Linear Equations - Computational Methods - Lecture Notes | ESM 2074, Study notes of Engineering

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Linear Equations

linear algebraic equationsMany problems give rise to

a 11 x 1

a 12 x 2

=

b 1

a 21 x 1

a 22 x 2

=

b 2

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In matrix form  

a 11

a 12

a 21

a 22

  (^) ·  

x 1

x 2  

b 1

b 2  

or

A

(^) x

b

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computing

inv(A)

is inefficient

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How to solve the system

Matlab

provides the

\

left

slash operator

x = A

\

b

A To help remember this we read

divides

b

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easy to solve  

a 11

a 21

a 22

  (^) ·  

x 1

x 2  

b 1

b 2  

equation forHere we can solve the first

x 1

x 1 =

b 1 /a

11

With

x 1 known,we can solve

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the second equation for

x 2

x 2 = (

b 2 −

(^) a

21

∗ (^) x

1 (^) ) (^) /a

22

We call this a

forward solution

procedure.

The

A

matrix is said to be

lower triangular

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the second equation for

x 1

x 1 = (

b 1 −

(^) a

12

∗ (^) x

2 (^) ) (^) /a

11

backward substitutionThis procedure is called a

The

A

matrix is said to be

upper triangular

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So how do we get such nice

matrices?

Let’s put in some numbers

x 1

• 3

x 2

= 1

x 1

• 5

x 2

= 9

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a scalar

m

2 and add

[

m

1 ∗ (^) 2 +

(^) m

2 ∗ (^) ( −

4)]

(^) x 1

+ [

m

1 ∗ (^) 3 +

(^) m

2 ∗ 5]

(^) x 2

[

m

1 ∗ (^) 1 +

(^) m

2 ∗ (^) 9]

sense that any (with the original two,in theThis new equation is consistent

x 1 , x

2 (^) ) that

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any choice ofalso satisfy the new equation for satisifies the original two,will

m

1 and

m

2 (^).

coefficient ofLet’s choose these so that the

x 1 is zero,that is

[

m

1 ∗ (^) 2 +

(^) m

2 ∗ (^) ( −

4)] = 0

For example,the choice

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In matrix form

 

2

4 11

  (^) ·  

x 1

x 2  

substitution procedure.can use the backwardlower traingular form and weIt’s clear the system is now in

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formalize this procedure.We will now proceed to

augmented matrixconvenient to form theTo explain the procedure it’s

by tacking

b

as an

n (^) + 1st column to

A

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rows 2zeroes in the first column for times row 1. This will place

,... n

Divide the second row by

a 22

the 2nd column for rows 3and proceed to place zeros in

,... n

Continue until

a nn

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triangular matrix.We have constructed an upper