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# Minors and Co-Factors 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 aij (aij means the element belonging ith row and jth column) then we obtain a second order determinant which is minor of aij and is denoted by Mij. In general, minor Mij of an element aij is the determinant excluding ith row and jth 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 aij is defined as Cij = (-1)i+j Mij.

Where, Cij = cofactor of aij.

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  = ∑mi=1  aij .Cij, j = 1, 2, ...... m.

= ∑mj=1  aij.Cij, 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 = = ∑mi=1  aij cij

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.

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