1. Chp8 Linear Algebraic Eqns-1
Engr/Math/Physics 25
Chp8 Linear Algebraic Eqns-1
Bruce Mayer, PE
Registered Electrical & Mechanical Engineer

2. 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

3. Linear Equations  Example
In Many Engineering Analyses (e.g.: ENGR36 & ENGR43) The Engineer Must Solve Several Equations in Several Unknowns; e.g.:
Contains 3 Unknowns (x,y,z) in the 3 Equations

4. Linear Systems - Characteristics
Examine the System of Equations
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, ew)
We notice These Characteristics that DEFINE Linear Systems
Transcendental Function  Functions which cannot be given by any algebraic expression involving only their variables and constants

5. 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 (Swap Eqns)
Scaling: Multiplying a row by a nonzero constant (Equalize CoEffs)
RePlacement: The row can be replaced by the sum of that row and a nonzero multiple of any other row (Add/Subtract Eqns)
ERO a.k.a “Pivoted Elimination”

6. ERO Example: Solve for x, y, z
Let’s do some MTH25  Solve on WhiteBoard by ERO’s
Fa16: Short on Time; Do on Screen
Did not complete lecture from 31Oct; picked up at Wherehouse

7. ERO Example - 1 Let’s Solve The System of Eqns
Next SCALE by using Eqn (1) as the PIVOT To Multiply
(2) by 12/6
(3) by 12/[−5]
INTERCHANGE, or Swap, positions of Eqns (1) & (2)

8. ERO Example - 2 The Scaling Operation
Note that the 1st 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)

9. ERO Example - 3 The Replacement Operation Yields
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 −11 or −9.8
For the best numerical accuracy choose the LARGEST pivot Or

10. ERO Example - 4 Our Reduced Sys
Since |−11| > |−9.8| we do NOT need to interchange (2)↔(3)
Scale by Pivot against Eqn-(3) Or

11. ERO Example - 5 Perform Replacement by Subtracting (3) – (2)
The Hard Part is DONE
Find y & x by BACK SUBSTITUTION
From Eqn (2)
Now Easily Find the Value of z from Eqn (3)

12. ERO Example - 6 Q.E.F. BackSub into (1) x = 2 y = −3
z = 5
Q.E.F.
Thus the Solution Set for Our Linear System

13. Importance of Pivoting
Computers use finite-precision arithmetic
A small error is introduced in each arithmetic operation, AND… error propagates
When the pivot element is very small, then the multipliers will be even smaller
Adding numbers of widely differing magnitude can lead to a loss of significance.
To reduce error, row interchanges are made to maximize the magnitude of the pivot element

14. 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

15. 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

16. A Singular (InConsistent) Sys
Consider 2-Eqns in 2-Unknowns
Perform Elimination by
Swapping Eqns
Mult (2) by 2/1
Subtract (2) – (1)

17. Singular System - Geometry
Plot This System on the XY Plane y
The Lines do NOT CROSS to Define a Solution Point
Singular Systems Have at least Two “PARALLEL” Eqns

18. ILL-Conditioned Systems
A small deviation in one or more of the CoEfficients causes a LARGE DEVİATİON in the SOLUTİON.
2% change in one of 3 CoEff leads to major change in the solution

19. ILL-Conditioned Systems - 2
Systems in Which a Small Change in a CoEfficient Produces Large Changes in the Solution are said to be STIFF
Essentially the Lines Have very nearly Equal SLOPES
Tilt Region
“Tilting” The Equations just a bit Dramatically Shifts the Solution (Crossing Point)

20. 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, I1-I4
Sources: 4 mA, 12 V
Resistors: 1 & 2 kΩ
Notice Mesh Currents I1 & I2 are Defined by SOURCES

21. Matrix Methods for LinSys - 3
Using Techniques from ENGR43 find
Recall Matrix Multiplication to Write the Equation system in Matrix Form
A & b are KNOWN; need to find x

22. Matrix Methods for LinSys - 3
Thus The (linear) Ckt Can be Described by
This Can Be Written in Std Math Notation
Where
A  CoEfficient Matrix
m-Rows x n-Colunms
b  Constraint Vector
x  Solution Vector
X & b have the same no. of cols as A. The no. of rows in A is the number of eqns.

