the inverse of the matrix

2018
Lecture №3
THE INVERSE OF
THE MATRIX
Dobroshtan Оlena

2
3 2
-4 5
 
  
 
A
3 2
det( ) (3)(5) (2)( 4)
-4 5
15 8 23
   
  
A

3
2 -3 5
-1 4 6
 
  
 
2,3A = A
2 1
7 -4
3 1
 
   
  
3,2B = B
2 1
2 -3 5
7 -4
-1 4 6
3 1
 
       
    
C

4
11 (2)(2) ( 3)(7) (5)(3) 4 21 15 2c         
12 (2)(1) ( 3)( 4) (5)(1) 2 12 5 19c         
21 ( 1)(2) (4)(7) (6)(3) 2 28 18 44c         
22 ( 1)(1) (4)( 4) (6)(1) 1 16 6 11c           
2 19
44 -11
 
  
 
C

5
2 -3 5
-1 4 6
 
  
 
2,3A = A
2 1
7 -4
3 1
 
   
  
3,2B = B
2 1 3 -2 16
2 -3 5
7 -4 18 -37 11
-1 4 6
3 1 5 -5 21
   
         
       
D = BA

Cofactor Method for Inverses
• Let A = (aij) be an nxn matrix
• Recall, the co-factor Cij of element aij is:
• Mij is the (n-1) x (n-1) matrix made by
removing the ROW i and COLUMN j of A

• Put all co-factors in a matrix – called the
matrix of co-factors:
C11 C12 C1n
C21
Cn1
C22
Cn2
C2n
Cnn

• Inverse of A is given by:
A-1 = (matrix of co-factors)T1
|A|
1
|A|
C11 C21 Cn1
C12
C1n
C22
C2n
Cn2
Cnn
=

Examples
• Calculate the inverse of A =
a
c
b
d
|M11| = d C11 = dM11 = d

Examples
a
c
b
d
|M12| = c C12 = -cM12 = c

Examples
a
c
b
d
|M21| = b C12 = -bM21 = b

Examples
a
c
b
d
|M22| = a C22 = aM22 = a

Examples
a
c
b
d
• Found that:
C22 = aC21 = -bC12 = -cC11 = d
• So,
|A|

Examples
a
c
b
d
• Found that:
C22 = aC21 = -bC12 = -cC11 = d
• So,
(ad-bc)

Examples
a
c
b
d
• Found that:
C22 = aC21 = -bC12 = -cC11 = d
• So,
A-1 = 1
(ad-bc)
C11
C21
C12
C22
T

Examples
a
c
b
d
• Found that:
C22 = aC21 = -bC12 = -cC11 = d
• So,
A-1 = 1
(ad-bc)
C11
C12
C21
C22

Examples
a
c
b
d
• Found that:
C22 = aC21 = -bC12 = -cC11 = d
• So,
A-1 = 1
(ad-bc)
d
C12
C21
C22

Examples
a
c
b
d
• Found that:
C22 = aC21 = -bC12 = -cC11 = d
• So,
A-1 = 1
(ad-bc)
d
C12
-b
C22

Examples
a
c
b
d
• Found that:
C22 = aC21 = -bC12 = -cC11 = d
• So,
A-1 = 1
(ad-bc)
d
-c
-b
C22

Examples
a
c
b
d
• Found that:
C22 = aC21 = -bC12 = -cC11 = d
• So,
A-1 = 1
(ad-bc)
d
-c
-b
a

Examples 3x3 Matrix
• Calculate the inverse of B =
1
1
1
2
• Find the co-factors:
2
2
1
43
C11 = 2|M11| = 2M11 = 2 2
43

