Linear equation
A linear equation is an algebraic equation in which each term is either a constant or the product of a constant and (the first power of) a single variable.Example : The equations
Following equations are not linear,
732,1321,73 4321 xxxxzxyyx
xyxzzyxyx sin,423,13
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• Linear equations can have one or more variables. Linear equations occur with great regularity in applied mathematics. While they arise quite naturally when modeling many phenomena, they are particularly useful since many non-linear equations may be reduced to linear equations by assuming that quantities of interest vary to only a small extent from some "background" state. Linear equations do not include exponents.
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Figure
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Linear equation
A linear equation in unknowns is an equation that can be put in the standard form
Where and are constants. The constant is called the coefficient of and is called the constant term of the equation. Example:
nxxxx ,........,, 321
).(............................2211 ibxaxaxa nn .,...., 21 naaa b ka
kx b
423 zyx
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Solutions
A solution of the linear equation
is a list of values for the unknowns of numbers
such that the equation is satisfied when we substitute
The set of all solutions of the equation is called its solution set
or sometimes the general solution of the equation.
n
nn kxkxkx .,........., 2211
bxaxaxa nn ........2211
nkkk ,......, 21
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Systems of linear equations
An arbitrary system of linear equations in unknowns can
be written as
Where are the unknowns and the subscripted
and denote constants. The system is called system.
If the system is called square. sb ,
)....(..........
........
........
........
2211
22222121
11212111
i
bxaxaxa
bxaxaxabxaxaxa
mnmnmm
nn
nn
m n
nxxxx ,........,, 321sa, nm
nm
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The system (i) is said to be homogeneous if all the constant
terms are zero. ie., if , otherwise the system is
non-homogeneous. Homogeneous system
I f at least one constant of the set is not zero then
the system (i) is called a non-homogeneous .
0.......21 mbbb
)....(..........
0........
0........0........
2211
2222121
1212111
ii
xaxaxa
xaxaxaxaxaxa
nmnmm
nn
nn
}.......,{ 21 mbbb
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Example
Homogeneous
Non-homogeneous
)......(..........002032
32
21
321
ixx
xxxxx
)......(..........0563134292
iizyxzyxzyx
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Solution
If all are known called a particular solution.
If all are not known called a general solution.
is a solution of the system i.e, if it satisfies
each equation of the system then this are called
particular solution.The set of all particular solutions of the system is called a
general solution.
nn kxkxkx .,........., 2211ski'ski'
nn Rkkk ,...., 21
nkkk ,...., 21
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Solution(contd.)
Every homogeneous system of linear equation is consistent,
since all such system have as solution.
This solution is called the zero or trivial solution. If
is a solution of the homogeneous system
nn Rkkk ,...., 21
)....(..........
0........
0........0........
2211
2222121
1212111
ii
xaxaxa
xaxaxaxaxaxa
nmnmm
nn
nn
0.,.........0,0 21 nxxx
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Solution(contd.)
If is a solution of the homogeneous
system and if at least one is not zero, it is called a non –zero
non-zero or non-trivial solution
nn Rkkk ,...., 21
ik
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Echelon Form and Free Variables
If the disappearance of the leading variable is increased one line
by another line is increased then the reduced system is called
the echelon form Example:
12
94632873
zzyzyx
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Echelon Form and Free Variables(contd.)
The variables which do not appear at the beginning are called
free variables Example:
Numbers of free variables and what are they:In the above example there are 3 equations with 4 variables
So 4-3=1 free variables.And leading variable is missing so is free variable.
ix
13
38973285242
wzwzywzyx
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Consistent & Inconsistent
A system of linear equations
is said to be consistent if no linear combination of its equations
is of the form
Otherwise the system is inconsistent.
14
0,0....00 21 bbxxx n
mjbxan
jjjij ,.....2,1,
1
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At a Glance
Linear equation Solution System of linear equation Solution of the System of linear equations Homogeneous and non-homogeneous System of
linear equations Solution of Homogeneous and non-
homogeneous System of linear equations Echelon form Free variables How many free variables and what are they?
