Adjugate Matrix

Definition (Adjugate Matrix)

In linear algebra, the adjugate matrix (historically also called the classical adjoint) of a square matrix $A$ is the transpose of its cofactor matrix. It is an extremely important construction because it allows us to express the inverse of a matrix and the derivatives of determinants analytically without needing to perform division.

Before defining the adjugate, we must define the building blocks: minors and cofactors.

Definition (Minors and Cofactors)

For any square matrix $A\in {\mathbb{F}}^{n\times n}$:

1. Submatrices and Minors

If we delete row $i$ and column $j$ from $A$, we obtain an $(n-1)\times (n-1)$ submatrix, denoted as $A_{\backslash i,\backslash j}$. The minor $M_{ij}$ associated with the entry $A_{ij}$ is defined as the determinant of this submatrix:

$$M_{ij}=\det (A_{\backslash i,\backslash j})$$

Since $M_{ij}$ is the determinant of an $(n-1)\times (n-1)$ matrix, it measures the “volume scaling” of the subspace map ignoring the $i$-th input direction and $j$-th output direction.

2. Cofactors

The cofactor $C_{ij}$ associated with the entry $A_{ij}$ is the minor signed by its position in the checkerboard pattern:

$$C_{ij}=(-1{)}^{i+j}{M}_{ij}=(-1{)}^{i+j}\det (A_{\backslash i,\backslash j})$$

The sign factor $(-1{)}^{i+j}$ alternates in a checkerboard grid:

$$\left[\begin{array}{cccc}+& -& +& \cdots \\ -& +& -& \cdots \\ +& -& +& \cdots \\ ⋮& ⋮& ⋮& \ddots \end{array}\right]$$

The cofactor $C_{ij}$ represents the rate of change of the determinant with respect to the entry $A_{ij}$, i.e., $\frac{\partial \det (A)}{\partial A_{ij}}=C_{ij}$ (see Jacobi’s Formula).

Definition (Adjugate Matrix Construction)

The adjugate matrix $\mathrm{adj}(A)$ is the transpose of the cofactor matrix $C$:

$$\mathrm{adj}(A)=C^{\top }$$

This means that the $(i,j)$-entry of $\mathrm{adj}(A)$ is the cofactor $C_{ji}$:

$$\mathrm{adj}(A{)}_{ij}=C_{ji}=(-1{)}^{i+j}{M}_{ji}=(-1{)}^{i+j}\det (A_{\backslash j,\backslash i})$$

Theorem (Adjugate Entries are Determinants of Submatrix)

By definition, the cofactor $C_{ji}$ is $(-1{)}^{j+i}\det (A_{\backslash j,\backslash i})$. The term $\det (A_{\backslash j,\backslash i})$ is the determinant of the $(n-1)\times (n-1)$ submatrix formed by deleting the $j$-th row and $i$-th column of $A$.

Because the entry at row $i$, column $j$ of the adjugate is exactly $C_{ji}$, every single element in $\mathrm{adj}(A)$ is a signed determinant of one of these $(n-1)\times (n-1)$ submatrices.

Example (2x2 Adjugate Matrix)

Let $A$ be a general $2\times 2$ matrix:

$$A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$$

We compute the minors and cofactors for each entry:

• For $A_{11}=a$: Delete row 1, col 1 $⟹A_{\backslash 1,\backslash 1}=[d]⟹M_{11}=d$. Cofactor $C_{11}=(-1{)}^{1+1}d=d$.
• For $A_{12}=b$: Delete row 1, col 2 $⟹A_{\backslash 1,\backslash 2}=[c]⟹M_{12}=c$. Cofactor $C_{12}=(-1{)}^{1+2}c=-c$.
• For $A_{21}=c$: Delete row 2, col 1 $⟹A_{\backslash 2,\backslash 1}=[b]⟹M_{21}=b$. Cofactor $C_{21}=(-1{)}^{2+1}b=-b$.
• For $A_{22}=d$: Delete row 2, col 2 $⟹A_{\backslash 2,\backslash 2}=[a]⟹M_{22}=a$. Cofactor $C_{22}=(-1{)}^{2+2}a=a$.

The cofactor matrix is:

$$C=\left[\begin{array}{cc}{C}_{11}& C_{12}\\ C_{21}& C_{22}\end{array}\right]=\left[\begin{array}{cc}d& -c\\ -b& a\end{array}\right]$$

Taking the transpose of $C$ yields the adjugate matrix:

$$\mathrm{adj}(A)=C^{\top }=\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$$

Observe how:

1. Every entry is the determinant of a $1\times 1$ submatrix (a scalar).
2. For invertible $A$, $A^{-1}=\frac{1}{ad-bc}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$, which perfectly matches the $2\times 2$ inverse formula!

