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3. Linear Programming
3.1 Linear Algebra
Square n-by-n Matrix
Solution of Simultaneous Linear
Equations
a11 x1  a12 x2 
 a1n xn  b1
a21 x1  a22 x2 
 a2 n xn  b2
an1 x1  an 2 x2 
 ann xn  bn
Two ways of solving them directly:
(1) determinants (Cramer's rule), which gives
the solution as a ratio of two n  n determinants;
(2) elimination.
Cramer’s Theorem
(Solution of Linear Systems by Determinants)
If a linear system of n equations in the same number of unknowns
x1, … , xn
a11 x1  a12 x2   a1n xn  b1
a21 x1  a22 x2 
 a2 n xn  b2
an1 x1  an 2 x2 
 ann xn  bn
has a nonzero coefficient determinant D = det A, the system has
precisely one solution. This solution is given by the formulas
Dn
D1
D2
x1 
, x2 
,
, xn 
(Cramer's rule)
D
D
D
where Dk is the determinant obtained from D by replacing in D the
kth column by the column with the entries b1, … , bn.
Gaussian Elimination
Forward Elimination
pivot
2u+ v+ w= 1
4u+ v
= -2
-2 u + 2 v + w = 7
Backward Substitution
(1)
(2)
(3)
Step 1: equation (2) +(– 2) x equation (1)
Step 2: equation (3) + (+1) x equation (1)
2u+ v+ w = 1
(4)
(-1) v - 2 w = - 4
(5)
pivot
3v+2w = 8
(6)
Step 3: equation (6) + (+3) x equation (5)
2u+ v+ w = 1
(7)
- v - 2 w = -4
(8)
- 4 w = -4
(9)
w=1
v=2
u = -1
Elementary Transformation of
Matrices – (i)
An elementary matrix of the first kind it an
n  n diagonal matrix formed by replacing the
ith diagonal element of identity matrix I with
a nonzero constant q. For example, if n  4, i  3
1
0
Q
0

0
0
1
0
0
det Q  q
0
0
q
0
0
0 
0

1
1
0
1
Q 
0

0
0 0
1 0
0 1/ q
0 0
0
0 
0

1
Elementary Transformation of
Matrices – (ii)
An elementary matrix of the second kind is an
n  n matrix R formed by interchanging anytwo rows
i and j of the identity matrix I. For example,
if n  4, i  1 and j  3
0
0
R
1

0
0 1 0
1 0 0 
0 0 0

0 0 1
det R  1
0
0
R 1  
1

0
0 1 0
1 0 0 
0 0 0

0 0 1
Elementary Transformation of
Matrices – (iii)
An elementary matrix of the third kind it an
n  n matrix S formed by inserting the a nonzero
constant s into the off-diagonal position (i, j ) of
the identity matrix I. For example, if n  4, i  3 and
j 1
1 0
0 1
S
s 0

0 0
det S  1
0 0
0 0 
1 0

0 1
1
0
S 1  
s

0
0 0 0
1 0 0 
0 1 0

0 0 1
Elementary Row Operation
Any row manipulation can be
accomplished by pre-multiplication of
elementary matrices!
QA n p : multiplication of all elements of the ith row
in A by a constant q;
RA n p : interchange of the ith and jth row in A;
SA n p : addition of a scalar multiple s of the jth row
to the ith row.
1
0
S1S 2  
 s1

0
1
0
S 21S11  
0

0
0 0 0  1 0
1 0 0  0 1
0 1 0  0 0

0 0 1  0 s2
0 0 0  1
1 0 0   0

0 1 0    s1

 s2 0 1   0
0 0  1 0
0 0   0 1

1 0   s1 0
 
0 1   0 s2
0 0 0  1
1 0 0   0

0 1 0    s1
 
0 0 1  0
0 0
0 0 
1 0

0 1
0 0 0
1 0 0 
0 1 0

 s2 0 1 
Elementary Column Operation
Any column manipulation can be
accomplished by post-multiplication of
elementary matrices!
A pn Q : multiplication of all elements of the ith column
in A by a constant q;
A pn R : interchange of the ith and jth column in A;
A pnS : addition of a scalar multiple s of the jth column
to the ith column.
Gaussian Elimination = Triangular Factorization
 2 1 1  u   1 






Ax  4 1 0 v  2  b

   
 -2 2 1   w   7 
 3 elimination steps
2 1 1   u   1 
Ux   0 1 2   v    4   c

   
 0 0 4   w  4 
upper triangular!!!
Step 1: 2nd equation + (  2)  1st equation  S 21A
 1 0 0
where,
S 21   2 1 0


