Explanation and practice problems of gauss elimination theorem made by RISHABH GUPTA

1. GAUSS
ELIMINATION
THEOREM

2. GAUSS ELIMINATION FOR SOLVING A
SYSTEM OF EQUATIONS
• Write the augmented matrix of the system of equations.
• Use elementary row operation to construct a row equivalent
matrix in row-echelon form.
• A matrix is row-echelon form when the lower left quadrant of the matrix
has all zero entries and in each row that is not all zeroes the first entry is
1.
• Write the system of equation corresponding to the matrix in row-
echelon form.
• Use back-substitution to find the solution of the system.
2
GAUSS ELIMINATION THEOREM

3. EXAMPLE -1
Q) Find the roots of the given set of linear equations using Gauss
elimination method
2𝑥 − 4𝑦 + 5𝑧 = 36
−3𝑥 + 5𝑦 + 7𝑧 = 7
5𝑥 + 3𝑦 − 8𝑧 = −31

4. EXAMPLE-1
4
GAUSS ELIMINATION THEOREM
Let us consider the set of linearly independent equations.
2𝑥 − 4𝑦 + 5𝑧 = 36
−3𝑥 + 5𝑦 + 7𝑧 = 7
5𝑥 + 3𝑦 − 8𝑧 = −31
Augmented matrix for the set is:
2 -4 5 36
-3 5 7 7
5 3 -8 -31

5. EXAMPLE-1
5
GAUSS ELIMINATION THEOREM
Step-1: Eliminate x from the 2nd and 3rd equation.
2𝑥 − 4𝑦 + 5𝑧 = 36
−3𝑥 + 5𝑦 + 7𝑧 = 7
5𝑥 + 3𝑦 − 8𝑧 = −31
2 -4 5 26
-3 5 7 7
5 3 -8 -31
2 -4 5 26
0 -1 14.5 61
0 13 -20.5 -121
2𝑥 − 4𝑦 + 5𝑧 = 36
−𝑦 + 14.5𝑧 = 61
13𝑦 − 20.5𝑧 = −121
𝑅`2=
3
2
𝑅1 + 𝑅2
𝑅`3= -
5
2
𝑅1 + 𝑅3

6. EXAMPLE-1
6
GAUSS ELIMINATION THEOREM
Step-2: Eliminate y from the 3rd Equation.
2 -4 5 26
0 -1 14.5 61
0 13 -20.5 -121
2𝑥 − 4𝑦 + 5𝑧 = 36
−𝑦 + 14.5𝑧 = 61
13𝑦 − 20.5𝑧 = −121
𝑅``3= 13𝑅2 + 𝑅3
2 -4 5 26
0 -1 14.5 61
0 0 168 672
2𝑥 − 4𝑦 + 5𝑧 = 36
−𝑦 + 14.5𝑧 = 61
168𝑧 = 672

7. EXAMPLE-1
7
GAUSS ELIMINATION THEOREM
Step-3:
𝑅```1= 0.5𝑅1
2 -4 5 26
0 -1 14.5 61
0 0 168 672
2𝑥 − 4𝑦 + 5𝑧 = 36
−𝑦 + 14.5𝑧 = 61
168𝑧 = 672
1 -2 2.5 18
0 1 -14.5 -61
0 0 1 4
𝑥 − 2𝑦 + 2.5𝑧 = 18
𝑦 − 14.5𝑧 = −61
𝑧 = 4
𝑅```2= −𝑅2
𝑅```3=
1
168
𝑅3

8. EXAMPLE-1
8
GAUSS ELIMINATION THEOREM
Step-4:
Now,
From Row 3, 𝑧 = 4
From Row 2, 𝑦 − 14.5𝑧 = −61 ⇒ 𝑦 − 14.5 4 = −61 ⇒ 𝑦 = −3
From Row 1, 𝑥 − 2𝑦 + 2.5𝑧 = 18 ⇒ 𝑥 − 2 −3 + 2.3 4 = 18 ⇒ 𝑥 = 2
1 -2 2.5 18
0 1 -14.5 -61
0 0 1 4
𝑥 − 2𝑦 + 2.5𝑧 = 18
𝑦 − 14.5𝑧 = −61
𝑧 = 4