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System of Linear Equation
System of Linear EquationNon-homogeneous systemof linear equation
Homogeneous systemof linear equation
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Non-homogeneous Linear Systems
System of linear Equations
Consistent
Unique Solution
In echelon form free variables does not
exist
More than one
solution
In echelon form free variables
exist
Inconsistent
No solution
In echelon form at least one equation will appear of the form
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Problems
Let the system
Solution: The given system is
18
3353220329
zyxzyxzyx
)....(..........1531129
~
)..(..........3353220329
iizyzyzyx
izyxzyxzyx
133
122
2LLLLLL
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19
)....(..........41129
~ iiizzyzyx
233 LLL
So from
Again from
Finally
The solution Is
43 zL
3.,.1182 yeiyL
2.,.9431 xeixL
)4,3,2(
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Similar problem: Solve
Solution: The given system is
20
2226213452113432
zyxzyxzyxzyx
)....(..........1531129
~
)..(..........3353220329
iizyzyzyx
izyxzyxzyx
133
122
2LLLLLL
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Similar problem: Find the solution of the following system of
linear equation by reducing it to the echelon form
(i)
(ii) no solution
21
123422252
zyxzyxzyxzyx )1,1,2(
14322252331345
tzyxtzyxtzyx
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Augmented Matrices
If we mentally keep track of the location of the the and
the a system of linear equations in unknowns can be
Abbreviated by writing only the rectangular array of members
This is called the augmented matrices for the system.
22
,'s ,'sx,'s m n
mmnmm
n
n
baaa
baaabaaa
..........
..........
..........
21
2122221
111211
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Example
The augmented matrices for the above system of equation
23
056313429211
0563134292
321
321
321
xxxxxxxxx
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Problem(Augmented Matrices)
Solve:
The augmented matrices for the above system of equation is
24
056313429211
0563134292
zyxzyxzyx
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.
~
~
25
056313429211
271130177209211
133
122
32RRRRRR
3100177209211
233 32 RRR
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Solve the following system
Solution: The given system is
26
42543568522232
tzyxtzyxtzyx
133
122
32LLLLLL
)(..........2442
1222232
~ iitzytzytzyx
).(..........42543568522232
itzyxtzyxtzyx
)(..........1222232~ iii
tzytzyx
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The above system (iii) shows that it has two equations of four
Unknowns, so the system has more than one solution. Here the
Number of free variables 4-2=2.And these areLet andWe get from
Thus we have the solution
27
aZ
)(..........1222232~ iii
tzytzyx
tz &bt
bayL 2212
baxL 21
bababatzyx ,,221,2,,, DKD
Solve the following system
Solution: The given system is
28
22551231222132132
zyxszyxszyxszyxszyx
22551231222132132
zyxszyxszyxszyxszyx
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~Here 4-3=1 free variable which is ,
LetWe get
The solution is 29
156135
132
szszyszyx
s as
651
1azL
671
618255636
25512aaaaayL
651
661531426
251
37111
a
aaaaaaxL
aaaa ,6
51,671,6
51
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Solve the following system
Solution: The given system is
30
826355223222143
wzyxwzyxwzyx
wzyx
826355223222143
wzyxwzyxwzyx
wzyx
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~
~
~
31
11131137911458143
wzywzywzwzyx
144
133
122
322
LLLLLLLLL
45811131137911143
wzwzywzywzyx
4583240447911143
wzwzwzywzyx
233 3LLL
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~
~
32
458810117911143
wzwzwzywzyx
33 41LL
233 3LLL
2025810117911143
wwzwzywzyx
54
2520
4 wL
0.
054108113
zie
zL
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~
The solution is
33
5175
4)0(112 yL
2510
55161
154.40.35
11
xL
54,0,5
1,2
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Problem: Show that the system
has (1) a unique solution if (III) more than one solution if (III) no solution if
Solution: The given system is
34
111
zyaxazyxzayx
1a2,1 aa
2a
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Solution: The given system is
35
).........(111
izyaxazyxzayx
133
122
aLLLLLL
).........(
111)1(0)1()1(~1
iiiazaaa
zayazayx
).........(1)1()1(0)1()1(~1
2
iiazaya
zayazayx
133 1 LaLL
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Discussion: Case -1: The system (iv) is echelon form. It has a unique
solutionif the coefficient of z of the third equation is not zero ie if
Case-II: It has more than one solution if . Since under this
condition there exists in (iv) two equations in three unknowns.