Example (3x3 Adjugate Matrix)

Let $A$ be the $3\times 3$ matrix:

$$A=\left[\begin{array}{ccc}1& 0& 2\\ -1& 3& 0\\ 0& 2& 1\end{array}\right]$$

We compute the cofactors $C_{ij}=(-1{)}^{i+j}\det (A_{\backslash i,\backslash j})$:

• Row 1:
  • $C_{11}=+\det \left[\begin{array}{cc}3& 0\\ 2& 1\end{array}\right]=+(3\cdot 1-0)=3$
  • $C_{12}=-\det \left[\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right]=-((-1)\cdot 1-0)=1$
  • $C_{13}=+\det \left[\begin{array}{cc}-1& 3\\ 0& 2\end{array}\right]=+((-1)\cdot 2-0)=-2$
• Row 2:
  • $C_{21}=-\det \left[\begin{array}{cc}0& 2\\ 2& 1\end{array}\right]=-(0\cdot 1-4)=4$
  • $C_{22}=+\det \left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]=+(1\cdot 1-0)=1$
  • $C_{23}=-\det \left[\begin{array}{cc}1& 0\\ 0& 2\end{array}\right]=-(1\cdot 2-0)=-2$
• Row 3:
  • $C_{31}=+\det \left[\begin{array}{cc}0& 2\\ 3& 0\end{array}\right]=+(0\cdot 0-6)=-6$
  • $C_{32}=-\det \left[\begin{array}{cc}1& 2\\ -1& 0\end{array}\right]=-(1\cdot 0-(-2))=-2$
  • $C_{33}=+\det \left[\begin{array}{cc}1& 0\\ -1& 3\end{array}\right]=+(1\cdot 3-0)=3$

The cofactor matrix is:

$$C=\left[\begin{array}{ccc}3& 1& -2\\ 4& 1& -2\\ -6& -2& 3\end{array}\right]$$

Taking the transpose of $C$ yields the adjugate matrix:

$$\mathrm{adj}(A)=C^{\top }=\left[\begin{array}{ccc}3& 4& -6\\ 1& 1& -2\\ -2& -2& 3\end{array}\right]$$

We can verify the identity $A\mathrm{adj}(A)=\det (A)I_3$. Cofactor expansion along the first row of $A$ gives:

$$\det (A)=A_{11}{C}_{11}+A_{12}{C}_{12}+A_{13}{C}_{13}=1(3)+0(1)+2(-2)=-1$$

Computing the product:

$$A\mathrm{adj}(A)=\left[\begin{array}{ccc}1& 0& 2\\ -1& 3& 0\\ 0& 2& 1\end{array}\right]\left[\begin{array}{ccc}3& 4& -6\\ 1& 1& -2\\ -2& -2& 3\end{array}\right]=\left[\begin{array}{ccc}-1& 0& 0\\ 0& -1& 0\\ 0& 0& -1\end{array}\right]=-1\cdot {\mathbf{I}}_{3\times 3}$$

This perfectly holds!

Theorem (The Adjugate Identity)

For any $n\times n$ matrix $A$, we have:

$$A\mathrm{adj}(A)=\mathrm{adj}(A)A=\det (A)I_n$$

where $I_n$ denotes the $n\times n$ identity matrix (the matrix with $1$s on the main diagonal and $0$s elsewhere).

Proof

Let $B=A\mathrm{adj}(A)$. The $(i,j)$-entry of $B$ is given by matrix multiplication:

$$B_{ij}=\sum _{k=1}^n{A}_{ik}\mathrm{adj}(A{)}_{kj}=\sum _{k=1}^n{A}_{ik}{C}_{jk}$$

We analyze two cases:

1. If $i=j$: $$B_{ii}=\sum _{k=1}^n{A}_{ik}{C}_{ik}$$ This is exactly the cofactor expansion of $\det (A)$ along row $i$. Thus, $B_{ii}=\det (A)$.
2. If $i\ne j$: $$B_{ij}=\sum _{k=1}^n{A}_{ik}{C}_{jk}$$ This expression represents the cofactor expansion along row $j$ of a matrix $A^{\prime }$ obtained by replacing row $j$ of $A$ with row $i$. Since $A^{\prime }$ has two identical rows (row $i$ and row $j$), its determinant is zero. Thus, $B_{ij}=0$ for $i\ne j$.

Therefore, $B=\det (A)I_n$. A completely symmetric argument for $\mathrm{adj}(A)A$ yields the same result.

