Let \(A\), \(B\), \(C\), \(D\) be commuting \(n \times n\) matrices over the field \(F\). Show that the determinant of the \(2n \times 2n\) matrix \[ \begin{bmatrix} A & B \\ C & D \end{bmatrix} \] is \(\det(AD - BC)\).
Solution
Let \[ M=\begin{bmatrix}A&B\\ C&D\end{bmatrix}. \] We first prove the formula when \(A\) is invertible. Since \(A,B,C,D\) commute, \(A^{-1}\) also commutes with \(B,C,D\). The block factorization \[ \begin{bmatrix}A&B\\ C&D\end{bmatrix} = \begin{bmatrix}A&0\\ C&I\end{bmatrix} \begin{bmatrix}I&A^{-1}B\\ 0&D-CA^{-1}B\end{bmatrix} \] gives \[ \det M =\det A\,\det(D-CA^{-1}B). \] Because the blocks commute, \[ D-CA^{-1}B=A^{-1}(AD-CB)=A^{-1}(AD-BC). \] Thus \[ \det M =\det A\,\det(A^{-1})\det(AD-BC) =\det(AD-BC). \]
Now remove the assumption that \(A\) is invertible. Work over the rational-function field \(F(t)\). The matrix \(A+tI\) is invertible over \(F(t)\) because \[ \det(A+tI) \] is a nonzero polynomial in \(t\) with leading term \(t^n\). The matrices \(A+tI,B,C,D\) still commute, so the already proved case yields \[ \det\begin{bmatrix}A+tI&B\\ C&D\end{bmatrix} = \det\bigl((A+tI)D-BC\bigr) \] as an identity in \(F(t)\). Both sides are in fact polynomials in \(t\), so this is an identity in \(F[t]\). Evaluating at \(t=0\) gives \[ \det\begin{bmatrix}A&B\\ C&D\end{bmatrix} =\det(AD-BC), \] as required.