(?? -?? )
}=
1
2
(8?? -2?? -3???? )
4(?? -?? )
???? ?=
1
2
(8?? -2?? -3???? )
4(?? -?? )
,?? =
2?? -4?? +????
4(?? -?? )
?? ?=
2?? -????
4(?? -?? )
,?? =
?? -2
?? -??
Many solutions:
?? ?=?? ,?? =2
4?? +???? ?=2??? =
2-????
4
?? ?=1-?? =1-(
2-????
4
)??? =
2+????
4
2?? ?=1-3?? ??? =(1-
6+3????
4
)??? =
1
2
(-2-3???? )
4
2.3 Solve the following system of simultaneous equations, using Gauss-Seidel
iterative method :
?? ?? +???? ?? -?? ?=-????
???? ?? +?? -?? ?? ?=????
?? ?? -?? ?? +???? ?? ?=????
(2012 : 20 Marks)
Solution:
Solving each equation for the unknown having the largest coefficient, the new equations
are :
?? =
1
20
(17-?? +2?? ) (??)
?? =
1
20
(-18-3?? +?? ) (???? )
?? =
1
20
(25-2?? +3?? ) (?????? )
Page 4
Edurev123
2. Solutions of System of Linear Equations
2.1 The equation ?? ?? +???? +?? =?? has two real roots ?? and ?? . Show that the
iterative method given by: ?? ?? +?? =-
(?? ?? ?? +?? )
?? ?? ,?? =?? ,?? ,?? ….. is convergent near ?? =?? ,
if |?? |>|?? |.
(2009 : 6 Marks)
Solution:
Let the iterations are given by
?? ?? +1
=-
(?? ?? ?? +?? )
?? ?? =?? (?? ?? ) (say) (??)
Now by the known theorem, if ?? (?? ) and ?? '
(?? ) are continuous in an interval about a root a
of the equation ?? =?? (?? ) and if |?? '
(?? )|<1 for all ?? in the internal, then the successive
approximations ?? 1
,?? 2
,…. are given by ?? ?? =?? (?? ?? -1
)(?? =1,2,…) converges to the root ??
provided that the initial approximation ?? 0
is chosen in the interval.
? The iterations converges to ?? if
|?? (?? )|=|
-?? ?? ?? 2
|<1 ???????? (??)
Note that ?? '
(?? ) is continuous near ' ?? ' if the iteration converges to ?? =?? . So, we require
|?? '
(?? )|?=|
?? ?? 2
|<1
|?? |<|?? |
2
?????????????????????????????????????????????????????????????????????????(???? )
Now, given that ?? and ?? are roots of the equation
?????????????????????????????????? 2
+???? +?? =0
So, ??? +?? =-?? ????????????????????????????????????????????????????? =?? ????????????????????????????????????????????????????????????????????????????????????????????(?????? )
From (ii) and (iii)
?????????????????????????????????????????????|???? |<|?? |
2
??????????????????????????|?? |(|?? |-|?? |)>0
???????????????????????????????????????????????|?? |>0
2.2 Given the system of equations
?? ?? +?? ?? =?? ?? ?? +?? ?? +?? =?? ?? ?? +?? ?? +???? =?? ???? +???? =??
State the solubility ?? uniqueness conditions for the system. Give the solution
when it exists.
(2010 : 20 Marks)
Solution:
From numerical analysis and computer programming.
The given system of equation can be written as
[
2 3 0 0 1
2 4 1 0 2
0 2 6 ?? 4
0 0 4 ?? ?? ]
Applying ?? 2
??? 2
-?? 1
, we get
[
2 3 0 0 1
0 1 1 0 1
0 2 6 ?? 4
0 0 4 ?? ?? ]
[
2 3 0 0 1
0 1 1 0 1
0 0 4 ?? 2
0 0 4 ?? ?? ]
[
2 3 0 0 1
0 1 1 0 1
0 0 4 ?? 2
0 0 0 ?? -?? ?? -2
]
Unique solution if
?? -?? ?=0 and ?? -2?0
?? ???? and ?? ?2
Infinite solution if
?? -?? ?=0 or ?? =?? ?? -2?=0 or ?? =2
?? ?=?? and ?? ?2 or ?? ??? and ?? =2
Unique solution:
(?? -?? )?? ?=?? -2
?? ??=
?? -2
?? -?? 4?? +???? ?=2
???? ?=
2-
?? (?? -2)
?? -?? 4
=
2?? -????
