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    The classical adjoint of a $2 \times 2$ matrix over a commutative ring

    E-HK-52-3

    Area
    Algebra
    Topics
    Determinants, Linear Algebra
    Status
    draft

    Let \(K\) be a commutative ring with identity. If \(A\) is a \(2 \times 2\) matrix over \(K\), the classical adjoint of \(A\) is the \(2 \times 2\) matrix \(\operatorname{adj} A\) defined by \[ \operatorname{adj} A = \begin{bmatrix} A_{22} & -A_{12} \\ -A_{21} & A_{11} \end{bmatrix}. \] If \(\det\) denotes the unique determinant function on \(2 \times 2\) matrices over \(K\), show that

    1. \((\operatorname{adj} A)A = A(\operatorname{adj} A) = (\det A)I\);

    2. \(\det(\operatorname{adj} A) = \det(A)\);

    3. \(\operatorname{adj}(A^t) = (\operatorname{adj} A)^t\).

    (\(A^t\) denotes the transpose of \(A\).)

    Solution

    Write \[ A=\begin{bmatrix}a&b\\c&d\end{bmatrix}, \qquad \operatorname{adj}A=\begin{bmatrix}d&-b\\-c&a\end{bmatrix}. \] Since \(K\) is commutative, \[ (\operatorname{adj}A)A = \begin{bmatrix} da-bc&0\\0&ad-bc \end{bmatrix} =(\det A)I, \] and the same multiplication in the opposite order gives \[ A(\operatorname{adj}A)=(\det A)I. \] This proves (a).

    For (b), \[ \det(\operatorname{adj}A)=da-(-b)(-c)=ad-bc=\det A. \]

    Finally, \[ \operatorname{adj}(A^t) = \operatorname{adj}\begin{bmatrix}a&c\\b&d\end{bmatrix} = \begin{bmatrix}d&-c\\-b&a\end{bmatrix} =(\operatorname{adj}A)^t. \]

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