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Augmented Matrix & System of Linear Equations
Linear Equation
y= mx (1)
y, x = variables
m = constant
Here, y is expressed in terms of x.
This is an equation of a line.
A Linear Equation in n variables:
a1x1 a2 x2 a3 x3 .... an xn b (2)
where are variables
are constants/scalars
Linear System
A Linear System of linear equations and unknowns is:
a11 x1 a12 x2 a13 x3 .... a1n xn b1
a21 x1 a22 x2 a23 x3 .... a2 n xn b2
a31 x1 a32 x2 a33 x3 .... a3n xn b3
....................................................... (3)
.......................................................
........................................................
am1 x1 am 2 x2 am3 x3 .... amn xn bm
where x1, x2 , x3 ,..., xn are variables/unknowns
aij’s and bij’s are constants.
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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
x - 3y 3 Eq1
2x y 8 Eq 2
Solution:
-2Eq1 + Eq2
2 x 6y 6
2x y 8
______________
+7y = 14 y = 2
From Eq 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
x - 3y 7 Eq1
2 x - 6y 7 Eq 2
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Solution:
2Eq - Eq2
2 x - 6y -7
2 x 6 y -14
_____________________
0 + 0 = -21
This makes no sense because 0 -21, hence there is no solution.
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:
a11 x - a12 y c1 l1
a21 x - a22 y c2 l2
There are three possibilities:
No solution one solution infinitely many solutions
[inconsistent] [consistent] [consistent]
Note:1. A system can have a unique solution (only one solution) only when the
number of unknowns is equal to the 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 can be inconsistent.
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Augmented Matrix
System of linear equations:
a11 x1 a12 x2 a13 x3 b1
a21 x1 a22 x2 a23 x3 b2
a31 x1 a32 x2 a33 x3 b3
can be written in the form of matrices product
a11 a12 a13 x1 b1
a a a23 x2 b2
21 22
a31 a32 a33 x3 b3
or we may write it in the form AX=b,
a11 a12 a13 x1 b1
where A= a21 a22 a23 , X = x , b = b2
2
a31 a32 a33 x3 b3
a11 a12 a13 b1
Augmented matrix is A b a21 a22 a23 b2
a31 a32 a33 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
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3 1 6 x 6
1 1 1 y 2
2 1 4 z 3
3 1 6 6
Augmented form is A b 1 1 1 2 .
2 1 4 3
Methods for solving System of Linear equations
Gaussian Elimination Method (get REF)
Gauss – Jordan Method (get RREF)
Gaussian Elimination Method
STEP 1. by using elementary row operations
a11 a12 a13 b1 1 A12 A13 B1
a a a23 b2 0 1 A23 B2
21 22
a31 a32 a33 b3 0 0 1 B3
STEP 2. Find solution by back – substitutions.
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 1 2 8
1 2 3 1
3 7 4 10
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STEP 1.
1 1 2 8 R12 (1) 1 1 2 8
1 2 3 1
R13( 3)
3 7 4 10
1 1 2 8
R 2 ( 1) 0 1 5 9
0 10 2 14
1 1 2 8
R 23(10) 0 1 5
9
0 0 52 104
1 1 2 8
0 1 5 9
R 3
0 0 1 2
Equivalent system of equations form is:
x1 + x2 + 2x3 =8
x2 - 5x3 = -9
x3 = 2
STEP 2. Back Substitution
x3 = 2
x2 = 5x3 -9 =10 – 9 =1
x1 = - x2 - 2x3 + 8 = -1 – 4 +8 = 3
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Solution is x1 = 3, x2 = 1, x3 = 2.
or written as x1, x2 , x3 3,1,2
Gauss – Jordan Method
a11 a12 a13 b1 1 0 0 B1
a a a23 b2 0 1 0 B2
21 22
a31 a32 a33 b3 0 0 1 B3
Example.4. Solve the system of linear equations by Gauss - Jordan
elimination method
x1 + x2 + 2x3 = 8
- x1 - 2x2 + 3x3 = 1
3x1 - 7 x2 + 4x3 = 10
Solution: Augmented matrix is
1 1 2 8
1 2 3 1
3 7 4 10
The first step is to reduce the matrix to REF
Step 1 of Gaussian Elimination
1 1 2 8 R12 (1) 1 1 2 8
1 2 3 1 0 1 5 9
R13( 3)
3 7 4 10 0 10 2 14
8
1 1 2 8
R 2 ( 1) 0 1 5 9
0 10 2 14
1 1 2 8
R 23(10) 0 1 5
9
0 0 52 104
1 1 2 8
1
0 1 5 9
R
3 52
0 0 1 2
R 31 2 1 1 0 4
0 1
R 32
0 0 1 2
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Example: 5. Use the Gauss – Jordan method to solve the following system
of linear equations
x y 2 z w 1
2 x y 2 z 2w 2
x 2 y 4z w 1
3x - 3w 3
Solution:
Gauss-Jordan 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
3 0 0 3 3
There is a leading entry ‘1’ in the first row, making all other entries in the
first column zero
1 1 2 1 1 1 1 2 1 1
2 1 2 2 2 0 1 2 0 0
1 2 4 1 1 0 1 2 0 0
3 0 0 3 3 0 3 6 0 0
1 0 1 1
0
0 1 2 0 0
0 0 0 0 0
0 0 0 0 0
is the reduced row echelon form.
Equivalent system of equations form is
x w 1
y 2z 0
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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 VARIABLES
x and y are leading variables and w and z are free variables.
Let z = and w = , where and are real numbers ,
x = -1 +w = -1 +
y = 2z = 2
z=
w = ,
We write the solution in this form:
x, y, z, w 1,0,0,0 0,2,1,0 1,0,0,1 , ,
or
Solution set = x, y, z, w 1,0,0,0 0,2,1,0 1,0,0,1 ,
There are infinite many solutions of the given system.
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 2 1 4 1
1 3 7 2 2
1 12 11 16 5
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Reducing it to row echelon form (using Gaussian - elimination method)
1 2 1 4 1 R12 1 1 2 1 4 1
1 3
7 2 2 0 5 6 6 1 etc.
R13 1
1 12 11 16 5 0 10 12 12 4
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.
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 2 14) z a 2
(i) infinitely many solutions?
(ii) No solution?
(iii) Exactly one solution?
Solution:
Augmented matrix is
1 2 3 4
3 1 5 2
4 1 a 14 a 2
2
Reducing it to row echelon form
12
1 2 3 4 1 2 3 4
3 1
5 2 0 7 14 10
4 1 a 2 14 a 2 0 7 a 2 2 a 14
1 2 3 4
0 1 2 10
7
0 7 a 2 2 a 14
1 2 3 4
0 1 2 10
7
0 0 a 16 a 4
2
writing in the equation form,
x 2 y 3z 4 1
y 2z 107 2
( a 2 16) z a 4 3
or equation 3 can be written as
(a 4) (a 4) z a 4
CASE I .
a4 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
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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
Equation 3 (1 4)(1 4) z 1 4
-15z -3
z 15
y 107 52 35
64
x 4 53 2( 64
35 ) 35
47
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
2 1 3 c
reducing it to reduced row echelon form we obtain its row-equivalent matrix
1 1 2 a
0 1 1 ba
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.
