Download General Principles

Lecture #10: System of Linear
Equations & Matrices
 Lecture #9:
1.
2.
3.
4.
Linear Equations: y=mx +b
Solution System: N.S., U.S., I.S.
Augmented Matrix
Solving a System of Linear Equations
 Today:
1.
2.
3.
4.
Echelon Form, Reduced Echelon Form
Gauss-Jordan Elimination Method
Homogeneous Linear Equations
Matrix Operations
Announcements:
• Review Class on Tuesday 28:
– Room: Ricketts 203
– Time: 6:30-8:00pm
• Exam #1 Next Wednesday: 9/29
Unit Vectors, Cartesian Vector Form.
System of linear equations
• The general form:
A11x1+A12x2+A13x3+…..A1nxn=B1
A21x1+A22x2+A23x3+…..A2nxn=B2
A31x1+A32x2+A33x3+…..A3nxn=B3
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Am1x1+Am2x2+Am3x3+…Amnxn=Bm
Rx=0
Ry=0
Rz=0
Matrix Form:
• Coefficient Matrix
ROW #
a11
a21
a31
am1
a12
a22
a32
am 2
a13  .......
a23  .......
a33  .......
am3  .......
Column #
a1n
a2 n
a3n
amn
Augmented Matrix:
• System of linear eqns.
• Augmented Matrix:
(array of numbers of the system of
eqns)
1x + y + 2z = 9
2x + 4y – 3z = 1
3x + 6y –5z = 0
Remember:
Rx=0 Ry=0
Rz=0
1 1
2
9
2 4 3 1
3 6 5 0
Solving a System of Linear Eqns.
• GOAL
– FIND the solution for x, y,z (TA, TB, TC, TD, TE)
• The idea is to replace a given system by a
system which has the same solution set, but it
is easier to solve.
Basic Operations to find
Unknown
• Multiply a row by a nonzero constant.
(the row you multiply by a number after adding the two rows will not
change)
• Interchange two rows.
• Add a multiple of one row to another row.
Gauss-Jordan Elimination
• Goal:
to reduce the augmented matrix into a form
simple enough such that system of equation are
solved by inspection.
Reduced row-echelom form
1. If row does not consist entirely of zeros,
then the first non-zero number in row is 1.
2. If a row consist of zeros, then they are
moved to the bottom of matrix.
3. In any two successive rows that do not
consist entirely of zeros, the leading 1 in
the lower row occur farther to the right of
above row.
4. Each column that contains a leading 1 has
zero everywhere.
IMPORTANT
• Reduced Row echelom
– Must have zeros above
and below each leading 1.
• Row-echelom form
– Must have zeros below
each leading 1.
1 0 0 9
0 1 0 1
1 4 0 9
0 1 0 1
0 0 1 0
0 0 1 0
1 0 0 9
0 1 0 1
1  10  12 9
0 0 1 7
0 0 0 0
0
0
0
1
0
0
0
1
0
1
7
1
Gauss-Jordan Elimination Method
• Step1: Locate the leftmost column that does not
consist entirely of zero.
• Step 2: Interchange the top row with another row, if
necessary, to bring a nonzero entry to the top
from step 1.
• Step 3: If the entry that is now at the top is a
constant, divide entire row by it.
• Step 4: Add multiples to top row to the rows below
such that all entries have 1 as leading term.
• Step 5: Cover top row and begin with step 1 applied
to submatrix.
Example #1
• For problem 3.22 find FAB, FAC, FAD
using Gauss-Jordan method.
F
x
0
 0 
 2 
 3
FAD 
  FAC    FAB 
0
7
45
44
 




 Fy  0
 6 
 6 
6
 FAD 
  FAC    FAB 
0
7
 44 
 45 
 Fz  0
 3 
 2 
 2
FAD 
  FAC    FAB 
  1250
7
 44 
 45 
Activity:#1
• For Problem in example #1 solve using
Gauss-Jordan Method.
-TA (0.766) + TB (0.866) = 1699
TA (0.643) + TB (0.500) = 2943
Homogeneous System of Linear
Equations
• Non-homogeneous
• Homogeneous
A11x1+A12x2+A13x3+…..A1nxn=B1
A21x1+A22x2+A23x3+…..A2nxn=B2
A31x1+A32x2+A33x3+…..A3nxn=B3
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Am1x1+Am2x2+Am3x3+…Amnxn=Bm
A11x1+A12x2+A13x3+…..A1nxn= 0
A21x1+A22x2+A23x3+…..A2nxn= 0
A31x1+A32x2+A33x3+…..A3nxn= 0
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Am1x1+Am2x2+Am3x3+…Amnxn= 0
The Constants B not equal to 0
The Constants B’s Equal to 0
Solutions in Homogeneous System
• Trivial Solution
X1 = 0
X2 = 0
X3 = 0
….
….
Xn = 0
• For same # equations and
same # unknowns
• Non trivial solution:
X1 = C1
X2 = C2
X3 = C3
….
….
Xn = C4
• When there is more unknowns
than equations.