Examples 3x3 Matrix
1
1
1
2
2
2
1
43
C12 = 0|M12| = 0M12 = 1 2
42

Examples 3x3 Matrix
1
1
1
2
2
2
1
43
C13 = -1|M13| = -1M13 = 1 2
32

Examples 3x3 Matrix
1
1
1
2
2
2
1
43
C21 = -1|M21| = 1M21 = 1 1
43

Examples 3x3 Matrix
1
1
1
2
2
2
1
43
C22 = 2|M22| = 2M22 = 1 1
42

Examples 3x3 Matrix
1
1
1
2
2
2
1
43
C23 = -1|M23| = 1M23 = 1 1
32

Examples 3x3 Matrix
1
1
1
2
2
2
1
43
C31 = 0|M31| = 0M31 = 1 1
22

Examples 3x3 Matrix
1
1
1
2
2
2
1
43
C32 = -1|M32| = 1M32 = 1 1
21

Examples 3x3 Matrix
1
1
1
2
• First find the co-factors:
2
2
1
43
C33 = 1|M33| = 1M33 = 1 1
21

Examples 3x3 Matrix
1
1
1
2
• Next the determinant: use the top row:
2
2
1
43
|B| = 1x |M11| -1x |M12| + 1x |M13|
= 2 – 0 + (-1) = 1

• Using the formula,
B-1 = (matrix of co-factors)T1
|B|
= (matrix of co-factors)T1
1
Examples 3x3 Matrix

Examples 3x3 Matrix
2
-1
0
2 -1
0
1
1-1
|B|
=
1
1
T

2
0
-1
2 -1
-1
0
1-1
Examples 3x3 Matrix
|B|
=
• Same answer obtained by Gauss-Jordan method

34
The inverse of A below is developed in the text.
1 2 -1
-1 1 3
3 2 1
 
   
  
A
1
-0.25 -0.2 0.35
0.5 0.2 -0.1
-0.25 0.2 0.15

 
   
  
A

35
The inverse of A below is developed in the text.
1 2 -1
-1 1 3
3 2 1
 
   
  
A
1
-0.25 -0.2 0.35
0.5 0.2 -0.1
-0.25 0.2 0.15

 
   
  
A

36
Simultaneous Linear Equations
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
....
....
: : :
....
m m
m m
m m mm m m
a x a x a x b
a x a x a x b
a x a x a x b
  
  
  

37
Matrix Form of Simultaneous
Linear Equations
11 12 1 1 1
21 22 2 2 2
1 2
....
....
: :: :
....
m
m
m mm m mm
a a a x b
a a a x b
x ba a a
     
     
     
     
     
         

38
Define variables as follows:
11 12 1
21 22 2
1 2
....
....
: :
....
m
m
m m mm
a a a
a a a
a a a
 
 
 
 
 
  
A
1
2
:
m
x
x
x
 
 
 
 
 
  
x
1
2
:
m
b
b
b
 
 
 
 
 
  
b

39
Matrix Solution Development
Ax b
-1 -1
A Ax A b
-1
A A = I
Ix = x

40
The general form and the final solution follow.
-1
x = A b
Ax = b

41
Use matrices to solve the simultaneous
equations below.
1 2 32 8x x x   
1 2 33 7x x x   
1 2 33 2 4x x x  

42
1
2
3
1 2 -1 8
-1 1 3 7
3 2 1 4
x
x
x
    
        
        
Ax = b

43
1
2
3
-0.25 -0.2 0.35 -8
0.5 0.2 -0.1 7
-0.25 0.2 0.15 4
x
x
x
     
           
         
x
-1
x = A b
2
-3
4
 
   
  
x

To show that matrices are inverses of one another,
show that the multiplication of the matrices is
commutative and results in the identity matrix.
Show that A and B are inverses.















23
35
53
32
BandA
































10
01
)2(5)3(3)3(5)5(3
)2(3)3(2)3(3)5(2
23
35
53
32
AB
































10
01
)5(2)3(3)3(2)2(3
)5)(3()3(5)3)(3()2(5
53
32
23
35
BA














dc
ba
BandALet
53
21
Use the equation AB = E
Write and solve the equation:


















10
01
53
21
dc
ba

the inverse of the matrix

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