Case-III: then the system has no solution. Since under
this condition the third equation of (iv) becomes the impossible
solution.
36
1a
2,1 aa
).........(
1210)1()1(~1
ivazaa
zayazayx
)2(1
111)1)(1(1)1)(1(1
aaaa
aaaaaa
021 aa
2a
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Problem: What relation may exist among the constants
such that the following system has a solution
Solution: The given system is
37
czyxbzyxazyx
72116232
1a2,1 aa
2a
cba ,,
)(..........104
25232
iiaczyabzy
azyx
133
122 2LLLLLL
)....(..........72116232
iczyxbzyxazyx
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38
233 2LLL
(iii) & (i) are equivalent. Now for having a solution the system
(iii) must be exist and it is possible if
Similar problem: Determine the relationship among the
constants under which the following system has a
solution
052 abc
)(..........520
252~32
iiiabc
abzyazyx
rzyxqzyxpzyx
852332
rqp ,,
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39
Problem: For what values of and the following system of
linear equations has (i) no solution (ii) more than one solution
(iii) a unique solution.
Solution: The given system is
zyxzyxzyx
210326
izyxzyxzyx
.........2
10326
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40
)....(..........
61426
~ iizyzyzyx
233 LLL Case-1: If and then the last equation of the
system (iii) a non-zero real number, which is impossible .In
This case the system has no solution. Case-II: If and then there are two equations in
three unknowns. Since in this case the last equation of this
system becomes
3
133
122
LLLLLL
)....(..........
103426
~ iiizzyzyx
1003
10
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41
Case-1II: If then system has three equations in
the three unknowns. In this case the system has a unique
solution. Similar problem: e
10,3
103426
~ z
zyzyx
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42
Problem: For what values of the following system of
linear equations has (i) a unique solution (ii) more than one
solution (iii) no solution.
Solution: The given system is
233321
zayxazyxzyx
a
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43
Solution: The given system is
Case-I:
Case-II: Case-III:
azaazayzyx
2)2)(3(1)2(1
023 aa 2,3 aa
2a3a
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Homogeneous Linear Systems
System of homogeneous linear Equations
Consistent
Zero Solution In echelon form free variables does not exist
Non-zero solution
In echelon form free variables exist
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Problems (Homogeneous)
45
2. Solve
Solution: Given,
~
~
).(..........04420220322
iitwztwztwzyx
058630320322
twzyxtwzyxtwzyx
)(..........058630320322
itwzyxtwzyxtwzyx
133
122
3LLLLLL
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46
~The system (iii) is echelon form in which equation with 5
variables. So there are 5-2=3 free variables and they are
Let
We have from
Thus the solution is
).(..........0220322 iii
twztwzyx
cbcbacbatwzyx ,,22,,752,,,,
twy &,ctbway &,
bczL 222
cbaxL 7521
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Problems (Homogeneous)
47
Solve: 1. 2.
3. 4.
02550250320432
szyxszyxszyxszyx
0,0,0,0
074202320222
twzyxtwzyxtwzyx
0,,,,42 bbaba
02305557
0202
tszyxtszyx
tszyxtszyx
aa,,0,0,0
0327016131140233207543
szyxszyxszyxszyx
bababa ,,17
2019,17133
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48
Solve:
The given system is
~ )..(..........
0354052305320354
i
szyxszyxszyxszyx
0354052305320354
szyxszyxszyxszyx
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49
The given system is
~
~
Let
)..(..........
0151515010101005550354
ii
szyszyszyszyx
133
133
122
432
LLLLLLLLL
)..(..........00354 iii
szyszyx
The system (iii) is echelon form in which equation with 4 variables. So there are 4-2= free variables and they are
sz&bsaz , )(2 abyL abxL 1
baababszyx ,,,,,, DKD