 0 0 1 
Step 2: 3rd equation + (+1)  1st equation  S31S 21A
 1 0 0
where,
S31   0 1 0


 1 0 1
Step 3: 3rd equation + (+3)  2nd equation  S32S31S 21A
where,
1 0 0 
S32  0 1 0


0 3 1 
U  S32S31S 21A 

 S32S31S 21Ax  S32S 31S 21b

Upper triangular
Lower triangular
c  S 32S 31S 21b 
 1 0 0
Let Sˆ  S 32S 31S 21   2 1 0 
 1 3 1 
 A  Sˆ 1U  S 1S 1S 1U  LU
21
31
32
 1 0 0
1 1
where L  S 211S31
S32   2 1 0 
 1 3 1 
Note that 2, -1 and -3 are the negative values
of the multipliers used in the 3 elimination steps.
Conclusions
If no pivots are zero, the matrix A can be written as
a product LU,i.e.,
Gaussian Elimination  Triangular Factorization
L is a lower triangular matrix with 1's on the main diagonal,
i.e., see next page.
U is an upper triangular matrix.
The nonzero entries of U are the coefficients of the equations
which appear after elimination and before back-substitution.
The diagonal entries of U are the pivots.
0
1

L  e21 1

 e31 e32
0

0

1 
Thus, eij is the quantity that multiply row j when
it is subtracted from row i to produce zero in the
(i, j ) entry.
Implications
Solve: Ax n  b n
n  1, 2,3,
(1) Obtain A  LU
(2) Solve Lc n  b n with forward substitution for c n
(3) Solve Ux n  c n with backward substitution for x n
Row Exchange
Ax  b
1
0
A
0

0
2
0
0
c
3
5
d
7
4
1
0
6 
 R 24 A  
0
6


8
0
2
c
0
0
3
7
d
5
4
8 
6

6
If c  0, the problem is incurable and the matrix is
called singular.
Elimination with Row Exchange
Assume A is nonsingular, then there exists a
permutation matrix R that reorders the rows of
A so that
RA  LU
Round Off Error
Consider
1.0 
1.0
A
;

1.0 1.0001
0.0001 1.0 
A  

1.0
1.0


First Point : Some matrices are extremely sensitive
to small changes, and others are not. The matrix A is
ill-conditioned (i.e. sensitive); A is well-conditioned.
A is "nearly" singular
Singular
matrix
1 
1 1
1
1 1  A  1 1.0001




x1  2
2
(1) Ax  b    
x2  0
2
 2 
x1  1
(2) Ax  b  


x2  1
 2.0001
No numerical methods can provide this
sensitivity to small perturbations!!!
Second Point: Even a well-conditioned
matrix can be ruined by a poor algorithm.
0.0001 1.0  x1  1.0 
Ax  
 



1.0  x2   2.0
 1.0
Correct solution:
10000
x1 
 1.00010001 (round off after 9th digit)
9999
9998
x2 
 0.99989998 (round off after 9th digit)
9999
If a calculator is capable of keeping only 3
digits, then Gaussian elimination gives the
wrong answer!!!
(0.0001) x1  x2  1
(A)
x1  x2  2
(B)
Eq. (B) - 10000  Eq.(A):
(1.0  0.000110000.0) x1  (1.0  1.0 10000.0) x2
=2.0  1.0 10000.0
1.0  1.0 10000.0  9999.0
0
 1.00  1.00E4  1.00E4
2.0  1.0 10000.0  9998.0
 2.00  1.00E4  1.00E4
x2  1.00 (not too bad)
Substituting into Eq.(A)
x1  0.00 (This is wrong)
Third Point
A computer program should compare each pivot
with all the other possible pivots in the same
column.
Choosing the largest of these candidates, and
exchanging the corresponding rows so as to
make this largest value pivot, is called partial
pivoting.
Solution of m Equations with n
Unknowns (m<n) – Example 1
A mn x  b
pivot  1
3 2
A   2 6 9 5 
 1 3 3 0 
 elementary row operation S31S 21
3
1 3 3 2 
0 0 3 1 

 pivot
0 0 6 2 
 elementary row operation S32
1 3 3 2 
U  0 0 3 1   Echelon Form!
0 0 0 0 
Echelon Form
Solution of m Equations with n
Unknowns (m<n) – Example 2
A m n x  b
pivot  1
3
A2 6

 1 2

3 2
9 5

3 0 
elementary row operation S31S 21
1 3 3 2 
0 0 3 1 


0 1 6 2 
 elementary row operation R 23
1 3 3 2 
U  0 1 6 2 


0 0 3 1 
CONCLUSION
To any m-by-n matrix A there corresponds
1.a square (i.e., m-by-m) permutation matrix R,
2.a square (i.e., m-by-m) lower triangular matrix
L with unit diagonal, and
3.an m-by-n echelon matrix U, such that
RA = LU
Homogeneous Solution for
Example 1
b0
Ax  0
 Ux  0
pivot
u
1 3 3 2     0
v






