Matrix Product with Adjugate Matrix

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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 $\adj {\mathbf A}$ be its adjugate matrix.


Then:

\(\ds \mathbf A \cdot \adj {\mathbf A}\) \(=\) \(\ds \map \det {\mathbf A} \cdot \mathbf I_n\)
\(\ds \adj {\mathbf A} \cdot \mathbf A\) \(=\) \(\ds \map \det {\mathbf A} \cdot \mathbf I_n\)


where $\map \det {\mathbf A}$ is the determinant of $\mathbf A$, and $\mathbf I_n$ denotes the unit matrix of order $n$.


Proof

Let $\mathbf A = \paren {a_{i j} }$.

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


Right Multiplication

We show that $\mathbf A \cdot \adj {\mathbf A} = \map \det {\mathbf A} \cdot \mathbf I_n$.

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

If $i = j$, expanding $\map \det {\mathbf A}$ along row $i$ shows that:

$\ds \map \det {\mathbf A} = \sum_{k \mathop = 1}^n a_{i k} A_{i k}$

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:

\(\ds 0_R\) \(=\) \(\ds \map \det {\mathbf A'}\) Square Matrix with Duplicate Rows has Zero Determinant
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 1}^n a'_{j k} A'_{j k}\) expanding $\map \det {\mathbf A'}$ along row $j$
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 1}^n a_{i k} A_{j k}\)


By definition of matrix product, element $\tuple {i, j}$ of $\mathbf A \cdot \adj {\mathbf A}$ is:

$\ds \sum_{k \mathop = 1}^n a_{i k} A_{j k} = \begin {cases} 0_R & \text {for} & i \ne j \\ \map \det {\mathbf A} & \text {for} & i = j \end {cases}$

Hence:

$\mathbf A \cdot \adj {\mathbf A} = \map \det {\mathbf A} \cdot \mathbf I_n$

$\Box$


Left Multiplication

We show that $\adj {\mathbf A} \cdot \mathbf A = \map \det {\mathbf A} \cdot \mathbf I_n$.

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

If $i = j$, expanding $\map \det {\mathbf A}$ along column $j$ shows that:

$\ds \map \det {\mathbf A} = \sum_{k \mathop = 1}^n a_{k j} A_{k j}$

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:

\(\ds 0_R\) \(=\) \(\ds \map \det {\mathbf A'}\) Square Matrix with Duplicate Columns has Zero Determinant
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 1}^n a'_{k i} A'_{k i}\) expanding $\map \det {\mathbf A'}$ along column $i$
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 1}^n a_{k j} A_{k i}\)


By definition of matrix product, element $\tuple {i, j}$ of $\adj {\mathbf A} \cdot \mathbf A$ is:

$\ds \sum_{k \mathop = 1}^n A_{k i} a_{k j} = \begin {cases} 0_R & \text {for} & i \ne j \\ \map \det {\mathbf A} & \text {for} & i = j \end {cases}$

Hence:

$\adj {\mathbf A} \cdot \mathbf A = \map \det {\mathbf A} \cdot \mathbf I_n$

$\blacksquare$


Also see


Sources

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