23. Determinants - 1
If we Solve a LinSys by Elimination we may do a Lot of work Before Discovering that the system is Singular or Very-Stiff
Determinants Can Alert us ahead of time to these Difficulties
Determinants are Defined only for SQUARE Arrays
The 2x2 Definition
D2 is Sometimes called the “Basic Minor”

24. Determinants - 2
Calculating Larger-Dimension DETs becomes very-Tedious very-Quickly
Consider a 3x3 Det
Example

25. Determinants - 3
A Determinant, no matter what its size, Returns a SINGLE Value
Matrix vs. Determinant
For Square Matrix A the Notation
MATLAB vs det
The det Calc is quite Painful, but MATLAB’s “det” Fcn Makes it Easy
For the D3ex
>> A = [-4,9,6; 7,13,-2; -3,11,5];
>> D3ex = det(A)
D3ex = 87 

26. Determinant Indicator - 1
The LARGER the Magnitude of the Determinant relative to the CoEfficients, The LESS-Stiff the System (approx.)
If det=0, then the System is SINGULAR

27. Determinant Indicator - 2
Consider this System
Check the “Stiffness”
Thus The system appears NON-Stiff
Find Solution by Elimination as

28. x = A\b MATLAB Left Division Read
MATLAB has a very nice Utility for Solving Well-Conditioned Linear Systems of the Form
The Syntax is Quite Simple
the hassle is entering the Matrix-A and Vector-b
x = A\b
Well Conditioned →
Square System → No. of Eqns & Unknowns are Equal
det  0
Read

29. Left-Div Example - 1
Consider a 750 kg Crate suspended by 3 Ropes or Cables
Using Force Mechanics from ENGR36 Find 3 Eqns in 3 Unknowns

30. Left-Div Example - 2 The MATLAB Command Window Session Or
0.8, , ;...
-0.36, , ];
>> w = [0; 9.81*750; 0]
>> T = A\w
T =
1.0e+003 *
2.6254
3.8157
2.4258 Or
TAB = kN
TAC = kN
TAD = kN

31. Matrix Inverse - 1
Recall The Matrix Formulation for n-Eqns in n-Unknowns
Note that the IDENTITY Matrix , I, Has Property
In Matrix Land
Use A-1 in Matrix Eqn
Order of A-1 is critical on both sides as MAT-Mult is NOT commutative
To Isolate x, employ the Matrix Inverse A-1 as Defined by

32. Matrix Inverse - 2 Thus the Matrix Shorthand for the Solution
In addition A-1 is, in general, Less Numerically Accurate Than Pivoted Elimination
However
Determining the Inverse is NOT Trivial (Ask your MTH6 Instructor)
is Symbolically Elegant and Will be Useful in that regard

33. Compare MatInv & LeftDiv
% Bruce Mayer, PE
% ENGR25 * 21Oct09 * Rev 08Apr13 * Rev 2Nov16
% file = Compare_MatInv_LeftDiv_0910 %
A = [3 -7 8; ; ]
b = [13; -29; 37]
Ainv = inv(A)
xinv = Ainv*b
xleft = A\b
% CHECK Both by b = A*x
CHKinv = A*xinv
CHKleft = A*xleft
Time for
Live
Demo

34. Matrix Inversion by Adjoint
All Done for Today
Given A, Find A-1
Matrix Inversion by Adjoint
The “Adjoint” of a matrix is the transpose of the matrix made up of the “CoFactors” of the original matrix.

35. Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu
Engr/Math/Physics 25
Appendix
Time For
Live Demo
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer