So
A11 = (–1)1 + 1  M11 = (–1)2 (3) = 3
A12 = (–1)1 + 2  M12 = (–1)3 (4) = – 4
A21 = (–1)2 + 1  M21 = (–1)3 (–2) = 2
A22 = (–1)2 + 2  M22 = (–1)4 (1) = 1
Rationalised 2023-24
DETERMINANTS     85
Example 10 Find minors and cofactors of the elements a11, a21 in the determinant
∆ =
11
12
13
21
22
23
31
32
33
a
a
a
a
a
a
a
a
a
Solution By definition of minors and cofactors, we have
Minor of a11 = M11 = 
22
23
32
33
a
a
a
a
 = a22 a33– a23 a32
Cofactor of a11 = A11 = (–1)1+1  M11 = a22 a33 – a23 a32
Minor of a21 = M21 = 
12
13
32
33
a
a
a
a
 = a12 a33 – a13 a32
Cofactor of a21 = A21 = (–1)2+1  M21 = (–1) (a12 a33 – a13 a32) = – a12 a33 + a13 a32
Remark Expanding the determinant ∆, in Example 21, along R1, we have
∆ = (–1)1+1 a11 
22
23
32
33
a
a
a
a
+ (–1)1+2 a12 
21
23
31
33
a
a
a
a
 + (–1)1+3  a13 
21
22
31
32
a
a
a
a
   = a11 A11 + a12 A12 + a13 A13, where Aij is cofactor of aij
    = sum of product of elements of R1 with their corresponding cofactors
Similarly, ∆ can be calculated by other five ways of expansion that is along R2, R3,
C1, C2 and C3 Hence ∆ = sum of the product of elements of any row (or column) with their
corresponding cofactors ANote  If elements of a row (or column) are multiplied with cofactors of any
other row (or column), then their sum is zero For example,
∆ = a11 A21 + a12 A22 + a13 A23
= a11 (–1)1+1 
12
13
32
33
a
a
a
a
+ a12 (–1)1+2 
11
13
31
33
a
a
a
a
+ a13 (–1)1+3 
11
12
31
32
a
a
a
a
= 
11
12
13
11
12
13
31
32
33
a
a
a
a
a
a
a
a
a
 = 0 (since  R1 and R2 are identical)
Similarly, we can try for other rows and columns