Transcript Slide 1

Recap
Row and Reduced Row Echelon
 Elementary Matrices


If m and n are positive integers, then
an m  n matrix is a rectangular
array in which each entry aij of the
matrix is a number. The matrix has m
rows and n columns.
a1,1
a2,1
a3,1


am ,1
a1,2
a1,3
a2, 2
a3,2
a2,3
a3,3
am,2
am,3
a1,n 
a2,n 
a3, n 


am ,n 
A real matrix is a matrix all of whose
entries are real numbers.
 i (j) is called the row (column)
subscript.
 An mn matrix is said to be of size (or
dimension) mn.
 If m=n the matrix is square of order n.
 The ai,i’s are the diagonal entries.

Given a system of equations we can
talk about its coefficient matrix and its
augmented matrix.
 These are really just shorthand ways of
expressing the information in the
system.
 To solve the system we can now use
row operations instead of equation
operations to put the augmented
matrix in row echelon form.

1. Interchange two rows.
2. Multiply a row by a nonzero constant.
3. Add a multiple of a row to another row.
Two matrices are said to be row
equivalent if one can be obtained
from the other using elementary row
operations.
 A matrix is in row-echelon form if:

› All rows consisting entirely of zeros are at
the bottom.
› In each row that is not all zeros the first
entry is a 1.
› In two successive nonzero rows, the
leading 1 in the higher row is further left
than the leading 1 in the lower row.
1. Write the augmented matrix of the
system.
2. Use elementary row operations to find a
row equivalent matrix in row-echelon
form.
3. Write the system of equations
corresponding to the matrix in rowechelon form.
4. Use back-substitution to find the solutions
to this system.

In Gauss-Jordan elimination, we
continue the reduction of the
augmented matrix until we get a row
equivalent matrix in reduced rowechelon form. (r-e form where every
column with a leading 1 has rest zeros)
1
0

0
0 0
1 0
0 1
a
b
c 

A system of linear equations in which
all of the constant terms is zero is
called homogeneous.
 All homogeneous systems have the
solutions where all variables are set to
zero. This is called the trivial solution.

New Stuff
Using Elementary Matrices

An n by n matrix is called an elementary
matrix if it can be obtained from In by a
single elementary row operation.

These matrices allow us to do row
operations with matrix multiplication.
Theorem: Let E be the elementary matrix
obtained by performing an elementary
row operation on In. If that same row
operation is performed on an m by n
matrix A, then the resulting matrix is given
by the product EA.
Three types of Elementary
Matrices

These correspond to the three types of
EROs that we can do:
› Interchanging rows of I -> Type I EM
› Multiplying a row of I by a constant -> Type II
EM
› Adding a multiple of one row to another ->
Type III EM
Type I EM

 0 1 0


1
0
0


0 0 1


E1 =
How is this created?
Eg. 1 Suppose

A=

 a11

 a21
a
 31
a12
a22
a32
E1A =
 0 1 0   a11


1
0
0

  a21
 0 0 1   a31




a22
a32
=
 a21

 a11
a
 31
a13 

a23 
a33 
a12
a22
a12
a32
a23 

a13 
a33 
What is AE1?
a13 

a23 
a33 
Type II EM



E2 =
E2A =
AE2 =
1 0 0


 0 1 0
 0 0 3


 1 0 0   a11

 a
 0 1 0   21
 0 0 3   a31


a22
a32
 a11

 a21
a
 31
1 0 0


 0 1 0
 0 0 3


a12
a22
a32
a22 

a23 
a33 
a12
a22 

a23 
a33 
=
=
a12
 a11

 a21 a22
 3a
 31 3a32
 a11

 a21
a
 31
a12
a 22
a32
a22 

a23 
3a33 
3a13 

3a23 
3a33 
Type III EM



E3 =
E3A =
AE3 =
 1 0 3


0
1
0


0 0 1


 1 0 3   a11 a12 a22 

 

0
1
0

  a21 a22 a23 
0 0 1  a


  31 a32 a33 
 a11

 a21
a
 31
a12
a22
a32
a22 

a23 
a33 
 1 0 3


0
1
0


0 0 1


=
 a11  3a31

 a21
 a
31

a12  3a32
=  a11 a12

 a21 a22
a
 31 a32
a22
a32
a13  3a33 

a23 
a33 
3a11  a13 

3a21  a23 
3a31  a33 

Let A and B be m by n matrices. Matrix B
is row equivalent to A if there exists a
finite number of elementary matrices E1,
E2, ... Ek such that
B = EkEk-1 . . . E2E1A.
Break it down

This means that B is row equivalent to A if
B can be obtained from A through a
series of finite row operations.

