Download 1.1solution of linear system

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System of Linear Equations and Matrices
1.1 Linear Equation
y= mx
.1
is an equation, in which variable y is expressed in terms of x and the constant
m , is called Linear Equation. In Linear Equation exponents of the variable is always
‘ one’.
Equation 1 is also called equation of line.
1.2Linear Equation in n variables:
a1x1  a2 x2  a3 x3  ....  an xn  b
.2
where x1 , x2 , x3 ,..., xn are variables and
a1 , a2 , a13 ,..., an and b are constants.
1.3 Linear System:
A Linear System of m linear equations and n unknowns is:
a11x1  a12 x2  a13 x3  ....  a1n xn  b1
a21x1  a22 x2  a23 x3  ....  a2 n xn  b2
a31x1  a32 x2  a33 x3  ....  a3n xn  b3
.......................................................
.3
.......................................................
........................................................
am1 x1  am 2 x2  am 3 x3  ....  amn xn  bm
where x1 , x2 , x3 ,..., xn are variables or unknowns and a’s and b’s
are constants.
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1.4 Solution:
Solution of the linear system (3) is a sequence of n numbers
s1, s2 , s3 ,..., sn , which satisfies system (3) when we substitute
x1  s1 , x2  s2 , x3  s3 ,..., xn  sn .
Example.1. Solve the system of equations
Solution:
x - 3y   3
1
2x  y  8
2
-2E1 + E2 
 2x  6 y  6
2x  y  8
______________
+7y = 14
 y=2
From equation 1
x = -3 +3y
x = -3 + 6 = 3
Solution is x = 3 and y = 2
Check
Substitute the solution in Equations 1 and 2
Equation 1  3 – 3(2) = 3 – 6 = -3
Equation 2  2(3) +2 = 6 + 2 = 8 .
Example.2. Solve the system of equations
Solution:
x - 3y   7
1
2x - 6 y  7
2
2E1 - E2 
2x - 6 y  - 7
 2 x  6 y  - 14
_____________________
0 + 0 = -21
This makes no sense as 0  -21, hence there is no solution.
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NOTE:
Inconsistent , the system of equations is inconsistent, if the system has
no solution.
Consistent, the system of equations is consistent if the system has at
least one solution.
Example: Inconsistent and consistent system of equations
For the system of linear equations which is represented by straight lines:
a1 x - b1 y  c1
 l1
a2 x - b 2 y  c2
 l2
There are three possibilities:
No solution
solutions
[inconsistent]
one solution
[consistent]
infinite many
[consistent]
Note:1. A system will have unique solution (only one solution)when number of
unknowns is equal to number of equations
Note:2. A system is over determined , if there are more equations then unknowns and it
will be mostly inconsistent.
Note:3. A system is under determined if there are less equations then unknowns and it
may turn inconsistent.
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1.5 Augmented Matrix
System of linear equations:
a11x1  a12 x2  a13 x31  b1
a21x1  a22 x2  a23 x31  b2
a31x1  a32 x2  a33 x31  b3
can be written in the form of matrices product
 a11 a12
a
 21 a22
a31 a32
a13   x1   b1 
a23   x2   b2 
a33   x3  b3 
or we may write it in the form AX=b,
 a11 a12
where A= a21 a22
 a31 a32
a13 
a23  , X =
a33 
 x1 
x 
 2
 x3 
 a11 a12
Augmented matrix is A : b  a21 a22
a31 a32
 b1 
, b = b2 
b3 
a13
a23
a33
b1 
b2 
b3 
Example: 4. Write the matrix and augmented form of the system of linear equations
3x – y + 6z = 6
x+y+ z=2
2x + y +4z = 3
Solution: Matrix form of the system is
3
1

2
1
1
1
6   x  6 
1   y   2
4  z  3
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 3  1 6 6
Augmented form is A : b  1 1 1 2  .


