Matrix Product with Adjugate Matrix

Theorem
Let $\mathbf A = \begin{bmatrix} a \end{bmatrix}_n$ be an invertible square matrix of order $n$ over a commutative ring with unity $\left({R, +, \circ}\right)$.

Let $\mathbf B = \begin{bmatrix} b \end{bmatrix}_n = \mathbf A^{-1}$ be the inverse of $\mathbf A$.

Then each entry $b_{ij}$ of $\mathbf B$ satisfies:


 * $b_{ij} = \dfrac {1} {\det \left({\mathbf A}\right) } A_{ji}$

where:


 * $\dfrac{ 1 }{\det \left({\mathbf A}\right)}$ is the multiplicative inverse of the determinant of $\mathbf A$
 * $A_{ji}$ is the cofactor of $a_{ji} \in \mathbf A$.

Determinant has Inverse
We show that the determinant $\det \left({\mathbf A}\right)$ has a multiplicative inverse in $R$.

Let $1_R$ denote the unity of $R$, and let $I_n$ denote the identity matrix of order $n$:

This shows that $\det \left({\mathbf B}\right) = \dfrac{1}{\det \left({\mathbf A}\right)}$.

Right Inverse
We show that $\mathbf A \mathbf B = \mathbf I_n$.

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 produtct, element $\left({i, j}\right)$ of $\mathbf A \mathbf B$ is:


 * $\displaystyle \sum_{k \mathop = 1}^n a_{ik} b_{kj} = \dfrac{1}{\det \left({\mathbf A}\right)} \sum_{k \mathop = 1}^n a_{ik} \mathbf A_{jk} = \left\{ { \begin{array}{ccc} 0_R & \text{for} & i \ne j \\ 1_R & \text{for} & i = j \end{array} }\right.$

Hence, $\mathbf A \mathbf B = \mathbf I_n$.

Left Inverse
We show that $\mathbf B \mathbf A = \mathbf I_n$.

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 produtct, element $\left({i, j}\right)$ of $\mathbf B \mathbf A$ is:


 * $\displaystyle \sum_{k \mathop = 1}^n b_{ik} a_{kj} = \dfrac{1}{\det \left({\mathbf A}\right)} \sum_{k \mathop = 1}^n a_{kj} \mathbf A_{ki} = \left\{ { \begin{array}{ccc} 0_R & \text{for} & i \ne j \\ 1_R & \text{for} & i = j \end{array} }\right.$

Hence, $\mathbf B \mathbf A = \mathbf I_n = \mathbf A \mathbf B$, so $\mathbf B$ is the inverse of $\mathbf A$.