R.Rosnawati


Any straight line in xy-plane can be
represented algebraically by an equation
of the form: ax + by = c
General form: define a linear equation in
the n variables x1 , x2 ,..., xn :
a1x1 a2x2 ...anxn  b
◦ Where a , a ,..., a , and b are real constants.
1
2
n
◦ The variables in a linear equation are sometimes
called unknowns.



1
x  3 y  7, y  x  3 z  1,
The equations
and
2
x1  2 x2  3x3  x4  7
are linear.
Observe that a linear equation does not involve any
products or roots of variables. All variables occur
only to the first power and do not appear as
arguments for trigonometric, logarithmic, or
exponential functions.
The equations sin x +cos x = 1, log x = 2
are not linear.


Find the solution of (a ) 4 x  2 y  1
Solution(a)
we can assign an arbitrary value to x and solve
for y , or choose an arbitrary value for y and solve
for x .If we follow the first approach and assign x
an arbitrary value ,we obtain 1
1
1
x  t1 , y  2t1 
◦ arbitrary numbers t
1,
◦ for example
t2
x  t2  , y  t2
2
2
4
are called parameter.
or
t1  3 yields the solution x  3, y 
11
2
as t 2 
11
2
◦ A solution of a linear equation is a sequence of n numbers
s1, s2, ..., sn such that the equation is satisfied. The set of
all solutions of the equation is called its solution set or
general solution of the equation


Find the solution of(b) x1  4 x2  7 x3  5.
Solution(b)
we can assign arbitrary values to any two
variables and solve for the third variable.
◦ for example
x1  5  4s  7t ,
x2  s ,
x3  t
◦ where s, t are arbitrary values



A finite set of linear equations
in the variablesx1 , x2 ,..., xn
a11x1  a12 x2  ...  a1n xn  b1
is called a system of linear
equations or a linear system . a21x1  a22 x2  ...  a2n xn  b2
A sequence of numbers
is called a
solution
the system.
s1 , s2 ,..., sof
n
A system has no solution is
said to be inconsistent ; if
there is at least one solution
of the system, it is called
consistent.




am1x1  am2 x2  ...  amn xn  bm
An arbitrary system of m
linear equations in n unknowns


Every system of linear equations has
either no solutions, exactly one
solution, or infinitely many solutions.
A general system of two linear
equations: (Figure1.1.1)
a1 x  b1 y  c1 (a1 , b1 not both zero)
a2 x  b2 y  c2 (a2 , b2 not both zero)
◦ Two lines may be parallel -> no
solution
◦ Two lines may intersect at only one point
-> one solution
◦ Two lines may coincide
-> infinitely many solution



The location of the +’s,
the x’s, and the =‘s can
be abbreviated by writing
only the rectangular array
of numbers.
This is called the
augmented matrix for the
system.
Note: must be written in
the same order in each
equation as the unknowns
and the constants must be
on the right.
a11 x1  a12 x2  ...  a1n xn  b1
a21 x1  a22 x2  ...  a2 n xn  b2




am1 x1  am 2 x2  ...  amn xn  bm
1th column
a11 a12 ... a1n b1 
a a ... a

b
2n
2 
 21 22
  
 


a
a
...
a
b
mn
m
 m1 m 2
1th row


The basic method for solving a system of linear equations
is to replace the given system by a new system that has
the same solution set but which is easier to solve.
Since the rows of an augmented matrix correspond to the
equations in the associated system. new systems is
generally obtained in a series of steps by applying the
following three types of operations to eliminate unknowns
systematically. These are called elementary row
operations.
1. Multiply an equation through by an nonzero constant.
2. Interchange two equation.
3. Add a multiple of one equation to another.
x  y  2z  9
2 x  4 y  3z  1
3x  6 y  5 z  0
1 1 2 9 
2 4  3 1


3 6  5 0
x y2
 2times
z 9
add
y  7 z  1 7
the2first
equation
3x  6 y  5 z  0
to the second
   
add - 2 times
the first row
to the second
  

add -3 times
the first equation
to the third
   
9 
1 1 2
0 2  7  17


3 6  5 0 
add - 3 times
the first row
to the third
   
x  y  2z 
the second
x  y  2 z  9 multiply
y  z  1
3 y  11by
z 0
2 y  7 z  17 equation
  2
3 y  11z  27
7
2
9
17
2
add - 3 times
the second equation
to the third
    
multily the second
9  add -3 times
9 
1 1 2
1 1 2
1
0 2  7  17 
0 1  7  17  the second row
row by
2
2 


2

to the third







0 3  11  27
0 3  11  27
x  y  2z 
9
x  y  2z 
9
Add -1 times the
second equation
to the first
Multiply
the third
y z 
equation
z by
3 -2
7
2
17
2
y  72 z   172    

   
 12 z   32
1 1 2
0 1  7
2

0 0  12
9 
 172 
 32 
Multily the third
row by - 2

1 1 2
0 1  7
2

0 0 1
9 
Add -1 times the
 172  second row
to the first

3  
x

11
2
z
35
2
y z
7
2
z
1 0 112

7
0
1

2

0 0 1
x
17
2
3


 
3 
35
2
17
2
y
Add - 11
times
2
the third equation
to 1the first and 72 times
 2third equation
the
zto
 the
3 second
    

Add - 11
times
2
the third row
to the first and 72
times the third row
to the second
   

1 0 0 1 
0 1 0 2 


0 0 1 3
 The solution x=1,y=2,z=3 is now evident.