2 1 4 3
1.5 Elementary Row operations:
Elementary row operations are steps for solving the linear system of equations:
I.
II.
III.
Interchange two rows
Multiply a row with non zero real number
Add a multiple of one row to another row
Note: Elementary row operations produce same results when operated either on a
system or on its augmented matrix form.
1.6 Methods for solving System of Linear equations
1.
2.
Gaussian Elimination Method
Gauss – Jorden Elimination Method
1.7 Gaussian Elimination Method
STEP 1. by using elementary row operations
 a11 a12
a
 21 a22
a31 a32
a13
a23
a33
b1  1
b2   0
b3  0
A12
1
0
A13 B1 
A23 B2 
1 B3 
STEP 2. Find solution by back – substitutions.
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Example:3. Solve the system of linear equations by Gaussion- elimination method
x1 + x2 + 2x3 = 8
- x1 - 2x2 + 3x3 = 1
3x1 - 7 x2 + 4x3 = 10
Solution:
Augmented matrix is
1 2 8
1
 1  2 3 1 


 3  7 4 10
STEP 1.

2 8 
1 1
0  1
5 9 

0  10  2 - 14

2
1 1
0 1  5

0 0  52

R1+R2, -3R1+R3
8 
- 9  -R2, 10R2+R3
- 104
8
1 1 2
0 1  5  9

 -R3/52
0 0 1
2 
Equivalent system of equations form is:
x1 + x2 + 2x3 = 8
x2 - 5x3 = -9
x3 = 2
STEP 2. Back Substitution
Solution is
x1 = 3,
x3 = 2
x2 = 5x3 -9 =10 – 9 =1
x1 = - x2 - 2x3 + 8 = -1 – 4 +8 = 3
x2 = 1,
x3 = 2.
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1.8 Gauss – Jorden Elimination Method
 a11 a12
a
 21 a22
a31 a32
a13
a23
a33
b1  1 0 0 B1 
b2   0 1 0 B2 
b3  0 0 1 B3 
Example.4. Solve the system of linear equations by Gauss - Jorden elimination method
x1 + x2 + 2x3 = 8
- x1 - 2x2 + 3x3 = 1
3x1 - 7 x2 + 4x3 = 10
Solution:
Augmented matrix is
1 2 8
1
 1  2 3 1 


 3  7 4 10





2 8 
1 1
0  1
5 9 

0  10  2 - 14
2
8 
1 1
0 1  5
- 9 

0 0  52 - 104
8
1 1 2
0 1  5  9


0 0 1
2 
1 1 0 4 
0 1 0 1 


0 0 1 2
1 0 0 3 
0 1 0 1 


0 0 1 2
R1+R2, -3R1+R3
-R2, 10R2+R3
-R3/52
-2R3+R1, 5R3+R2
-R2+R1
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Equivalent system of equations form is:
x1 = 3
x2 = 1
x3 = 2 is the solution of the system.
1.9 Row Echelon Form
A form of a matrix, which satisfies following conditions, is row echelon form
i.
ii.
iii.
iv.
‘1’ (leading entry) must be in the beginning of each row,
‘1’ must be on the right of the above leading entry,
Below the leading entry all values must be zero,
A row containing all zero values must be in the bottom.
Examples:
(i)
1 2 3 4
0 1 2 3


0 0 1 2
1
0
(ii) 
0

0
2 3 4
1 2 3
0 1 2

0 0 0
0 1 2 3 4
(iii) 0 0 0 1 2


0 0 0 0 1
1.10 Reduced Row Echelon Form
A form of a matrix, which satisfies following conditions, is row echelon form
i.
ii.
iii.
iv.
‘1’ (leading entry) must be in the beginning of each row,
‘1’ must be on the right of the above leading entry,
All entries in the column containing leading entry must be zero,
A row containing all zero values must be in the bottom.
Examples
1 0 0 3 
(i) 0 1 0 2 ,
0 0 1 1 
1 0 0
(ii) 0 1 0 , (iii)


0 0 1
0
0

0

0
1  2 0 1
0 0 1 3
0 0 0 0

0 0 0 0
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Example:5. Use Gauss – Jorden method to solve the system of linear system
x  y  2 z  w  1
2 x  y  2 z  2 w  2
 x  2 y  4z  w  1
3x
- 3w   3
Solution: Gauss-Jorden method is same as to reduce the augmented matrix to reduced
row echelon from.
Augmented matrix is
 1  1 2  1  1
 2 1  2  2  2