4(?? -?? )
?? +?? ?=1??? =1-?? =1-
2?? -????
4(?? -?? )
=
2?? -4?? +????
4(?? -?? )
2?? +3?? ?=1??? =
1
2
(1-3?? )
?? ?=
1
2
{1-3×
(2?? -4?? +???? )
4(?? -?? )
}=
1
2
(8?? -2?? -3???? )
4(?? -?? )
???? ?=
1
2
(8?? -2?? -3???? )
4(?? -?? )
,?? =
2?? -4?? +????
4(?? -?? )
?? ?=
2?? -????
4(?? -?? )
,?? =
?? -2
?? -??
Many solutions:
?? ?=?? ,?? =2
4?? +???? ?=2??? =
2-????
4
?? ?=1-?? =1-(
2-????
4
)??? =
2+????
4
2?? ?=1-3?? ??? =(1-
6+3????
4
)??? =
1
2
(-2-3???? )
4
2.3 Solve the following system of simultaneous equations, using Gauss-Seidel
iterative method :
?? ?? +???? ?? -?? ?=-????
???? ?? +?? -?? ?? ?=????
?? ?? -?? ?? +???? ?? ?=????
(2012 : 20 Marks)
Solution:
Solving each equation for the unknown having the largest coefficient, the new equations
are :
?? =
1
20
(17-?? +2?? ) (??)
?? =
1
20
(-18-3?? +?? ) (???? )
?? =
1
20
(25-2?? +3?? ) (?????? )
Choosing ?? =0,?? =0 to start with, we get,
?? (1)
=
1
20
(17-0+0)=0.85 (first approximation)
Putting ?? =?? (1)
=0.85,?? =0 in (ii)
?? (1)
=
1
20
(-18-2.55)=-1.0275
Putting ?? =?? (1)
=0.85,?? =?? (1)
=-1.0275 in (iii), we get
?? (1)
=
1
20
(25-1.7-3.0825)=1.0109
?? (2)
=
1
20
(17+1.0275+2.0218)=1.0025
?? (2)
=
1
20
(-18-3.0075+1.0109)=-0.9998
?? (2)
=
1
20
(25-2.0050-2.9994)=0.9998
?? (3)
=
1
20
(17+0.9998+1.9996)=0.99997
?? (3)
=
1
20
(-18-2.99991+0.9998)=-1.0000
?? (3)
=
1
20
(25-1.99994-3.0000)=1.0000
?? (4)
=
1
20
(17+1.0000+2.0000)=1.0000
?? (4)
=
1
20
(-18-3.0000+1.0000)=-1.0000
?? (4)
=
1
20
(25-2.0000-3.0000)=1.0000
Since ?? (4)
,?? (4)
,?? (4)
are sufficiently close to ?? (3)
,?? (3)
,?? (3)
respectively, so that values
1.0000,-1.0000,1.0000 can be taken as the solution of the given system.
2.4 Solve the system of equations :
?? ?? ?? -?? ?? =?? -?? ?? +?? ?? ?? -?? ?? =?? -?? ?? +?? ?? ?? =??
Using Gauss-Seidel iteration method (Perform three iterations).
(2014 : 15 Marks)
Solution:
The given system of equations can be written as
Page 5
Edurev123
2. Solutions of System of Linear Equations
2.1 The equation ?? ?? +???? +?? =?? has two real roots ?? and ?? . Show that the
iterative method given by: ?? ?? +?? =-
(?? ?? ?? +?? )
?? ?? ,?? =?? ,?? ,?? ….. is convergent near ?? =?? ,
if |?? |>|?? |.