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Corollary (Matrix Inverse Formula)

If $\det (A)\ne 0$, then $A$ is invertible and:

$$A^{-1}=\frac{1}{\det (A)}\mathrm{adj}(A)$$

Proof

Divide the adjugate identity (Theorem 1) by the scalar $\det (A)$:

$$A\left(\frac{1}{\det (A)}\mathrm{adj}(A)\right)={\mathbf{I}}_{3\times 3}$$

By uniqueness of the matrix inverse, the formula holds.

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Theorem (Determinant of the Adjugate)

For any $n\times n$ matrix $A$:

$$\det (\mathrm{adj}(A))=\det (A{)}^{n-1}$$

Proof

Taking the determinant of both sides of the adjugate identity $A\mathrm{adj}(A)=\det (A)I_n$:

$$\det (A)\det (\mathrm{adj}(A))=\det (\det (A)I_n)=\det (A{)}^n\det (I_n)=\det (A{)}^n$$

• Case 1 ($\det (A)\ne 0$): We can divide both sides by $\det (A)$ to get: $$\det (\mathrm{adj}(A))=\det (A{)}^{n-1}$$
• Case 2 ($\det (A)=0$): If $\det (A)=0$, then $A\mathrm{adj}(A)=0$. If $\mathrm{adj}(A)$ were invertible, we could multiply by its inverse to get $A=0$, which would imply $\mathrm{adj}(A)=0$ (for $n>1$), a contradiction. Thus $\mathrm{adj}(A)$ must be singular, so $\det (\mathrm{adj}(A))=0$. Thus, $0=0^{n-1}$ holds (assuming $n>1$).

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Theorem (Adjugate of a Product)

For any two $n\times n$ matrices $A$ and $B$:

$$\mathrm{adj}(AB)=\mathrm{adj}(B)\mathrm{adj}(A)$$

Proof

• If $A$ and $B$ are invertible: $$\mathrm{adj}(AB)=\det (AB)(AB{)}^{-1}=\det (A)\det (B)B^{-1}{A}^{-1}=\left(\det (B)B^{-1}\right)\left(\det (A)A^{-1}\right)=\mathrm{adj}(B)\mathrm{adj}(A)$$
• General Case: Since invertible matrices are dense in the set of all matrices, the identity extends to all matrices by continuity.

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Application (Invariant Derivatives in Continuum Mechanics)

In Stress-Strain Relation, we differentiate the characteristic polynomial $P_{\mathbf{C}}(z)=\det (z\mathbf{I}-\mathbf{C})$ with respect to the right Cauchy-Green deformation tensor $\mathbf{C}$.

Applying Jacobi’s Formula, we start with the matrix derivative:

$$\frac{\partial \det (A)}{\partial A}=\det (A)A^{-\top }$$

Setting $A=z\mathbf{I}-\mathbf{C}$, the chain rule gives:

$$\frac{\partial P_{\mathbf{C}}(z)}{\partial \mathbf{C}}=\frac{\partial \det (z\mathbf{I}-\mathbf{C})}{\partial \mathbf{C}}=-\det (z\mathbf{I}-\mathbf{C})(z\mathbf{I}-\mathbf{C}{)}^{-\top }$$

Using the inverse formula ${\mathbf{A}}^{-1}=\frac{\mathrm{adj}(\mathbf{A})}{\det (\mathbf{A})}$, we can rewrite the transpose-inverse term:

$$(z\mathbf{I}-\mathbf{C}{)}^{-\top }={\left(\frac{\mathrm{adj}(z\mathbf{I}-\mathbf{C})}{\det (z\mathbf{I}-\mathbf{C})}\right)}^{\top }$$

Substituting this back into the derivative:

$$\begin{array}{rl}\frac{\partial P_{\mathbf{C}}(z)}{\partial \mathbf{C}}& =-P_{\mathbf{C}}(z){\left(\frac{\mathrm{adj}(z\mathbf{I}-\mathbf{C})}{\det (z\mathbf{I}-\mathbf{C})}\right)}^{\top }\\ & =-P_{\mathbf{C}}(z){\left(\frac{\mathrm{adj}(z\mathbf{I}-\mathbf{C})}{P_{\mathbf{C}}(z)}\right)}^{\top }\\ & =-\mathrm{adj}(z\mathbf{I}-\mathbf{C}{)}^{\top }\end{array}$$

The determinant terms cancel out completely. Since $\mathrm{adj}(z\mathbf{I}-\mathbf{C})$ is a matrix polynomial in $z$, we can equate powers of $z$ on both sides to compute the derivatives of individual invariants $\frac{\partial I_k}{\partial \mathbf{C}}$ without needing to perform division by singular determinants.