If we then take two augmented matrices
(A|b) and (B|c) and they are row
equivalent, then Ax = b and Bx=c must
be equivalent series
Break it down
If A is row equivalent to B, B is row
equivalent to A
 If A is row equivalent to B and B is row
equivalent to C then A is row equivalent
to C

Example

Compute the inverse of A for A =
4 3 1 0 0
1



1

2
0
|
0
1
0


2
2 3 0 0 1 

3 1 0 0
1 4


0
2
3
|
1
1
0


0  6  3  2 0 1


3
1
1 
 1 4 3 1 0 0   1 4 0 2
2
2


 0 2 3 |1 1 0   0 2 0 | 12  12  12 

 0 0 6 1 3 1   0 0 6 1
3
1 

 

4 3
 1


  1  2 0
 2
2 3 

Example
1
1  1 0 0 1
1
1 
1 0 0 1

2
2
2 
2
2
2
1
1   0 1 0 | 1
1
1 
0 2 0 | 1
2
2
2 
4
4
4

1
1 
3
1  0 0 1 1
0 0 6 1
6
2
6

 
Now, solve the system:
x1 4 x2  3x3  12
 x1  2 x2  12
2 x1  2 x2  3x3  8
Example

We can employ the format Ax = b so x=A-1b

We just calculated A-1 and b is the column
vector  12 
  12 
 8 



So we can easily find the values of x by
multiplying the two matrices
Determinants
Keys to calculating Inverses
Require square matrices
 Each square matrix has a determinant
written as det(A) or |A|
 Determinants will be used to:

› characterize on-singular matrices
› express solutions to non-singular systems
› calculate dimension of subspaces


If A and B are square
then AB  A B
It is not difficult to
appreciate that
AT  A
 If A has a row (or
column) of zeros
then
A 0

If A has two identical
rows (or columns)
then
1 7 1
2
0
2 0
3
5
3
C1  C2

If B is obtained from A by ERO,
Ri  R j (Ci  C j ) interchanging two
rows (or columns) then
B A

If B is obtained from A by ERO where row
(or column) of A were multiplied by a
scalar k, then B  k A

If B is obtained from A by ERO where a
multiple of a row (or column) of A were
added to another row (or column) of A
then A  B
A
a11
a12
a21
a22
 a11 a22  a12 a21
That Is the determinant is equal to the
product of the elements along the
diagonal minus the product of the
elements along the off-diagonal.
  3 2

A  
 1 5
A  (3)(2)  (1)(6)  0
Note: The matrix A is said to be
invertible or non-singular if det(A)≠ 0. If
det(A) = 0, then A is singular.
For a 3 by 3 matrix

Using EROs on rows 2 and 3
 a11

 a21
a
 31
a12
a22
a32
a13 

a23 
a33 


 a11

 0

 0


a12
a11a22  a21a12
a11
a11a32  a31a12
a11


a13

a11a23  a21a13 

a11

a11a33  a31a13 

a11

For a 3 by 3 matrix

The matrix will be row equivalent to I iff:
a11a22  a21a12
a11
a11
a11a32  a31a12
a11
a11a23  a21a13
a11
0
a11a33  a31a13
a11
For a 3 by 3 matrix

This implies that the
Det(A) =
a11a22a33  a11a32a23  a12a21a33  a12a31a23  a13a21a32  a13a31a22

Use EROs to find:
3
2
1
5
5
4
0
2
6

STEP 1: Apply C1  C3
from property 5 this
gives us B   A
1 2 3
 2 4 5
6 0
5

STEP 2: Convert
matrix to Echelon
form
1
R2  R2  2R1
R3  R3  6R1
2
3
0
0
1
0  12 23

