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

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Condition (linear algebra)

  • 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)