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Gauss elimination & Gauss Jordan method

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Solve the following equations.

𝟐𝒙 + 𝟑𝒚 = 𝟔

𝒙 + 𝟐𝒚 = 𝟐

A𝐧𝐬𝐰𝐞𝐫

𝒙 = 𝟔 & 𝒚 = −𝟐

Methods of Solution of Linear

Equations

Methods of Solution of Linear

Equations… Continue…

Row Echelon

Method

Reduce Row

Echelon Method

Get the Upper

Triangle

Get the Identity

Matrix

Methods of Solution of Linear

Equations…Continue…

Row Echelon

Method

Reduce Row

Echelon Method

Get the Upper

Triangle

Get the Identity

Matrix

Gauss Elimination

Method

Gauss Jordan

Method

Why I’m

Happy Today?

Don’t 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

Equations…Continue…

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

Equations…Continue… Are you able to get Augmented matrix from set of equations?

2𝑥 + 𝑦 − 𝑧 = 8 −3𝑥 − 𝑦 + 2𝑧 = −11 −2𝑥 + 𝑦 + 2𝑧 = −3

2𝑥 + 𝑦 − 𝑧 = 3 𝑦 + 𝑧 = 10 𝑥 + 2𝑧 = 9

𝑥 − 𝑦 + 2 = 0 𝑦 + 𝑧 = 7 𝑥 + 2𝑧 = 𝑦

2 1 −1−3 −1 2−2 1 2

|||

8−11−3

2 1 −10 1 11 0 2

|||

3109

1 −1 00 1 11 −1 2

||| −270

Set of Equations Augmented matrix

Example of Gauss Elimination Method

Ex.1 Solve the following equations by Gauss Elimination or

Backward Substitution.

2𝑦 + 𝑧 = −8 𝑥 − 2𝑦 − 3𝑧 = 0 −𝑥 + 𝑦 + 2𝑧 = 3

Solution:

Augmented Matrix

0 2 11 −2 −3

−1 1 2

|||

−803

Augmented Matrix

0 2 11 −2 −3

−1 1 2

|||

−803

Interchange 𝑅1& 𝑅2

1 −2 −30 2 1

−1 1 2

|||

0−83

𝑅3 + 𝑅1

1 −2 −30 2 10 −1 −1

|||

0−83

Interchange 𝑅2& 𝑅3

1 −2 −30 −1 −10 2 1

|||

03

−8

1 −2 −30 −1 −10 2 1

|||

03

−8

𝑅3 + 𝑅2(2)

1 −2 −30 −1 −10 0 −1

|||

03

−2

Therefore, 𝑥 − 2𝑦 − 3𝑧 = 0 … 1

−𝑦 − 𝑧 = 3 … 2 −𝑧 = −2 ⇒ 𝒛 = 𝟐 … (𝟑)

By using (3) in (2) −𝑦 − 2 = 3 ⇒ −𝑦 = 5 ⇒ 𝒚 = −𝟓 … (𝟒)

By using (3), (4) in (1) 𝑥 − 2 −5 − 3 2 = 0

⇒ 𝑥 + 10 − 6 = 0 ⇒ 𝒙 = −𝟒 … (𝟓)

Thus the solution of given equations

2𝑦 + 𝑧 = −8 𝑥 − 2𝑦 − 3𝑧 = 0 −𝑥 + 𝑦 + 2𝑧 = 3

are 𝒙 = −𝟒, 𝒚 = −𝟓

& 𝒛 = 𝟐

Example of Gauss Jordan Method

Ex.1 Solve the following equations by Gauss Jordan

Method.

2𝑦 + 𝑧 = −8 𝑥 − 2𝑦 − 3𝑧 = 0 −𝑥 + 𝑦 + 2𝑧 = 3

Solution:

Augmented Matrix

0 2 11 −2 −3

−1 1 2

|||

−803

Augmented Matrix

0 2 11 −2 −3

−1 1 2

|||

−803

Interchange 𝑅1& 𝑅2

1 −2 −30 2 1

−1 1 2

|||

0−83

𝑅3 + 𝑅1

1 −2 −30 2 10 −1 −1

|||

0−83

Interchange 𝑅2& 𝑅3

1 −2 −30 −1 −10 2 1

|||

03

−8

1 −2 −30 −1 −10 2 1

|||

03

−8

𝑅3 + 𝑅2(2)

1 −2 −30 −1 −10 0 −1

|||

03

−2

𝑅1 − 𝑅2(2)

1 0 −10 −1 −10 0 −1

|||

−63

−2

𝑅2 − 𝑅3 & 𝑅1 − 𝑅3

1 0 00 −1 00 0 −1

|||

−45

−2

Therefore, 𝒙 = −𝟒, 𝒚 = −𝟓 & 𝒛 = 𝟐

Thus the solution of given equations

2𝑦 + 𝑧 = −8 𝑥 − 2𝑦 − 3𝑧 = 0 −𝑥 + 𝑦 + 2𝑧 = 3

are 𝒙 = −𝟒, 𝒚 = −𝟓

& 𝒛 = 𝟐

1 0 00 −1 00 0 −1

|||

−45

−2

Multiply 𝑅2 & 𝑅3 both by (−1)

1 0 00 1 00 0 1

|||

−4−52

Which Method is Better for solution

purpose?

