The minor of an element of a determinant is again a determinant (of lesser order) formed by excluding the row and column of the element. For example take the following determinant

        Δ =

        If we leave the row and column passing through a_ij (a_ij means the element belonging i^th row and j^th column) then we obtain a second order determinant which is minor of a_ij and is denoted by M_ij. In general, minor M_ij of an element a_ij is the determinant excluding i^th row and j^th column. Thus we have 9 minors corresponding to 9 elements of above determinant Δ. Here we are illustrating some minors of the determinant Δ.

(i)     The minor of element a₁₁ = M₁₁ =

(ii)    The minor of element a₂₂ = M₂₂ =

(iii)    The minor of element a₃₁ = M₃₁ = and so on.

        Cofactor of an element a_ij is defined as C_ij = (-1)^i+j Mij.

        Where, C_ij = cofactor of a_ij.

Illustration:

        Find the minors and cofactors of along second column.

Solution:

        Minors along second column i.e. elements 2, 5 and 8 are

        And cofactors of the corresponding elements are

        C₁₂ = (-1)¹⁺² (-6) = 6

        C₂₂ = (-1)²⁺² (-12) = -12

        C₃₂ = (-1)³⁺² (-6) = 6 respectively.

Evaluation of a Determinant

        The determinant of order m can be evaluated as

        Δ  = ∑^m_i=1  a_ij .C_ij, j = 1, 2, ...... m.

            = ∑^m_j=1  a_ij.C_ij, i = 1, 2, ...... m.

        i.e. the determinant can be evaluated by multiplying the elements of a single row or a column with their respective co-factors and then adding them.

For e.g.

Cofactor of a = d(-1)¹⁺¹ = d

Cofactor of b = c(-1)²⁺¹ = -c

So, the value of the determinant is (ad - bc).

Illustration:

   Expand the following determinant.

   Δ = = ∑^m_i=1  a_ij c_ij

Solution:

.·. Δ = a.  a(ek - f²) -b (dk - if) +c (dj - ie)

       = a M₁₁ - b M₁₂ + c M₁₃ = a C₁₁ + b C₁₂ + c C₁₃

Note :  Though in the example, elements of first row and their cofactors are considered, the value of the determinant can be evaluated from any row and column.

Sarrus Rule:     Sarrus give a rule for a determinant of order 3. Write down the three rows of a determinant and rewrite the first two rows. The three diagonals sloping down the right give the three positive terms and the three diagonals sloping down to the left give the three negative terms.

.·. Δ = P - N.