Solve the following equations.
ππ + ππ = π
π + ππ = π
Aπ§π¬π°ππ«
π = π & π = βπ
Methods of Solution of Linear
Equations
Traditional Method Matrix Method
Methods of Solution of Linear
Equationsβ¦ Continueβ¦
Traditional Method Matrix Method
Row Echelon
Method
Reduce Row
Echelon Method
Get the Upper
Triangle
Get the Identity
Matrix
Methods of Solution of Linear
Equationsβ¦Continueβ¦
Row Echelon
Method
Reduce Row
Echelon Method
Get the Upper
Triangle
Get the Identity
Matrix
Gauss Elimination
Method
Gauss Jordan
Method
Derived at AD1850 Derived in AD1880
Why Iβm
Happy Today?
Donβt think Jordan was Smarter than
Gauss.!!
Related Field
1.Number Theory,
2. algebra,
3. statistics,
4. analysis,
5. differential geometry,
6. geodesy,
7. geophysics,
8. electrostatics,
9. astronomy,
10. Matrix theory and
11. optics
Related Field
1. geodesy
Methods of Solution of Linear
Equationsβ¦Continueβ¦
Now all we need is to construct a matrix from
given problem or set of equations.
This matrix is known as Augmented matrix.
Methods of Solution of Linear
Equationsβ¦Continueβ¦ Are you able to get Augmented matrix from set of equations?
2π₯ + π¦ β π§ = 8 β3π₯ β π¦ + 2π§ = β11 β2π₯ + π¦ + 2π§ = β3
2π₯ + π¦ β π§ = 3 π¦ + π§ = 10 π₯ + 2π§ = 9
π₯ β π¦ + 2 = 0 π¦ + π§ = 7 π₯ + 2π§ = π¦
2 1 β1β3 β1 2β2 1 2
|||
8β11β3
2 1 β10 1 11 0 2
|||
3109
1 β1 00 1 11 β1 2
||| β270
Set of Equations Augmented matrix
Example of Gauss Elimination Method
Ex.1 Solve the following equations by Gauss Elimination or
Backward Substitution.
2π¦ + π§ = β8 π₯ β 2π¦ β 3π§ = 0 βπ₯ + π¦ + 2π§ = 3
Solution:
Augmented Matrix
0 2 11 β2 β3
β1 1 2
|||
β803
Augmented Matrix
0 2 11 β2 β3
β1 1 2
|||
β803
Interchange π 1& π 2
1 β2 β30 2 1
β1 1 2
|||
0β83
π 3 + π 1
1 β2 β30 2 10 β1 β1
|||
0β83
Interchange π 2& π 3
1 β2 β30 β1 β10 2 1
|||
03
β8
1 β2 β30 β1 β10 2 1
|||
03
β8
π 3 + π 2(2)
1 β2 β30 β1 β10 0 β1
|||
03
β2
Therefore, π₯ β 2π¦ β 3π§ = 0 β¦ 1
βπ¦ β π§ = 3 β¦ 2 βπ§ = β2 β π = π β¦ (π)
By using (3) in (2) βπ¦ β 2 = 3 β βπ¦ = 5 β π = βπ β¦ (π)
By using (3), (4) in (1) π₯ β 2 β5 β 3 2 = 0
β π₯ + 10 β 6 = 0 β π = βπ β¦ (π)
Thus the solution of given equations
2π¦ + π§ = β8 π₯ β 2π¦ β 3π§ = 0 βπ₯ + π¦ + 2π§ = 3
are π = βπ, π = βπ
& π = π
Example of Gauss Jordan Method
Ex.1 Solve the following equations by Gauss Jordan
Method.