Why?

or

Gaussian Elimination Method

Gauss Jordan Method

or

Can we extend these methods one

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 𝑡 = 5,7, 7.5 and 8 seconds.

𝒗 𝒕 = 𝒂𝒕𝟐 + 𝒃𝒕 + 𝒄 , 𝟐 ≤ 𝒕 ≤ 𝟖

Problem: Given is not linear equation.

Question: Can we treat it as a linear?

𝒗 𝒕 = 𝒂𝒕𝟐 + 𝒃𝒕 + 𝒄 , 𝟐 ≤ 𝒕 ≤ 𝟖

In this equation 𝑣 𝑡 = 𝑎𝑡2 + 𝑏𝑡 + 𝑐 … 1 ; 𝑡 & 𝑣 are given so

don’t worry about it. Time (𝑡)

In second (unit)

Velocity (𝑣)

In 𝐾.𝑚.

𝑠𝑒𝑐𝑜𝑛𝑑 (unit)

𝑡1 = 2 𝑣1 = 1

𝑡2 = 4 𝑣2 = 2

𝑡3 = 6 𝑣3 = 4

So, from 1st equation we have

𝑣1 = 𝑎𝑡12 + 𝑏𝑡1 + 𝑐 … 2

𝑣2 = 𝑎𝑡22 + 𝑏𝑡2 + 𝑐 … 3

𝑣3 = 𝑎𝑡32 + 𝑏𝑡3 + 𝑐 … 4

Do you think equations 2′ , 3′ & 4′ are linear equations?

1 = 4𝑎 + 2𝑏 + 𝑐 … 2′

2 = 16𝑎 + 4𝑏 + 𝑐 … 3′

4 = 36𝑎 + 6𝑏 + 𝑐 … 4′

Will You Generate Matrix from this…?

4 2 116 4 136 6 1

|||

124

Have you got this…!

Okay, then solve…

Answer is 𝒂 = 𝟎. 𝟏𝟐𝟓 𝒃 = −𝟎. 𝟐𝟓𝟎 𝒄 = 𝟏. 𝟎𝟎𝟎

1 = 4𝑎 + 2𝑏 + 𝑐 … 2′ 2 = 16𝑎 + 4𝑏 + 𝑐 … 3′

4 = 36𝑎 + 6𝑏 + 𝑐 … 4′

From the above values 𝑎 = 0.125

𝑏 = −0.250

𝑐 = 1.000 & equation_(1) 𝑣 𝑡 = 𝑎𝑡2 + 𝑏𝑡 + 𝑐 we have

𝒗 𝒕 = 𝟎. 𝟏𝟐𝟓𝒕𝟐 − 𝟎. 𝟐𝟓𝒕 + 𝟏 … 𝟓

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 𝑡 = 5,7, 7.5 and 8 seconds.

From equation_(5) 𝑣 𝑡 = 0.125𝒕2 − 0.25𝒕 + 1

Put 𝑡 = 5 ⇒ 𝑣 5 = 0.125 × 𝟓2 − 0.25 × 𝟓 + 1 ⇒ 𝒗 𝟓 = 𝟐. 𝟖𝟕𝟓 𝒌𝒎

𝒔

𝑡 = 7 ⇒ 𝑣 7 = 0.125 × 𝟕2 − 0.25 × 𝟕 + 1 ⇒ 𝒗 𝟕 = 𝟓. 𝟑𝟕𝟓 𝒌𝒎

𝒔

𝑡 = 7.5 ⇒ 𝑣 7.5 = 0.125 × 𝟕. 𝟓2 − 0.25 × 𝟕. 𝟓 + 1 ⇒ 𝒗 𝟕. 𝟓 = 𝟔. 𝟏𝟓𝟔 𝒌𝒎

𝒔

𝑡 = 8 ⇒ 𝑣 8 = 0.125 × 𝟖2 − 0.25 × 𝟖 + 1 ⇒ 𝒗 𝟖 = 𝟕. 𝟎𝟎 𝒌𝒎

𝒔

Would you balance the following

chemical reactions for me?

𝐶𝐻4 + 𝑂2 → 𝐶𝑂2 + 𝐻2𝑂

𝑁𝑎𝐻𝐶𝑂3 + 𝐻2𝑆𝑂4 → 𝑁𝑎2𝑆𝑂4 + 𝐶𝑂2 + 𝐻2𝑂

𝑍𝑛(𝐶2𝐻3𝑂2)2+𝐻𝐵𝑟 → 𝑍𝑛𝐵𝑟2 + 𝐻𝐶2𝐻3𝑂2

𝐻2 + 𝑂2 → 2𝐻2𝑂

Can We Generate Matrix for each

reaction?

We will see in next lecture…

If Yes then “HOW???”

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 “How???”