 1 2  4 1
1 


0  3  3
3 0
There is a leading entry ‘1’ in the first row, making all other entries in the first
column zero
1  1 2  1  1
0 1  2 0
0 

(-2R1+R2)/3 , R1+R3 , -3R1+R4

0 1  2 0
0


0
0 3  6 0
1 0 0  1  1
0 1  2 0 0 
 R2+R1 , -R2+R3 , -3R2+R4

0 0 0
0 0


0 0
0 0 0
is reduced row echelon form
Equivalent matrix form is
x  w  1
y  2z  0
there are four variables x, y, w and z in the example, variables appearing as
leading entries are called LEADING VARIABLES, and other variables are
FREE VARIABLE
x and y are leading variables and w and z are free variables.
Let z = s and w = t , where s and t are real numbers ,
x = -1 +w = -1 + t
y = 2z
= 2s
z=s
w = t,
There are infinite many solutions of the given system.
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SYSTEM WITH NO SOLUTION
Example: 6 . Solve the system of linear equations
x  2y 
z  4u  1
x  3 y  7 z  2u  2
x  12 y  11z  16u  5
Solution:
Augmented matrix is:
1
 4 1
1  2
1 3
7
2 2

1  12  11  16 5
Reducing it to row echelon form (using Gaussian - elimination method)
1
 4 1
1  2

 0 5
6
6 1
0  10  12  12 4
1  2 1  4 1 
 0 5 6 6
1 
0 0 0 0  3
R2- R1, R3-R1
-R3+2R2
Last equation is
0 x  0 y  0 z  0u  3
but
0  -3
hence there is no solution for the given system of linear equations.
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Conditions on Solutions
Example:7. For which values of ‘a’ will be following system
x  2 y  3z
4
3x  y  5 z
2
4 x  y  (a  14) z  a  2
infinitely many solutions?
No solution?
Exactly one solution?
2
(i)
(ii)
(iii)
Solution:
Augmented matrix is
3
4 
1 2
3  1
5
2 

4 1 a 2  14 a  2
Reducing it to reduced row echelon form
3
4 
1 2

 0  7  14
 10 
0  7 a 2  2 a  14
R2-3R1, R3-4R1
3
4 
1 2
10 
 17 R2, R3-R2
 0 1
2
7 
0 0 a 2  16 a  4
writing in the equation form,
x  2 y  3z
y  2z
4
1
 107
2
(a 2  16) z  a  4
or equation 3 can be written as
(a  4) (a  4) z  a  4
3
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CASE I .
a4
 0z  0
x  2 y  3z  4
y  2 z  107
as number of equations are less than number of unknowns, hence the system has
infinite many solutions,
let
z=t
y  107  2t
x  4  3t  4t  207  t  87
where ‘t’ is any real number.
CASE II
a  4

0z  -8 , but 0  -8 , hence, there is no solution.
CASE III
a  4, a  4, let
a 1
Equatins .3.  (1  4)(1  4) z  1  4
- 15z  -3
z
1
5
y  107  52 
64
35
x  4  53  2( 64
)  47
35
35
the system will have unique solution when a  4 and a  -4
and for a=1 the solution is
x
47
35
,y 
64
35
and
z  15 .
NOTE: (i) a=-4, no solution,
(ii) a=4, infinite many solutions and
(iii)a  4, a  -4, exactly one solution .
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Example:8. What conditions must a, b, and c satisfy in order for the system of
equations
x  y  2z  a
x
z b
2 x  y  3z  c
to be consistent.
Solution:
The augmented matrix is
1 1 2 a 
1 0 1 b  reducing it to reduced row echelon form


2 1 3 c 
2
a 
1 1

 0  1  1 b  a  R2-R1, R3-2R1
0  1  1 c  2a 
a 
1 1 2

 0  1  1
b  a  R3-R1
0 0 0 c  a  b
The system will be consistent if only if c – a - b = 0
or c = a + b
Thus the required condition for system to be consistent is
c = a + b.