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MODUL–1
System of Linear Equation
The Theory of linear equation plays an important and motivating role in the subject of
linear algebra. In fact, many problem in linear algebra are equivalent to studying a
system of linear equations, e.g. finding the kernel of a linear mapping and characterizing
the subspace spanned by a set of a vectors. Thus the techniques introduced in this
chapter will be applicable to the more abstract treatment given later. On the other hand ,
some of the results of the abstract treatment will give us new insights in to the structure
of “concrete” system of linear equations.
For simplicity , we assume that all equations in this chapter are over the real field R. We
emphasize that the results and techniques also hold for equations over the comlec field
C or over any arbitrary field K.
LINEAR EQUATION
By a linear equation over the real field R, we mean an expression of the form ;
a1x1 + a2x2 + . . . .+ anxn = b
where the ai , b € R and the xi are indeterminants ( or: unknowns or variables). The
scalars ai are called the coefficients of the xi respectively, and b is called the constant
term of simply constantof the equation.
A set of valueas for the unknown, say : x1 = k1, x2 = k2, . . . . , xn = kn
Is a solution of (1) if the statement obtained by substituting ki for xi, such that :
a1k1 + a2k2 + . . . .+ ankn = b
is true, This set of values in then said to satisfy the equations. If there is no ambiguity
about the position of the unknowns in the equation, then we denote this solution by
simply the n tuple
u = (k1, k2, . . . , kn)
Example 2.1. Consider the equation : x + 2y – 4z + w = 3
The 4 tuple , u = (3, 2, 1, 0) is a solution of the equation since
3 + 2*2 – 4 * 1 + 0 = 3
Is a true statement , however, the 4 tuple v = (1, 2, 4, 5) is not a solution
of the equation since, because : 1 + 2 * 2 – 4 *4 + 5 = - 6 and - 6 not equal to 3
Is not a true statement.
2008
1
Matematika III
Ir. Sonny Koeswara, M.Sc
Pusat Bahan Ajar dan eLearning
http://www.mercubuana.ac.id
Step – 1 : Interchange equations so that the first unknown x1 has a non zero coefficient
in the first equation, that is, so that a11 ≠ 0
Step – 2 : For each I > 1 , apply the operation : Li - - ai1L1 + a11Li
That is, replace the ith linear equation Li by the equation obtained by multiplying the first
equation L1 by – ai1, multiplying the ith equation Li by a11, and then adding
Example 2.3. Consider the following system of linear equations :
2x + 4y – z + 2v + 2w
=1
3x + 6y + z – v + 4w
=-7
4x + 8y + z + 5v - w
=3
We eliminate the unknown x from the second and third equations by
applying the following operations :
L2 - - 3 L1 + 2L2 and L3 - - 2L1 + L3
- 6x – 12y + 3z – 6v – 6w
=-3
6x + 12y + 2z – 2v + 8w
= - 14
-----------------------------------------------5z – 8 v + 2w
= - 17 . . . . . L2
- 4x – 8y + 2z – 4v – 4w
=-2
4x + 8y + z + 5v – w
= 3
--------------------------------------------3z + v – 5w
= 1 . . . . . . L3
Thus the original system has been reduced to the following equivalent system :
2x + 4y – z + 2v + 2w
=1
5z – 8v = 2W
= - 17
3z + v – 5w
=1
Observe thet y has also been eliminated from the second and third equations. Here the
unknown z plays the role of the unknown xj2 above
We note that the above equations, excluding the first, form a subsystem which has fewer
equations and fewer unknowns than the original system (*) We also note that :
If an equation 0x1 + . . . + 0xn = b , b ≠ 0 occurs, then the equation can be
(i)
deleted without affecting the solution.
(ii)
If an equation 0x1 + . . . . + 0xn = 0 occurs, then the equation can be deleted
without affecting the solution.
2008
2
Matematika III
Ir. Sonny Koeswara, M.Sc
Pusat Bahan Ajar dan eLearning
http://www.mercubuana.ac.id
5x – 3y – z
= 16
We reduce the following system by applying the operations:
L2 : - 3 L1 + 2L2
and L3 : - 5 L1 + 2 L3
- 6x – 3y + 9z
- 10 x – 5 y + 15 z
= - 15
6x – 4y + 4z
10x - 6y – 2z
= 10 +
- 7 y + 13 z
= - 5 ---- L2
- 11 y + 13 z
= - 25
= 32 +
= 7 --- L3
From L2 and L3 ,we obtained : ---- 4 y = - 12 ---- y = - 3 , x = 1, and z = - 2, that is
the unique solution
x + 2y – 3z
2.)
=6
2x – y + 4z
=2
4x + 3y – 2z
= 14
L2 : - 2 L1 + L2 and
L3 : - 4L1 + L3
- 2x – 4y + 6z
2x – y + 4z
- 5y + 10z
And : - 4x – 8y + 12 z
= - 12
=
2
= - 10 ------ L2
= - 24
4x + 3y – 2z
= 14 +
- 5y + 10 z
= - 10 ------- L3
From L2 and L3, we find it : - 5y + 10 z = - 10 - or : - y + 2z = - 2
Let z = a , we obtained : y = 2a + 2 , substitusi to x + 2y – 3z = 6 , we find it : x = - a + 2
Now , x = 2 – a ,
y = 2 + 2a
and z = a that is consisitent with more than one solution
We reduce the following system by applying operations :
3.) x + 2y – 3z
=-1
3x – y + 2z
= 7
5x + 3y – 4z
= 2
We reduce the following system by applying operations :
L2 : - 3 L1 + L2 and L3 : - 5 L1 + L3
- 3x – 6y + 9z = 3
- 5x – 10y + 15z
=5
3x – y + 2z = 7 +
5x + 3y – 4z
=2 +
- 7y + 11z = 10
- 7y + 11z
=7
Because 0 ≠ 3,
So The Sysytem of Linear Equation Not Consistent , and not have solution
2008
3
Matematika III
Ir. Sonny Koeswara, M.Sc
Pusat Bahan Ajar dan eLearning
http://www.mercubuana.ac.id