1
2
3
R2  R3  0  12 23
0
0
1
Therefore
the same
as:
1
2
 3 2 1is
5 4 2
5 0 6
3
 0  12 23  1(12)(1)  12
0 0
1
matrix is now in echelon
form so we can multiply
elements of main
diagonal to get
determinant

1 x
x2
Factorize the determinants of 1 y
2
1 z
1 2 4

What is
1 1 1
?
1 3 9
y
z2

We see that y – x is a factor of row 2 and
z – x is a factor of row 3 so we factor
them out from:
x
x2
yx
y2  x2
zx
z2  x2
1
R2  R2  R1 0
R3  R3  R1 0

And we get:
( y  x)(z  x)
1 x
x2
0 1
yx
0 1
zx
R3  R3  R2
( y  x)(z  x)
1 x
x2
0 1
yx
0 0
zy

The matrix is now in echelon form so we
can multiply elements of main diagonal
to get determinant and then multiply by
factors to get:
1 x
x2
1 y
y2
1 z
z2
=
( y  x)(z  x)(z  y)
1 2 4

Now, the matrix 1
1 1 corresponds to
1 3 9
x  2
y 1
z  3

Since
1 x
x2
1 y
y2
1 z
z2
= ( y  x)(z  x)(z  y)

Then
1 2 4
1 1 1
1 3 9
= (1  (2))(3  (2))(3 1)  3(1)(4)  12

Cofactor expansion is one method used
to find the determinant of matrices of
order higher than 2.

If A is a square matrix, then the minor Mi,j
of the element ai,j of A is the determinant
of the matrix obtained by deleting the ith
row and the jth column from A.

Consider the matrix
  3 2 1


A    5 4 2 .
 5 0 6


The minor of the entry “0” is found by
deleting the row and the column
associated with the entry “0”.

The minor of the entry “0” is
3 1
5 2
 3(2)  1(5)  1
Note: Since the 3 x 3 matrix A has 9
elements there would be 9 minors
associated with the matrix.

The cofactor Ci,j = (-1)i+jMi,j.
(1) i  j

 1, i  j even


 1, i  j odd
Since
we can think of the
cofactor of as nothing more than its signed
minor.

Find the minor and cofactor of the entry
  3 2 1
“2” for


A    5 4 2
 5 0 6



We first need to delete the row and
column corresponding to the entry “2”

The Minor of 2 is
5 2
 5(6)  5(2)  40
5 6
The minor corresponds to row 1 and
column 2 so applying the formula, we
have cij  (1)12  40  1(40)  40
 So the cofactor of the entry “2” is 40.


Theorem: Let A be a square matrix of
order n. Then for any i,j,
n

Columns:
det (A)  A   ai , jCi, j
j 1
and
n
 Rows: det (A)  A   ai , jCi, j .
i1
  3 2 1


A    5 4 2
 5 0 6



Given

Cofactor is found for the first entry in
column 1 “-3”
4 2
c11  

find det(A).
0 6
 24  0  24
Cofactor is found for the second entry in
column 1 “-5” c   2 1  12  0  12
21
0
6

Cofactor is found for the third entry in
column 1 “5”
c31

2

4
1
 44  0
2
The cofactors are then multiplied by the
corresponding entry and summed.
A  3(c11 )  5c21   5c31 
 3( 2 4)  5(1 2)  5(0)
 7 2  6 0
 1 2


Using row 2 - expansion we fix row 2 and
find the minors for each entry in row 2
then apply the sign corresponding to
each entries position to find the
cofactors. The cofactors are then
multiplied by the corresponding entry
and summed.
A  5(c22 )  4c22   2c23 
 5 * 
2 1
0 6
 4*
3 1
5
6
 2*
3 2
5
 5 * (12)  4 * (23)  2 * (10)
 60  92  20
 12
0

It is easy to show that
a11
a12
a13
a14
0
0
a 22
0
a 23
a33
a 24
 a11 a 22 a33 a 44
a34
0
0
0
a 44