(2009 : 6 Marks)
Solution:
Let the iterations are given by
?? ?? +1
=-
(?? ?? ?? +?? )
?? ?? =?? (?? ?? ) (say) (??)
Now by the known theorem, if ?? (?? ) and ?? '
(?? ) are continuous in an interval about a root a
of the equation ?? =?? (?? ) and if |?? '
(?? )|<1 for all ?? in the internal, then the successive
approximations ?? 1
,?? 2
,…. are given by ?? ?? =?? (?? ?? -1
)(?? =1,2,…) converges to the root ??
provided that the initial approximation ?? 0
is chosen in the interval.
? The iterations converges to ?? if
|?? (?? )|=|
-?? ?? ?? 2
|<1 ???????? (??)
Note that ?? '
(?? ) is continuous near ' ?? ' if the iteration converges to ?? =?? . So, we require
|?? '
(?? )|?=|
?? ?? 2
|<1
|?? |<|?? |
2
?????????????????????????????????????????????????????????????????????????(???? )
Now, given that ?? and ?? are roots of the equation
?????????????????????????????????? 2
+???? +?? =0
So, ??? +?? =-?? ????????????????????????????????????????????????????? =?? ????????????????????????????????????????????????????????????????????????????????????????????(?????? )
From (ii) and (iii)
?????????????????????????????????????????????|???? |<|?? |
2
??????????????????????????|?? |(|?? |-|?? |)>0
???????????????????????????????????????????????|?? |>0
2.2 Given the system of equations
?? ?? +?? ?? =?? ?? ?? +?? ?? +?? =?? ?? ?? +?? ?? +???? =?? ???? +???? =??
State the solubility ?? uniqueness conditions for the system. Give the solution
when it exists.
(2010 : 20 Marks)
Solution:
From numerical analysis and computer programming.
The given system of equation can be written as
[
2 3 0 0 1
2 4 1 0 2
0 2 6 ?? 4
0 0 4 ?? ?? ]
Applying ?? 2
??? 2
-?? 1
, we get
[
2 3 0 0 1
0 1 1 0 1
0 2 6 ?? 4
0 0 4 ?? ?? ]
[
2 3 0 0 1
0 1 1 0 1
0 0 4 ?? 2
0 0 4 ?? ?? ]
[
2 3 0 0 1
0 1 1 0 1
0 0 4 ?? 2
0 0 0 ?? -?? ?? -2
]
Unique solution if
?? -?? ?=0 and ?? -2?0
?? ???? and ?? ?2
Infinite solution if
?? -?? ?=0 or ?? =?? ?? -2?=0 or ?? =2
?? ?=?? and ?? ?2 or ?? ??? and ?? =2
Unique solution:
(?? -?? )?? ?=?? -2
?? ??=
?? -2
?? -?? 4?? +???? ?=2
???? ?=
2-
?? (?? -2)
?? -?? 4
=
2?? -????
4(?? -?? )
?? +?? ?=1??? =1-?? =1-
2?? -????
4(?? -?? )
=
2?? -4?? +????
4(?? -?? )
2?? +3?? ?=1??? =
1
2
(1-3?? )
?? ?=
1
2
{1-3×
(2?? -4?? +???? )
4(?? -?? )
}=
1
2
(8?? -2?? -3???? )
4(?? -?? )
???? ?=
1
2
(8?? -2?? -3???? )
4(?? -?? )
,?? =
2?? -4?? +????
4(?? -?? )
?? ?=
2?? -????
4(?? -?? )
,?? =
?? -2
?? -??
Many solutions:
?? ?=?? ,?? =2
4?? +???? ?=2??? =
2-????
4
?? ?=1-?? =1-(
2-????
4
)??? =
2+????
4
2?? ?=1-3?? ??? =(1-
6+3????
