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Transcript
Handout 5 MAT285
Jan.-April 2010
______________________________________________________________________
Definition of Rank of a Matrix
The number of non-zero rows of a matrix A in row- echelon form ( or reduced
row echelon form ) is called the rank of matrix A is denoted by rank A or r(A).
Example 1
Determine the rank of the following matrix:
4 10
A

1 2 
Solution
We need to perform ERO until we are able to produce row –echelon form or
reduced row-echelon form.
Example 2
Determine the rank of the following matrix:
1 2 
A

 2 4
Solution
We need to perform ERO until we are able to produce row –echelon form or
reduced row-echelon form.
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Handout 5 MAT285
Jan.-April 2010
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Example 3 ( Please refer Handout 4, Example 3 page 41 )
Determine the rank of the following augmented matrix:
1 1 2 9 
2 4  3 1


3 6  5 0
The rank of the augmented matrix =…………………………
Theorem ( A System is consistent )
The system of linear equations Ax =b is consistent if and only if the rank of
matrix A(coefficent matrix) is equal to the rank of the augmented matrix [ A : b ]
Based on Example 3 above ,
rank A = ______________
rank [ A : b ]= ________________
Conclusion: _______________________________________
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Handout 5 MAT285
Jan.-April 2010
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Theorem ( Linearly dependent and independent systems )
If the augmented matrix [A: b] in row echelon form or in reduced row echelon
form has a zero row, the system of linear equations is linearly dependent.
Otherwise it is linearly independent.
Example 4
Solve the following system of linear equations using the Gauss Elimination
method
x  2y  z  1
2x  y  z  2
4 x  3 y  3z  4
2x  y  3z  5
Solution
1 2
2  1

4 3

2  1
1
1 1

1 2 ERO 0

 
3 4
0


3 5
0
2 1
1
1
5
0 1
0
0
1

0
3

2
0 
By doing back substitution, z = 3/2, y = -3/10, x = 1/10.
This system of linear equations has a unique solution.
Exercise
Is this system of linear equations linear dependent or linearly independent ?
Answer: ____________________
rank A = ______________
rank [ A : b ]= ________________
Is this system consistent or inconsistent?
Answer: ___________________
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Handout 5 MAT285
Jan.-April 2010
______________________________________________________________________
Types of Solutions Of Non- Homogenous Equations Ax = b
3 possibilities will arise:
Case 1 ( no solution or inconsistent)
rank [ A ]  rank [ A : b]
The system of equations is inconsistent and has no solution.
Example 5
Suppose that the augmented matrix for a system of linear equations has been
reduced by row operations to the given reduced row – echelon form.
 1 0 0 0
0 1 2 0


0 0 0 1




Rank (A) =……………………….
Rank [A: b ] = ………………………………..
Conclusion: ___________________________________
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Handout 5 MAT285
Jan.-April 2010
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Case 2 (unique solution)
rank [ A ]  rank [ A : b]  n where n is the number of unknowns.
The system of equations is consistent and has a unique solution.
Example 6
Refer to Example 4 page 49
Solve the following system of linear equations using the Gauss Elimination
method
x  2y  z  1
2x  y  z  2
4 x  3 y  3z  4
2x  y  3z  5
Solution
1 2
2  1

4 3

2  1
1
1 1

1 2 ERO 0

 
3 4
0


3 5
0
2 1
1
1
5
0 1
0
0
1

0
3

2
0 
rank A = ______________
rank [ A : b ]= ________________
number of unknowns= n= ____________
Conclusion: __________________________________
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Handout 5 MAT285
Jan.-April 2010
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Case 3 (infinitely many solutions)
rank [ A ]  rank [ A : b]  n where n is the number of unknowns.
The system is consistent and has infinitely many solutions.
Example 7
Suppose that the augmented matrix for a system of linear equations has been
reduced by row operations to the given reduced row – echelon form.
1 0 0 4  1
0 1 0 2 6 


0 0 1 3 2 
Rank (A) =……………………….
Rank [A: b ] = ………………………………..
number of unknowns= n= ____________
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Handout 5 MAT285
Jan.-April 2010
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Example 8 (Oct 2008 )
Consider the non homogenous system
xyz  2
y  2z  4
x  y  k 2 z  2k
where k is a real number.
Determine the value(s) of k so that the system has
a) has no solution
b) has infinitely many solutions
c) has a unique solution.
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Handout 5 MAT285
Jan.-April 2010
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EIGENVALUES AND EIGENVECTORS
Motivation ( Linear system of the form Ax  x or
I  Ax  0 )
Example 9 ( page 107, Linear Algebra )
The linear system
x1  3x 2  x1
4x1  2x 2  x 2
can be written in matrix form as
 x1 
1 3  x1 
 4 2  x     x 

  2
 2
which is of form Ax  x with
A =……………………………………..
and matrix x = ……………………………………..
This system can be written as
 x  1 3  x 
0
1
 1   

 

 x2  4 2  x2  0
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Handout 5 MAT285
Jan.-April 2010
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or
1 0  x1  1 3  x1  0
  
  
0 1  x2  4 2  x2  0

or
  1  3   x1  0
  4   2  x   0

  2  
which is of form (I  A) x  0 with
  1  3 
I  A  

  4   2
Problem
How can we determine those values of  for which the system (I  A) x  0 has
a nontrivial solution?
Those values of  is called an ………………… A.
If  is an eigenvalue of A, then the nontrivial solutions are called the
eigenvectors of A corresponding to  .
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Handout 5 MAT285
Jan.-April 2010
______________________________________________________________________
The system I  Ax  0 has a nontrivial solutions if and only if detI  A   0
This is called the characteristic equation of A.
The eigenvalues of A can be found by solving this equation for A.
Example 10 ( adapted from Engineering Mathematics, K.A.Stroud)
4  1
Find the eigenvalues of the matrix A  

2 1 
Solution
I  A  .......... .......... ....
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Handout 5 MAT285
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Ans:   2 or   3
Example 11 ( page 108, Linear Algebra )
1 3
Find the eigenvalues and corresponding eigenvectors of the matrix A  

 4 2
Solution
________________________________________________________________________
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Handout 5 MAT285
Jan.-April 2010
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Example 12( Oct 2008 )
0
1
4

Given A   1  6  2 .Find
 5
0
0 
a) all the eigenvalues of A.
b) b) the eigenvector corresponding to the largest eigenvalue of A.
 t 
 3 
Ans:   6,5,1. The eigenvector corresponding to 5 is  t 
 11

 t 
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