Let
11
12
13
21
22
23
31
32
33
A =
a
a
a
a
a
a
a
a
a










Rationalised 2023-24
 88
MATHEMATICS
Then
11
12
13
21
22
23
31
32
33
A
A
A
A =Transposeof A
A
A
A
A
A
adj










11
21
31
12
22
32
13
23
33
A
A
A
= A
A
A
A
A
A










Example 12 
2
3
Find 
 A for A = 1
4
adj






Solution We have  A11 = 4, A12 = –1, A21 = –3, A22 = 2
Hence
adj A =
11
21
12
22
A
A
4
–3
 =
A
A
–1
2












Remark For a square matrix of order 2, given by
A =
11
12
21
22
a
a
a
a






The adj A can also be obtained by interchanging a11 and a22 and by changing signs
of a12 and a21, i e ,
We state the following theorem without proof Theorem 1 If A be any given square matrix of order n, then
A(adj A) = (adj A) A = A I ,
where I is the identity matrix of order n
Verification
Let
A = 
11
12
13
21
22
23
31
32
33
a
a
a
a
a
a
a
a
a










,  then adj  A = 
11
21
31
12
22
32
13
23
33
A
A
A
A
A
A
A
A
A










Since sum of product of elements of a row (or a column) with corresponding
cofactors is equal to |A| and otherwise zero, we have
Rationalised 2023-24
DETERMINANTS     89
A (adj A) = 
A
0
0
0
A
0
0
0
A










 = A  
1
0
0
0
1
0
0
0
1










 = A  I
Similarly, we can show  (adj A) A = A  I
Hence A (adj A) = (adj A) A = A  I
Definition 4 A square matrix A is said to be singular if A  = 0