9. EXAMPLE -2
Q) Find the roots of the given set of linear equations using Gauss
elimination method
2𝑥 + 𝑦 + 𝑧 = 10
3𝑥 + 2𝑦 + 3𝑧 = 18
𝑥 + 4𝑦 + 9𝑧 = 16

10. EXAMPLE-2
10
GAUSS ELIMINATION THEOREM
Let us consider the set of linearly independent equations.
2𝑥 + 𝑦 + 𝑧 = 10
3𝑥 + 2𝑦 + 3𝑧 = 18
𝑥 + 4𝑦 + 9𝑧 = 16
Augmented matrix for the set after pivoting is:
2 1 1 10
3 2 3 18
1 4 9 16
1 4 9 16
2 1 1 10
3 2 3 18

11. EXAMPLE-2
11
GAUSS ELIMINATION THEOREM
Step-1: Eliminate x from the 2nd and 3rd equation.
𝑥 + 4𝑦 + 9𝑧 = 16
2𝑥 + 𝑦 + 𝑧 = 10
3𝑥 + 2𝑦 + 3𝑧 = 18
1 4 9 16
2 1 1 10
3 2 3 18
1 4 9 16
0 -7 -17 -22
0 -10 -24 -30
𝑥 + 4𝑦 + 9𝑧 = 16
−7𝑦 − 17𝑧 = 22
−10𝑦 − 24𝑧 = −30
𝑅`2= 𝑅2 − 2𝑅1
𝑅`3= 𝑅3 − 3𝑅1

12. EXAMPLE-2
12
GAUSS ELIMINATION THEOREM
Step-2: Eliminate y from the 3rd Equation.
𝑥 + 4𝑦 + 9𝑧 = 16
−7𝑦 − 17𝑧 = 22
−10𝑦 − 24𝑧 = −30
1 4 9 16
0 -7 -17 22
0 -10 -24 -30
1 4 9 16
0 -7 -17 22
0 0 2 10
𝑥 + 4𝑦 + 9𝑧 = 16
−7𝑦 − 17𝑧 = 22
2𝑧 = 10
𝑅``3= 7𝑅3 − 10𝑅2

13. EXAMPLE-2
13
GAUSS ELIMINATION THEOREM
Step-3:
Now,
From Row 3, 2𝑧 = 10 ⇒ 𝑧 = 5
From Row 2, −7𝑦 − 17𝑧 = 22 ⇒ −7𝑦 − 17 5 = 22 ⇒ 𝑦 = −9
From Row 1, 𝑥 + 4𝑦 + 9𝑧 = 16 ⇒ 𝑥 + 4 −9 + 9 5 = 16 ⇒ 𝑥 = 7
1 4 9 16
0 -7 -17 22
0 0 2 10
𝑥 + 4𝑦 + 9𝑧 = 16
−7𝑦 − 17𝑧 = 22
2𝑧 = 10

14. Practice Problems
• Solve the following
system of equations
using gauss
elimination method:
• Find the roots of the
given set of
equations by using
gauss elimination
method:
Presentation title 14
• Solve the given set
of equations using
gauss elimination
method:
𝑥 + 𝑦 + 𝑧 = 2
𝑥 + 2𝑦 + 3𝑧 = 5
2𝑥 + 3𝑦 + 4𝑧 = 11
𝑥 + 𝑦 + 𝑧 = 9
2𝑥 + 5𝑦 + 7𝑧 = 52
2𝑥 + 𝑦 − 𝑧 = 0
𝑥 + 3𝑦 + 6𝑧 = 12
𝑥 + 4𝑦 + 5𝑧 = 14
𝑥 + 6𝑦 + 7𝑧 = 10

15. Thank you
22BCE10213-Rishabh Gupta
22BCE10233-Manav Rawat
22BCE10419-Ashish Rawat
22BCE10545-Shashank Pandey
22BCE11324-Ayush Tripathi