Matrix Product with Adjugate Matrix

Theorem
Let $R$ be a commutative ring with unity.

Let $\mathbf A \in R^{n \times n}$ be a square matrix of order $n$.

Let $\operatorname{adj}(\mathbf A)$ be its adjugate matrix.

Then:
 * $\mathbf A \cdot \operatorname{adj}(\mathbf A) = \det \left({\mathbf A}\right) \cdot \mathbf I_n$
 * $\operatorname{adj}(\mathbf A) \cdot \mathbf A = \det \left({\mathbf A}\right) \cdot \mathbf I_n$

where $\det \left({\mathbf A}\right)$ is the determinant of $\mathbf A$.

Proof
Let $\mathbf A = (a_{ij})$.

Let $A_{ij}$ denote the cofactor of $a_{ij} \in \mathbf A$.

Right Multiplication
We show that $\mathbf A \cdot \operatorname{adj}(\mathbf A) = \det \left({\mathbf A}\right) \cdot \mathbf I_n$.

Let $i, j \in \left\{ {1, \ldots, n}\right\}$.

If $i = j$, expanding $\det \left({\mathbf A}\right)$ along row $i$ shows that:


 * $\displaystyle \det \left({\mathbf A}\right) = \sum_{k \mathop = 1}^n a_{ik} \mathbf A_{ik}$

If $i \ne j$, define $\mathbf A'$ as the matrix obtained by replacing row $j$ of $\mathbf A$ with row $i$ of $\mathbf A$.

Then $\mathbf A' = \begin{bmatrix} a' \end{bmatrix}_n$ has two identical rows, so:

By definition of matrix product, element $\left({i, j}\right)$ of $\mathbf A \cdot \operatorname{adj}(\mathbf A)$ is:


 * $\displaystyle \sum_{k \mathop = 1}^n a_{ik} A_{jk} = \begin{cases}

0_R & \text{for} & i \ne j \\ \det \left({\mathbf A}\right) & \text{for} & i = j \end{cases}$

Hence, $\mathbf A \cdot \operatorname{adj}(\mathbf A) = \det \left({\mathbf A}\right) \cdot \mathbf I_n$.

Left Multiplication
We show that $\operatorname{adj}(\mathbf A) \cdot \mathbf A = \det \left({\mathbf A}\right) \cdot \mathbf I_n$.

Let $i, j \in \left\{ {1, \ldots, n}\right\}$.

If $i = j$, expanding $\det \left({\mathbf A}\right)$ along column $j$ shows that:


 * $\displaystyle \det \left({\mathbf A}\right) = \sum_{k \mathop = 1}^n a_{kj} \mathbf A_{kj}$

If $i \ne j$, define $\mathbf A'$ as the matrix obtained by replacing column $i$ of $\mathbf A$ with column $j$ of $\mathbf A$.

Then $\mathbf A' = \begin{bmatrix} a' \end{bmatrix}_n$ has two identical columns, so:

By definition of matrix product, element $\left({i, j}\right)$ of $\operatorname{adj}(\mathbf A) \cdot \mathbf A$ is:


 * $\displaystyle \sum_{k \mathop = 1}^n A_{ki} a_{kj} = \begin{cases}

0_R & \text{for} & i \ne j \\ \det \left({\mathbf A}\right) & \text{for} & i = j \end{cases}$

Hence, $\operatorname{adj}(\mathbf A) \cdot \mathbf A = \det \left({\mathbf A}\right) \cdot \mathbf I_n$.

Also see

 * Matrix is Invertible iff Determinant has Multiplicative Inverse