Transcript Matrix multiplication, cofactor and inverse

 a11 a12

 a 21 a 22
a
 31 a 32
a13 

a 23 
a 33 
 b11 b12

 b 21 b 22
b
 31 b32
b13 

b 23 
b33 
=
Matrix
Multiplication
a11b11  a12b 21  a13b31
a11b12  a12 b 22  a13b33
a11b13  a12 b 23  a13b33
a 21b11  a 22b 21  a 23b31
a 21b12  a 22 b 22  a 23b33
a 21b13  a 22b 23  a 23b33
a 31b11  a 32b 21  a 33b31
a 31b12  a 32b 22  a 33b33
a 31b13  a 32b 23  a 33b33
Cofactor
Aij
a11 a12
a 21 a 22
a13
a 23
a 31 a 32
a 33
of the (i,j)th entry
(-1)i+j (determinant obtained by deleting the
ith row and jth column of the matrix)
1a1 a 22 a a 23
 (1) 11 13
1132   a a 32 a a 33
21
33
AA
1 2
a 21 a 23
A12  (1) a a
a11 31 a13 33
A22   a a
1331a 21 33a 22
A13  (1) a 31 a 32
   


   

The sign of (-1)i+j is given by 
   
next
a11
a12
a 21 a 22
a 31 a 32
a13
a 23 
a 33
Determinant
+a11a22a33 +a21a32a13 +a31a12a23
– a31a22a13 – a21a12a33 – a11a32a23
direct
expansion
expanded
along
a 21 a13
a 22 a 23
a 21 a 22 st
= a11 A11+ a12 A12+ a13A13  a11
1 row
a13
– a12
+
a 31 a 33
a 32 a 33
a 31 a 32
a
a
a
a
a
a
= a21 A21+ a22 A22+ a23A23  a 21 12 13  a 22 11 13  a 23 11 12 2nd row
a 32 a 33
a 31 a 33
a 31 a 32
a12 a13
a11 a13
a11 a12 rd
= a31 A31+ a32 A32+ a33A33  a 31
3 row
i+j

a

a
32
33by
The
sign
of
(-1)
is
given
a 22 a 23
a 21 a 23
a 21 a 22

= a11 A11+ a21 A21+ a31A31
= a12 A12+ a22 A22+ a32A32
= a13 A13+ a23 A23+ a33A33
  

a 22 a 23
a 21 a 23
a 21 a 22
 a11
 a12
a13



a 32 a 33  a 31 a 33  a 31 a 32
a 21 a 23  
a11 
a13
a11 a13
 a
a
a
12
22
32
first
column
second
column
a 31 a 33
a 31 a 33
a 21 a 23
a12 a13
a11 a12
a11 a12
third
 a 31
 a 23
 a 33
a 22 a 23
a 31 a 32
a 21 a 22
column
A
1
Inverse
1

(A ij ) t
det A
Proof
 a11
1
1 
t
A
(A ij ) 
 a 21
det A
det A 
 a 31
 a11
1 

 a 21
det A 
 a 31
a12
a 22
a 32
a12
a 22
a 32
a13  A11

a 23  A12
a 33  A13
a13  A11 A12

a 23  A 21 A 22
a 33  A 31 A 32
A 21 A 31 

A 22 A 32 
A 23 A 33 
A13 

A 23 
A 33 
t
 a11A11  adetA

A22+a13A23
12 A12  a13A13 a11A21+a120


0
1

A12+a23A13 a 21A 21  adetA
a21A11+a220

22 A 22  a 23A 23
0
det A 
a 31A31  a 32A32  a 33A33 
0

0
detA
=I.
Explanation
Similarly, it can be proved that
Explanation
1
(A ij ) t A  I. A-1 hence found.
det A
END
a11
a12
a 21 a 22
a 31 a 32
a13
a 23 
a 33
Proof
+a11a22a33 +a21a32a13 +a31a12a23
– a31a22a13 – a21a12a33 – a11a32a23
=a11(a22a33-a32a23) – a12(a21a33-a31a13)+ a13(a21a32-a31a22)
a 21 a13a 21 aa2321 a 22
a 22 a 23
a 21
 a11
– a12 – a12 + a13
a13
+
a 31 a 33a 31 aa3331 a 32
a 32 a 33
a 31
= a11 A11 + a12 A12 + a13A13
a 22
a 32
(expanded along first row)
   


   

The sign of (-1)i+j is given by 
   
back
a11 a12
a 21 a 22
a13
a 23
a 31 a 32
a 33
a11What
A11+about
a12 A12+ a13A13 = det A
a11 A21+ a12 A22+ a13A23?
a21 A21+ a22 A22+ a23A23 = det A
Proof2
0
a31 A31+ a32 A32+ a33A33 = det A
Note that
ax21
A + ay12A22 + a13
z A23=
23
11 21 a22
along 2nd row
Expand along first row
Expand along second row
Expand along third row
a11 a12
y 12
aax21
11 aa22
a 31 a 32
a21 A11+ a22 A12+ a23A13?
along 1st row
a13
az13
23
back to
inverse
a 33
a 21 a 22
a 23
 a 21 a 22
a 31 a 33
a 23
a 33
=0