4
)??? =
1
2
(-2-3???? )
4
2.3 Solve the following system of simultaneous equations, using Gauss-Seidel
iterative method :
?? ?? +???? ?? -?? ?=-????
???? ?? +?? -?? ?? ?=????
?? ?? -?? ?? +???? ?? ?=????
(2012 : 20 Marks)
Solution:
Solving each equation for the unknown having the largest coefficient, the new equations
are :
?? =
1
20
(17-?? +2?? ) (??)
?? =
1
20
(-18-3?? +?? ) (???? )
?? =
1
20
(25-2?? +3?? ) (?????? )
Choosing ?? =0,?? =0 to start with, we get,
?? (1)
=
1
20
(17-0+0)=0.85 (first approximation)
Putting ?? =?? (1)
=0.85,?? =0 in (ii)
?? (1)
=
1
20
(-18-2.55)=-1.0275
Putting ?? =?? (1)
=0.85,?? =?? (1)
=-1.0275 in (iii), we get
?? (1)
=
1
20
(25-1.7-3.0825)=1.0109
?? (2)
=
1
20
(17+1.0275+2.0218)=1.0025
?? (2)
=
1
20
(-18-3.0075+1.0109)=-0.9998
?? (2)
=
1
20
(25-2.0050-2.9994)=0.9998
?? (3)
=
1
20
(17+0.9998+1.9996)=0.99997
?? (3)
=
1
20
(-18-2.99991+0.9998)=-1.0000
?? (3)
=
1
20
(25-1.99994-3.0000)=1.0000
?? (4)
=
1
20
(17+1.0000+2.0000)=1.0000
?? (4)
=
1
20
(-18-3.0000+1.0000)=-1.0000
?? (4)
=
1
20
(25-2.0000-3.0000)=1.0000
Since ?? (4)
,?? (4)
,?? (4)
are sufficiently close to ?? (3)
,?? (3)
,?? (3)
respectively, so that values
1.0000,-1.0000,1.0000 can be taken as the solution of the given system.
2.4 Solve the system of equations :
?? ?? ?? -?? ?? =?? -?? ?? +?? ?? ?? -?? ?? =?? -?? ?? +?? ?? ?? =??
Using Gauss-Seidel iteration method (Perform three iterations).
(2014 : 15 Marks)
Solution:
The given system of equations can be written as
?? 1
=
1
2
(7+?? 2
)
?? 2
=
1
2
(1+?? 1
+?? 3
)
?? 3
=
1
2
(1+?? 2
)
}
(??)
By the Gauss-Seidel method, system (i) can be written as
?? 1
?? +1
=
1
2
(7+?? 2
(?? )
)
?? 2
?? +1
=
1
2
(1+?? 2
(?? +1)
+?? 3
?? )
?? 3
?? +1
=
1
2
(1+?? 2
(?? +1)
)
where, ?? =0,1,2,3…
Now taking,??? (0)=0, we obtain the following iterations
?? =0 :
?? ?
(1)
=
1
2
(7+0)=
7
2
=3.5
?? 2
(1)
=
1
2
(1+3.5+0)=
4.5
2
=2.25
?? 3
(1)
=
1
2
(1+2.25)=
1
2
(3.25)=1.625
?? =1?:
?? 1
(2)
=
1
2
(7+?? 2
(1)
)=
1
2
(7+2.25)=
9.25
2
=4.625
?? 2
(2)
=
1
2
(1+?? 1
(2)
+?? 3
(1)
)=
1
2
(1+4.625+1.625)=3.625
?? 3
(2)
=
1
2
(1+?? 2
(2)
)=
1
2
(1+4.625+1.625)=3.625
?? 3
(2)
=
1
2
(1+?? 2
(2)
)=
1
2
(1+3.625)=2.3125
?? =2 :
?? 1
(3)
=
1
2
(7+?? 2
(2)
)=
1
2
(7+3.625)=5.3125
?? 2
(3)
=
1
2
(1+?? 1
(4)
+?? 3
(2)
)=
1
2
(1+5.3125+2.3125)=4.3125
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