Here I summarized the conditions on a system of linear equation 1) Consistent 2) Non-Consistent

1. Instructor:
By SAYED GHAYOOR ALI SHAH
Electrical Engineering Department

2. By SAYED GAHYOOR ALI SHAH
2
 A system of linear equations (or a linear system) is a collection of one or more linear
equations involving the same variables.
There are three possibilities for a linear system:
A system of linear equations has
1. No solution, or
2. Exactly one solution, or
3. Infinitely many solutions.
This whole system is classified in two things
1) CONSISTENT:
A system of linear equations is said to be consistent if it has either one solution or
infinitely many solutions.
2) INCONSISTENT:
A system is inconsistent if it has no solution.

3. By SAYED GAHYOOR ALI SHAH
3
Here for some idea lets see the examples:
0x + 0y + 0z = 0
like 0 = 0
Let, we have;
0x + 0y + 0z = a – 2
Or 0 = a – 2
Here we can determine value of variable a as,
a = 2
And a  – 2
(Infinitely many solutions)
(Many solutions) Consistent
( No solutions) Inconsistent

4. By SAYED GAHYOOR ALI SHAH
4
Determine if the following system is consistent:
x2 – 4x3 = 8
2x1 – 3x2 + 2x3 = 1
4x1 – 8x2 + 12x3 = 1
SOLUTION:
The augmented matrix is:
0 1 −4 8
2 −3 2 1
4 −8 12 1
By elementary row operation;
To obtain an x1 in the first equation, interchange rows 1 and 2:
2 −3 2 1
0 1 −4 8
4 −8 12 1
To eliminate the term 4x1 in the third equation, add -2 times row 1 to row 3:
(1)

5. By SAYED GAHYOOR ALI SHAH
5
2 −3 2 1
0 1 −4 8
0 −2 8 −1
Next, use the term x2 in the second equation to eliminate the -2x2 term from the third
equation. Add 2 times row 2 to row 3:
2 −3 2 1
0 1 −4 8
0 0 0 15
The augmented matrix is now in triangular form. To interpret it correctly, go back to
equation notation:
2x1 – 3x2 + 2x3 = 1
x2 - 4x3 = 8
0 = 15
the equation is never true. The original system is inconsistent (i.e., has no solution).

6. By SAYED GAHYOOR ALI SHAH
6
Solve the following system:
x2 + 4x3 = - 5
x1 + 3x2 + 5x3 = - 2
3x1 + 7x2 + 7x3 = 6
SOLUTION:
The augmented matrix is:
=
0 1 4 −5
1 3 5 −2
3 7 7 6
R1 R2
=
1 3 5 −2
0 1 4 −5
3 7 7 6
R3 + (-3)R1
=
1 3 5 −2
0 1 4 −5
0 −2 8 12
2R2 + R3
(2)

7. By SAYED GAHYOOR ALI SHAH
7
=
1 3 5 −2
0 1 4 −5
0 −2 8 12
-4R2 + R3
=
1 3 5 −2
0 1 4 −5
0 0 0 22
Here 0x3 = 22
0 = 22 ( 0  22 )
The original system is inconsistent ( has no solution)