matrices lecturer 3

1.6 Rank of Matrix: we defined the rank of any matrix a that the order of
the largest square sub-matrix of a whose determinant not zero (det of submatrix
‡ 0)
Example: Let A =
? ? ?
?
?
? ? ?
?
?
7 8 9
4 5 6
1 2 3
find the rank of A
1 × 9×5 + 2 × 6 × 7 + 3 × 4 × 8 – 3 × 5 × 7 – 1 × 6 × 8 – 2 × 4 × 9 = 0
Since A of order 3 Rank ‡ 3
Since 3 0
4 5
1 2
= ? ? the rank ‡ 2
1.7 Minor of matrix: Let A =
? ? ? ? ?
?
?
? ? ? ? ?
?
?
an an ann
a a a n
a a a n
1 2
21 22 2
11 12 1
(4)
Is the square matrix of order n then the determinant of any square submatrix
of a with order (n-1) obtained by deleting row and column is called
the minor of A and denoted by Mij.
1.8 Cofactor of matrix: Let A be square matrix in (4) with mij which is the
minors of its. Then the Cofactor of a defined by Cij = (-1) i+j Mij
Example: Let A =
? ? ?
?
?
? ? ?
?
?
?
?
6 1 0
4 5 7
2 4 1
find the minor and the cofactor of element 7.
Solution: The minor of element 7 is
M23 = det 22
6 1
2 4
6 1
2 4
=
?
?
= ?
??
?
? ??
?
?
?
i.e (denoted by take the square sub-matrix by deleting the second rows and
third column in A).
the Cofactor of 7 is
13
22
6 1
2 4
23 ( 1) ( 1) 2 3
23
2 3 = ?
?
?
C = ? + M = ? +
1.9 Adjoint of matrix: Let matrix A in (4) then the transposed of matrix of
cofactor of this matrix is called adjoint of A, adjoint A = transposed matrix
of Cofactor.
The inverse of matrix: Let A be square matrix. Then inverse of matrix
{Where A is non-singular matrix} denoted by A-1 and A-1 = ( )
det
1 adj A
A
1.0 method to find the inverse of A: To find the inverse of matrix we must
find the following:
(i) the matrix of minor of elements of A.
(ii) the Cofactor of minor of elements of A
(iii) the adjoint of A .
then A-1 = adjA
A
1
Example: let A =
? ? ?
?
?
? ? ?
?
?
? ?
?
3 1 1
1 2 3
2 3 4
Find A-1
(1) Minors of A is Mij =
? ? ?
?
?
? ? ?
?
?
? ?
? ?
17 10 1
7 10 11
1 10 7
(2) Cofactor of A is (-1) Mij =
? ? ?
?
?
? ? ?
?
?
?
?
17 10 1
7 10 11
1 10 7
(3) Adj of A =
? ? ?
?
?
? ? ?
?
?
?
?
7 11 1
10 10 10
1 7 17
.
(4) det = 60
14
A-1 = .
7 11 1
10 10 10
1 7 17
60
1
? ? ?
?
?
? ? ?
?
?
?
?
1.11 Properties of Matrix Multiplication:
1 – (KA) B = K (AB) = A (KB) K is any number
2 – A (BC) = (AB) C
3 – (A + B) C = AC + BC
4 – C (A + B) = CA + CB
5 – AB ‡ BA (in general)
For example: Let A = ?
??
?
? ??
?
0 0
1 0 and B = ?
??
?
? ??
?
1 1
0 1
A B = ?
??
?
? ??
?
= ?
??
?
? ??
?
? ??
?
? ??
?
0 0
0 1
1 0
0 1
0 0
1 0
B A = ?
??
?
? ??
?
= ?
??
?
? ??
?
? ??
?
? ??
?
1 0
0 0
0 0
1 0
1 0
0 1
A B = ‡ B A
6 – A B = 0 but not necessarily A = 0 or B = 0
For Example: A = ?
??
?
? ??
?
+ ?
? +
= ?
??
?
? ??
?
1 1
1 1
,
2 2
1 1
B
A B =
0, 0
0 0
0 0
1 1
1 1
2 2
1 1
? ?
? ??
?
? ??
?
= ?
?? ?
? ??
?
?
?
? ??
?
? ??
?
A B
But
A B = 0
7 -
? ? ?
?
?
? ? ?
?
?
=
? ? ?
?
?
? ? ?
?
?
0 0 1
0 1 0
1 0 0
0 0
0 0
0 0
C
C
C
C
8 – A I = I A = A where I is identity matrix
15
9 – (A B) T = BT AT
10 – A-1 A = A.A-1 = I
1.12 Cramer’s Rule
Let the system of linear question as
( )
2
1
21 1 22 2
11 1 12 2 i
a a b
a a b
?
? ? ?
+ =
+ =
? ?
? ?
The system (i) can put in the form:
( )
2
1
2
1
11 22
11 12
ii
b
b
a a
a a
? ?
??
?
? ??
?
=
? ? ?
?
?
? ? ?
?
?
? ? ?
?
?
? ? ?
?
?
?
?
If D = 0
21 22
11 12
?
a a
a a
Then the system (ii) has a unique solution, and Cramer’s rule state that it
may be found from the formulas:
D
a b
a b
X
D
b a
b a
21 2
11 1
,
2 22
1 12
1 2 ? = =
Example: solve the system
3 9 1 2 X ? ? =
X1 + 2X2 = - 4
So, the system can put in th