AcademyDeterminants
Academy
Adjugate Matrix
Level 1 - Linear Algebra topic page in Determinants.
Principle
The adjugate matrix is built from cofactors. It gives a formula for the inverse of an invertible square matrix and explains why inverse entries depend on determinants.
For large matrices, row reduction is usually faster for computation, but the adjugate formula is important theoretically.
Notation
\(C_{ij}\)
cofactor of entry a_{ij}
\(C\)
the cofactor matrix
\(\operatorname{adj}(A)\)
the adjugate of A
\(A^{-1}\)
the inverse of A
\(\det A\)
determinant of A
The Core Method
First form the cofactor matrix \(C=(C_{ij})\). Then transpose it:
Adjugate
\[\operatorname{adj}(A)=C^T\]
The key identity is
Adjugate identity
\[A\operatorname{adj}(A)=\operatorname{adj}(A)A=(\det A)I\]
If \(\det A\ne0\), divide by \(\det A\) to get the inverse:
Inverse by adjugate
\[A^{-1}=\frac{1}{\det A}\operatorname{adj}(A)\]
For a \(2\) by \(2\) matrix, this reduces to the familiar formula.
Worked Cases
Question
Find \(\operatorname{adj}(A)\) for \(A=\begin{pmatrix}a&b\\c&d\end{pmatrix}\).
Answer
Compute cofactors. \(C_{11}=d\), \(C_{12}=-c\), \(C_{21}=-b\), and \(C_{22}=a\). So the cofactor matrix is \(C=\begin{pmatrix}d&-c\\-b&a\end{pmatrix}\). The adjugate is the transpose: \(\operatorname{adj}(A)=C^T=\begin{pmatrix}d&-b\\-c&a\end{pmatrix}\).
Examples
Question
Why does the adjugate formula require \(\det A\ne0\)?
Answer
The formula is \(A^{-1}=\frac{1}{\det A}\operatorname{adj}(A)\). If \(\det A=0\), this divides by zero, which is not defined. This agrees with the invertibility test: zero determinant means no inverse exists.