Minor, Cofactor, Adjoint, and Inverse

Minor Matrix

The minor of the element in the \( i^{\text{th}} \) row and \( j^{\text{th}} \) column of a square matrix \( A \) is denoted by \( M_{ij} \) and defined as:

\( M_{ij} = \left| A_{ij} \right| \)

Where \( A_{ij} \) is the submatrix formed by deleting the \( i^{\text{th}} \) row and \( j^{\text{th}} \) column from the original matrix \( A \), and \( \left| A_{ij} \right| \) is its determinant.

Cofactor Matrix

The cofactor of the element in the \( i^{\text{th}} \) row and \( j^{\text{th}} \) column of a square matrix \( A \) is given by:

\( \text{CO}_{ij} = (-1)^{i+j} \cdot M_{ij} \)

Where \( M_{ij} \) is the minor of the element, i.e., the determinant of the submatrix formed by deleting the \( i^{\text{th}} \) row and \( j^{\text{th}} \) column from \( A \).

Adjoint

The adjoint (or adjugate) of a square matrix \( A \) is the transpose of the matrix of cofactors:

\( \text{Adj}(A) = [\text{CO}_{ij}]^T \)

Inverse of a Matrix

If matrix \( A \) is invertible (i.e., \( \det(A) \ne 0 \)), then the inverse of \( A \) is given by:

\( A^{-1} = \frac{\text{Adj}(A)}{\det(A)} \)

Inverse of a Product

The inverse of a product of two invertible matrices is the product of their inverses in reverse order:

\( (AB)^{-1} = B^{-1} A^{-1} \)

Proof:

Start with the identity:

\( (AB)(AB)^{-1} = I \)

Pre-multiply both sides by \( A^{-1} \):

\( A^{-1}(AB)(AB)^{-1} = A^{-1} \Rightarrow B(AB)^{-1} = A^{-1} \)

Now pre-multiply both sides by \( B^{-1} \):

\( B^{-1}B(AB)^{-1} = B^{-1}A^{-1} \Rightarrow (AB)^{-1} = B^{-1}A^{-1} \)

Adjoint of a Product

The adjoint of the product of two matrices equals the product of their adjoints in reverse order:

\( \text{Adj}(AB) = \text{Adj}(B) \cdot \text{Adj}(A) \)

Why this works:

From the inverse formula:

\( (AB)^{-1} = \frac{\text{Adj}(AB)}{\det(AB)} \)

We also have:

\( (AB)^{-1} = B^{-1} A^{-1} \), and \( \det(AB) = \det(A) \cdot \det(B) \)

Substitute the inverse formulas:

\( A^{-1} = \frac{\text{Adj}(A)}{\det(A)} \), and \( B^{-1} = \frac{\text{Adj}(B)}{\det(B)} \)

Multiply the adjoints accordingly:

\( \text{Adj}(B) \cdot \text{Adj}(A) = \det(A) \cdot \det(B) \cdot B^{-1} \cdot A^{-1} \)

Thus, from the original inverse formula:

\( \text{Adj}(AB) = \text{Adj}(B) \cdot \text{Adj}(A) \)