Ux  0 0 3 1
 0

  w  
0 0 0 0    0
 y
u, w : basic variables (correspond to the pivots)
v, y : free variables
 Express basic variables in terms of free variables!
1
3w+y=0  w   y
3
u  3v  3w  2 y  0  u  3v  y
 3v  y 
 3
 v 
1
  v 
x
 y / 3 
0


 
 y 
0
 1 
 0 

y
 1 / 3


 1 
 All solutions are linear combinations of
-3
 -1 
1
 0 
  and 

0
-1/3
 


0
 1 
 Within the 4-D space of all possible x, the solution
of Ax = 0 form a 2-D subspace.
 the nullspace of A
Subspace
A subspace of a vector space is a subset that satisfies
two requirements:
1. If we add any two vectors x and y in the subspace,
the sum x+y is still in the subspace.
2. If we multiply any vector x in the subspace by any
scalar c, the multiple cx is still in the subspace.
Note that the zero vector belongs to every subspace.
Conclusions
• Every homogeneous system Ax=0, if it has
more unknowns than equations (n>m), has
infinitely many nontrivial solutions.
• The dimension of nullspace is the number of
free variables (which is larger than than or
equal to n-m>0).
• The nullspace is a subspace of Rⁿ.
Inhomogeneous Solution of Example 1
b0
Ax  LUx  b
Ux  L1b  c
u 
b1
1 3 3 2    

v
Ux  0 0 3 1      b2  2b1   c
 w
0 0 0 0    b3  2b2  5b1 
 y
Note that the equations are inconsistent unless
 5   b1 
b3  2b2  5b1   2   b2   0
 1   b3 
In other words, the set of attainable vectors b
is not the whole of the 3-D space - it lies on a
plane that is perpendicular to the vector [5 -2 1]T .
Ax  a1 a 2
a3 a 4  x  b
u 
 1 3 3 2     b1 
 2 6 9 5   v   b 

  w  2 
 1 3 3 0     b3 
 y
ua1  va 2  wa3  ya 4  b
1
3
 3
 2   b1 










u  2   v  6   w 9   y  5   b2 
 1
 3
3
 0   b3 
Conclusion
The system Ax=b is solvable if and
only if the vector b can be expressed
as a linear combination of the
columns of A, i.e.
n
Ax   x j a j  b
j 1
Note also that
a 2  3a1
1
a 4  a1  a3
3
b  ua1  va 2  wa3  ya 4
1 

 ua1  v  3a1   wa3  y  a1  a3 
3 

1 

  u  3v  y  a1   w  y  a3
3 

u
w
1
 3
 u   2   w 9 
 1
3
Conclusions
 Ax  b can be solved iff b lies in the plane
that is spanned by a1 and a3.
 The plane is a subspace of R m called column
space of the matix A.
 The equation Ax = b can be solved iff b lies
in the column space.
u 
 1 3 3 2    1 
v  



Ax   2 6 9 5 
 5   b
 w
 1 3 3 0    5
 y
u 
1 3 3 2    1 
v  



Ux  0 0 3 1 
 3  c
 w
0 0 0 0    0 
 y
1
3w  y  3  w  1  y
3
u  3v  3w  2 y  1  u  2  y  3v
 u   2 
 3 
 1 
v  0 
1
 0 

x      v  y
 w  1 
0
 1/ 3
   
 


y
0
0
1
   
 


particular soln
homogeneous solution
Ax=0
CONCLUSIONS
• Suppose the m-by-n matrix A is reduced by
elementary operations and row exchanges to a
matrix U in echelon form.
• Let there be r nonzero pivots; the last m-r rows
of U are zero. Then there will be r basic
variables and n-r free variables, corresponding
to the columns of U with and without pivots
respectively.
• Note that r  m  n
CONCLUSIONS
• The nullspace, formed of solutions to Ax=0,
has the n-r free variables as the independent
parameters, i.e., its dimension is also n-r. If
r=n ( i.e.,  m ), there are no free variables and
the null space contains only x=0.
• Solution always exists for every right side b iff
r=m<n. In this case, since U has no zero rows,
Ux=c can be solved by back-substitution.
CONCLUSIONS
• If r<m<n, then U will have m-r zero rows and
there are m-r constraints on b in order for
Ax=b to be solvable. If the particular solution
exists, then every other solution differs from it
by a vector in the nullspace of A.
• The number r is called the rank of the matrix A.