2π¦ + π§ = β8 π₯ β 2π¦ β 3π§ = 0 βπ₯ + π¦ + 2π§ = 3
Solution:
Augmented Matrix
0 2 11 β2 β3
β1 1 2
|||
β803
Augmented Matrix
0 2 11 β2 β3
β1 1 2
|||
β803
Interchange π 1& π 2
1 β2 β30 2 1
β1 1 2
|||
0β83
π 3 + π 1
1 β2 β30 2 10 β1 β1
|||
0β83
Interchange π 2& π 3
1 β2 β30 β1 β10 2 1
|||
03
β8
1 β2 β30 β1 β10 2 1
|||
03
β8
π 3 + π 2(2)
1 β2 β30 β1 β10 0 β1
|||
03
β2
π 1 β π 2(2)
1 0 β10 β1 β10 0 β1
|||
β63
β2
π 2 β π 3 & π 1 β π 3
1 0 00 β1 00 0 β1
|||
β45
β2
Therefore, π = βπ, π = βπ & π = π
Thus the solution of given equations
2π¦ + π§ = β8 π₯ β 2π¦ β 3π§ = 0 βπ₯ + π¦ + 2π§ = 3
are π = βπ, π = βπ
& π = π
1 0 00 β1 00 0 β1
|||
β45
β2
Multiply π 2 & π 3 both by (β1)
1 0 00 1 00 0 1
|||
β4β52
Which Method is Better for solution
purpose?
Why?
Traditional
or
Gaussian Elimination Method
Gauss Jordan Method
or
Can we extend these methods one
step ahead?
Rocket Velocity
The upward velocity of a rocket is
given at three different times
The velocity data is approximated by a polynomial as:
Find: The Velocity at π‘ = 5,7, 7.5 and 8 seconds.
π π = πππ + ππ + π , π β€ π β€ π
Problem: Given is not linear equation.
Question: Can we treat it as a linear?
Let me Help you,β¦
π π = πππ + ππ + π , π β€ π β€ π
In this equation π£ π‘ = ππ‘2 + ππ‘ + π β¦ 1 ; π‘ & π£ are given so
donβt worry about it. Time (π‘)
In second (unit)
Velocity (π£)
In πΎ.π.
π πππππ (unit)
π‘1 = 2 π£1 = 1
π‘2 = 4 π£2 = 2
π‘3 = 6 π£3 = 4
So, from 1st equation we have
π£1 = ππ‘12 + ππ‘1 + π β¦ 2
π£2 = ππ‘22 + ππ‘2 + π β¦ 3
π£3 = ππ‘32 + ππ‘3 + π β¦ 4
Do you think equations 2β² , 3β² & 4β² are linear equations?
1 = 4π + 2π + π β¦ 2β²
2 = 16π + 4π + π β¦ 3β²
4 = 36π + 6π + π β¦ 4β²
Will You Generate Matrix from this�
4 2 116 4 136 6 1
|||
124
Have you got thisβ¦!
Okay, then solveβ¦
Answer is π = π. πππ π = βπ. πππ π = π. πππ
1 = 4π + 2π + π β¦ 2β² 2 = 16π + 4π + π β¦ 3β²
4 = 36π + 6π + π β¦ 4β²
From the above values π = 0.125
π = β0.250
π = 1.000 & equation_(1) π£ π‘ = ππ‘2 + ππ‘ + π we have
π π = π. πππππ β π. πππ + π β¦ π
Equation_(5) Shows the Rocket Velocity equation in the time interval
of [2 8] seconds.
Now can you answer for these:
Find The Velocity at π‘ = 5,7, 7.5 and 8 seconds.
From equation_(5) π£ π‘ = 0.125π2 β 0.25π + 1
Put π‘ = 5 β π£ 5 = 0.125 Γ π2 β 0.25 Γ π + 1 β π π = π. πππ ππ
π
π‘ = 7 β π£ 7 = 0.125 Γ π2 β 0.25 Γ π + 1 β π π = π. πππ ππ
π
π‘ = 7.5 β π£ 7.5 = 0.125 Γ π. π2 β 0.25 Γ π. π + 1 β π π. π = π. πππ ππ
π
π‘ = 8 β π£ 8 = 0.125 Γ π2 β 0.25 Γ π + 1 β π π = π. ππ ππ
π
Would you balance the following
chemical reactions for me?
πΆπ»4 + π2 β πΆπ2 + π»2π
πππ»πΆπ3 + π»2ππ4 β ππ2ππ4 + πΆπ2 + π»2π
ππ(πΆ2π»3π2)2+π»π΅π β πππ΅π2 + π»πΆ2π»3π2
π»2 + π2 β 2π»2π
Can We Generate Matrix for each
reaction?
We will see in next lectureβ¦
If Yes then βHOW???β
Think & Come with pre-reading
What will be in next Lecture?
Can we solve Gauss Elimination &
Gauss Jordan Method in Mat-lab?
We will see in coming Lab.
If Yes then βHow???β
Think & Come with pre-reading
For Lab. β¦
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