Minors and Co-factors

Let A = [a_ij] be a square matrix of order n. Then, the minor M_ij of a_ij in A is the determinant obtained by deleting the i^th row and the j^th column in which element a_ij lies. It is denoted by M_ij​ of A.

Let A = [a_ij] be a square matrix of order n. Then, the cofactor C_ij (or A_ij) of a_ij in A is (−1)^i+j times M_ij, where M_ij is the minor of a_ij in A.

∴ C_ij = (−1)^i+j M_ij

To Find the Minor:

1. Select the element \[a_{ij}\] whose minor you want.

2. Delete the i-th row and j-th column of the matrix.

3. Calculate the determinant of the remaining \[(n - 1) \times (n - 1)\] matrix.

4. The resulting value is \[M_{ij}\], the minor of \[a_{ij}\].

To Find the Cofactor

1. Find the minor \[M_{ij}\] of \[a_{ij}\] as above.

2. Compute the sign factor \[(-1)^{i+j}\].

3. Multiply: \[C_{ij} = (-1)^{i+j} M_{ij}\].

Find minors and cofactors of all the elements of the determinant \[\begin{vmatrix} 1 & -2 \\ 4 & 3 \end{vmatrix}\]

Solution: Minor of the element \[a_{ij}\] is \[M_{ij}\]

Here \[a_{11} = 1\]. So \[M_{11} = \text{Minor of } a_{11} = 3\]

\[M_{12} = \text{Minor of the element } a_{12} = 4\]

\[M_{21} = \text{Minor of the element } a_{21} = -2\]

\[M_{22} = \text{Minor of the element } a_{22} = 1\]

Now, cofactor of \[a_{ij}\] is \[A_{ij}\]. So

\[A_{11} = (-1)^{1+1} M_{11} = (-1)^{2} (3) = 3\]

\[A_{12} = (-1)^{1+2} M_{12} = (-1)^{3} (4) = - 4\]

\[A_{21} = (-1)^{2+1} M_{21} = (-1)^{3} (-2) = 2\]

\[A_{22} = (-1)^{2+2} M_{22} = (-1)^{4} (1) = 1\]

• Minor \[M_{ij}\]: determinant of the matrix obtained by deleting row i and column j.

• Cofactor \[C_{ij}\]: \[C_{ij} = (-1)^{i+j}M_{ij}\].

• Determinant expansion along row i: \[|A| = \sum_{j=1}^{n} a_{ij}C_{ij}\].

• Determinant expansion along column j: \[|A| = \sum_{i=1}^{n} a_{ij}C_{ij}\].

• Determinant value is the same for any choice of row or column for expansion.

• Mixed row/column property: \[\sum_{j=1}^{n} a_{ij}C_{kj} = 0\] for \[i \neq k\].