Introduction
In linear algebra, the adjugate (or classical adjoint) of an n × n square matrix A is the transpose of its cofactor matrix. The relationship between a matrix, its adjoint, and its determinant forms the foundation for matrix inversion and the computation of higher-order adjoint determinants.
Part 1: Proving that A · (adj A) = |A| I_n
Let \(A = [a_{ij}]\) be an \(n \times n\) square matrix, and let \(C_{ij}\) denote the cofactor of the entry \(a_{ij}\). By definition, the adjugate matrix \(\text{adj } A\) is the transpose of the cofactor matrix:
\(\text{adj } A = [C_{ji}]\)
Let \(B = A \cdot (\text{adj } A)\). The \((i, j)\)-th entry \(b_{ij}\) of the product matrix is given by the row-by-column product:
\(b_{ij} = \sum_{k=1}^{n} a_{ik} C_{jk}\)
From the standard properties of determinants:
- Case 1 (When \(i = j\)): \(\sum_{k=1}^{n} a_{ik} C_{ik} = |A|\), which represents the Laplace cofactor expansion of the determinant \(|A|\) along the \(i\)-th row.
- Case 2 (When \(i \neq j\)): \(\sum_{k=1}^{n} a_{ik} C_{jk} = 0\), which represents the sum of the products of elements of the \(i\)-th row with the cofactors of the corresponding elements of a different (\(j\)-th) row, equivalent to the determinant of a matrix with two identical rows.
Therefore, the entry \(b_{ij} = |A|\) when \(i = j\), and \(b_{ij} = 0\) when \(i \neq j\). Writing this in matrix form:
\(A \cdot (\text{adj } A) = \text{diag}(|A|, |A|, \dots, |A|) = |A| I_n\)
Part 2: Proving that |adj (adj A)| = |A|^{(n-1)^2}
Taking the determinant on both sides of the identity \(A \cdot (\text{adj } A) = |A| I_n\):
\(|A \cdot (\text{adj } A)| = \big| |A| I_n \big|\)
Using the property that the determinant of a product of matrices equals the product of their determinants, \(|AB| = |A||B|\), and that scaling an \(n \times n\) identity matrix by a scalar \(c\) yields \(|c I_n| = c^n\):
\(|A| \cdot |\text{adj } A| = (|A|)^n\)
Since \(A\) is non-singular, \(|A| \neq 0\). Dividing both sides by \(|A|\) gives:
\(|\text{adj } A| = |A|^{n-1}\)
Now, substitute \(A\) with \(\text{adj } A\) into this derived expression:
\(|\text{adj } (\text{adj } A)| = |\text{adj } A|^{n-1}\)
Substituting \(|\text{adj } A| = |A|^{n-1}\) into the right-hand side:
\(|\text{adj } (\text{adj } A)| = \left(|A|^{n-1}\right)^{n-1}\)
Applying the exponential rule \((x^a)^b = x^{ab}\):
\(|\text{adj } (\text{adj } A)| = |A|^{(n-1)(n-1)} = |A|^{(n-1)^2}\)
Conclusion
The relation A · (adj A) = |A| I_n not only facilitates matrix inversion via A^(-1) = (adj A) / |A|, but also serves as the recursive building block to compute higher-order iterates of adjugate determinants, yielding |adj(adj A)| = |A|^{(n-1)^2}.