Linear Equations - Computational Methods - Lecture Slides, Slides of Calculus for Engineers

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Engr/Math/Physics 25

Chp8 Linear

Algebraic Eqns-

Learning Goals

• Define Linear Algebraic Equations
• Solve Systems of Linear Equations by Hand using - Gaussian Elimination (Elem. Row Ops) - Cramer’s Method
• Distinguish between Equation System Conditions: Exactly Determined, OverDetermined, UnderDetermined
• Use MATLAB to Solve Systems of Eqns

Linear Systems - Characteristics

• Examine the System of Equations

x y z

x y z

x y z

 We notice These Characteristics that DEFINE Linear Systems

 ALL the Variables are Raised EXACTLY to the Power of ONE (1)  COEFFICIENTS of the Variables are all REAL Numbers  The Eqns Contain No Transcendental Functions (e.g. ln, cos, e w^ )

Gaussian Elimination – ERO’s

• A “Well Conditioned” System of Eqns can be Solved by Elementary Row Operations (ERO): - Interchanges : The vertical position of two rows can be changed - Scaling : Multiplying a row by a nonzero constant - Replacement : The row can be replaced by the sum of that row and a nonzero multiple of any other row

ERO Example - 2

• The Scaling Operation

( ) [ ]

( ) [ 5 2 6 14 ] 5

3 12

6 3 4 41 6

2 12

1 12 5 7 26

− + + = −

− + =

• − = −

x y z

x y z

x y z

2 12 6 8 82

1 12 5 7 26

− − = −

− + =

• − = −

x y z

x y z

x y z

 Note that the 1 st Coeffiecient in the Pivot Eqn is Called the Pivot Value

• The Pivot is used to SCALE the Eqns Below it  Next Apply REPLACEMENT by Subtracting Eqs
• (2) – (1)
• (3) – (1)

ERO Example - 3

• The Replacement Operation Yields

2 0 11 15 108

1 12 5 7 26

− − = −

− + =

• − = −

x y z

x y z

x y z

Or

 Note that the x-variable has been ELIMINATED below the Pivot Row

• Next Eliminate in the “y” Column  We can use for the y-Pivot either of − or −9.
• For the best numerical accuracy choose the LARGEST pivot

2 11 15 108

1 12 5 7 26

− − = −

− + =

• − = −

y z

y z

x y z

ERO Example - 5

• Perform Replacement by Subtracting (3) – (2)

2 11 15 108

1 12 5 7 26

− = −

− + =

• − = −

z

y z

x y z

 Now Easily Find the Value of z from Eqn (3)

z = 116. 531 23. 306 = 5

 The Hard Part is DONE  Find y & x by BACK SUBSTITUTION  From Eqn (2)

33 11 3

11

108 75 11

108 15

= − = −

−

= − −

= −

y

y z

ERO Example - 6

• BackSub into (1)

2 12

24 12

35 15 26

12

7 5 26

12 5 7 26

= + − = =

⇒ = − −

• − = −

x

x z y

x y z

 Thus the Solution Set for Our Linear System

( ) ( ) ( ) 3 5 2 6 14

x y z

x y z

x y z

 x = 2  y = − 3  z = 5

Gaussian Elimination Summary

• INTERCHANGE Eqns Such that the PIVOT Value has the Greatest Magnitude
• SCALE the Eqns below the Pivot Eqn using the Pivot Value ratio’ed against the Corresponding Value below
• REPLACE Eqns Below the Pivot by Subtraction to leave ZERO Coefficients Below the Pivot Value

Poorly Conditioned Systems

• For Certain Systems Guassian Elimination Can Fail by - NO Solution → Singular System - Numerically Inaccurate Results → ILL-Conditioned System
• In a SINGULAR SYSTEM Two or More Eqns are Scalar Multiples of each other
• In ILL-Conditioned Systems 2+ Eqns are NEARLY Scalar Multiples of each other

Singular System - Geometry

• Plot This System on the XY Plane ( ) ( ) 2 2 4 5

x y

x y

 The Lines do NOT CROSS to Define a A Solution Point

 Singular Systems Have at least Two “PARALLEL” Eqns

y

ILL-Conditioned Systems

• A small deviation in one or more of the CoEfficients causes a LARGE DEVİATİON in the SOLUTİON.

1. 48 0. 99 1. 47

x y

x y 1

y

x

x y

x y 0

y

x

Matrix Methods for LinSys - 1

• Consider the Electrical Ckt Shown at Right

 The Operation of this Ckt May be Described in Terms of the

• Mesh Currents, I 1 -I 4
• Sources: 4 mA, 12 V
• Resistors: 1 & 2 kΩ

 Notice Mesh Currents I 1 & I 2 are Defined by SOURCES

Matrix Methods for LinSys - 3

• Using Techniques from ENGR43 find

 Recall Matrix Multiplication to Write the Equation system in Matrix Form

I I I mA

I I I mA

I I I

I mA

0 2 12

0 3 2 8

0 0

0 0 0 4

2 3 4

2 3 4

1 2 3

1

− − + = −

• • − =

• − + =

• • • =

























−

= 















































− −

−

−

12

8

0

4

0 1 1 2

0 1 3 2

1 1 1 0

1 0 0 0

4

3

2

1

I

I

I

I

A x^ b