# Minor and Co-factors of determinants

Minor: Let A = [aij] be a square matrix of order n. Then the minor Mij of aij in A is the determinant of the square sub-matrix of order (n – 1) obtained by leaving ith row and jth column of A.

Let $$A=\left[ \begin{matrix}{{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\{{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\{{a}_{31}}&{{a}_{32}} & {{a}_{33}} \\\end{matrix} \right]$$.

M₁₁ = Minor of a₁₁ = $$\left|\begin{matrix}{{a}_{22}} & {{a}_{23}} \\{{a}_{32}}&{{a}_{33}} \\\end{matrix}\right|$$.

M₁₂ = Minor of a₁₂ = $$\left| \begin{matrix}{{a}_{21}} & {{a}_{23}} \\{{a}_{31}} & {{a}_{33}} \\\end{matrix} \right|$$.

M₁₃ = Minor of a₁₃ = $$\left|\begin{matrix}{{a}_{21}} & {{a}_{22}} \\{{a}_{31}} &{{a}_{32}} \\\end{matrix} \right|$$.

M₂₁ = Minor of a₂₁ = $$\left| \begin{matrix}{{a}_{12}} & {{a}_{13}}\\{{a}_{32}} & {{a}_{33}}\\\end{matrix} \right|$$.

Similarly for the elements a22, a23, a31, a32 and a33.

Cofactor: Let A = [aij] be a square matrix of order n. Then the cofactor of Cij of aij in A is equal to (- 1)i+j times the determinant of the sub-matrix of order (n – 1) obtained by leaving ith row and jth column of A.

Cij = cofactor of aij in A.

= (- 1) i+j Mij, where Mij is minor of aij in A.

{{C}_{ij}}=\left\{\begin{align}&{{M}_{ij}},\,\,\,\,\,\,if\,\,\,i+j\,\,\,is\,\,\,even \\&-{{M}_{ij}},\,\,\,\,\,if\,\,\,\,i+j\,\,\,is\,\,\,odd \\\end{align} \right..

Ex: If $$A=\left[\begin{matrix}1&2&3\\-3&2&-1\\2&-4&3\\\end{matrix}\right]$$, then

C₁₁ = (-1)¹⁺¹ M₁₁ = M₁₁ = $$\left| \begin{matrix}2 & -1 \\-4 & 3 \\\end{matrix}\right|$$ = 2

C₁₂ = (-1)¹⁺² M₁₂ = – M₁₂= $$-\left|\begin{matrix}-3&-1\\2 &3 \\\end{matrix}\right|$$ = – (-9 + 2) = 7

C₁₃ = (-1)¹⁺³ M₁₃ = M₁₃= $$\left| \begin{matrix} -3 & 2\\2 & -4\\\end{matrix} \right|$$ = 8

C₂₃ = (-1)²⁺³ M₂₃ = – M₂₃= $$-\left| \begin{matrix}1&2\\2&-4\\\end{matrix} \right|$$ = 8 etc.