If A is square and is in Echelon form then
is the product of the entries on the
(main) diagonal.
1
0
0
0
0 0
2 0
0 1
0 0
2
0
 1* 2 * 1* 0  0
0
0
Using CRAMER’S RULE we can apply this
method to finding the solution to a
system of linear systems that have the
same number of variables as equations
 There are two cases to consider

Consider the square system AX
= B where A is n x n.
 If A  0 then the system has
either I) No solution or ii) Many
solutions

If A1 is formed from A by
replacing column 1 of A with
column B and
 I. A1  0 , then the system has
NO solution
 II. |A| ≠ 0, then the system has
a unique solution

Consider the square system AX
= B where A is n x n.
 If |A|≠ 0 then the system has
a unique solution


The unique solution is obtained
by using the Cramer’s rule.

Where Ai is found from A by replacing
column i of A with B.
X1 
A1
X2 
A2
X3 
A3
A
A
A
etc

Use Cramer’s rule to write down the
solution to the system
5 x1  3x2  7
4 x1  5 x2  3
5 x1  3x2  7
4 x1  5 x2  3
A  2512  37
A  2512  37
A1 
A2 
7
3
3
5
5
7
4 3
 26
 43
X1 
A1
A

26
37
 43
X2 

A
37
 26  43
,


37 
 27
A2
T
adjA

(
C
)
 The adjoint of A (adj.A)) isij
: , the transpose of the matrix of
cofactors (a matrix of signed minors).
 If A  0 then the inverse of A exists and
A 1 

1
adjA
A
If A  0, A has no inverse.

Find inverse of

Det(A) =
3  2 3


A  1 2 1
1 3 0


a11a22a33  a11a32a23  a12a21a33  a12a31a23  a13a21a32  a13a31a22
= (3*2*0) – (3*3*1) – (-2*1*0)+(-2*1*1)
+(3*1*3) –(3*1*2)
=0 – 9 – 0 – 2 + 9 – 6
= -8
adjA (Cij )T
 2
 
 3
 2
 
 3
 2
 
 2
1
0
3
0
3
1

1 1
1
3

1
3

1
0
3
0
3
1
1 2 


1 3 
3 2 


1 3 
3 2


1 2 
T
0 
3 1
  3 9 8




  9  3  11   1  3 0 
8 0
 0  11 8 
8 



T
1
A  adjA
A
1
8
3 9

1
1
A   1 3 0 
8 

0
11

1



a.
b.
c.
Given that
 1 4 2


A   1 1  1
 2 4 5 


show:
A2  4 A  3I  0
 0
A
A  13A  12I
3
d.
e. Solve
A 1 
1
( AI  A)
3
T
T
x

(
x
,
x
,
x
)
b

(
1
,
1
,
1
)
1 2 3 and
A x  b where
 1 4 2  1 4 2   1 16 8 


 

2
A   1 1  1 1 1  1   4 1  4 
  2 4 5   2 4 5    8 16 17 


 

 1 16 8   1 4 2   1 0 0   0 0 0 

 
 
 

2
A  4 A  3I   4 1  4   4 1 1  1  3 0 1 0    0 0 0   0
  8 16 17    2 4 5   0 0 1   0 0 0 

 
 
 

A 2  4 A  3I  0
A 2  4 A  3I
A( A  4 I )  3I
A( A  4 I )   3I

Taking the determinant

Implying that and therefore; exists
A A  4 I   3I
Since, A  4 A  3I  0

Multiplying through by A
2

A3  4 A 2  3 A  0
A3  4 A 2  3 A
( A 2  4 A  3I )
A3  4(4 A  3I )  3 A
A3  16A  12I  3 A
A3  13A  12I

Multiplying through by A-1
A 2  4 A  3I  0
( AA) A 1  4 AA1  3 A 1  0
( AA1  0)
A  4 I  3 A 1  0
3 A 1  4 I  A
A 1 
1
( 4 I  A)
3
1
A  (4 I  A)
3
 4 0 0   1 4 2 
 3 4 2 
 
 1 

1 
1
A   0 4 0    1 1  1    1 3
1
3 
3





0
0
4

2
4
5
2

4

1
 




1

Since A x  b
1
xA b
 3  4 2 1   3    1
  1    
1
x    1 3 1 1   3    1 
3
3   



 2  4  11   3    1