Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I found this interesting formula on Wikipedia on blocked determinant of a square matrix: $$\begin{vmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{vmatrix}= \begin{vmatrix} A_{11}A_{22}-A_{12}A_{21} \end{vmatrix}$$ provided that $A_{21}$ and $A_{22}$ commute.

Does anyone know of the obvious generalization of this formula? That is, is determinant of a matrix $A$ that is blocked into $n^2$ matrices $A_{11}, A_{12}, \cdots, A_{nn}$ equal to determinant of a "formal determinant" (computed using the Leibniz formula) by treating $A_{11}, A_{12} ,\cdots, A_{nn}$ as if they are scalars? (Here of course we assume that the blocking is done such that multiplications in the Leibniz formula make sense) Since the original formula involved conditions that $A_{21}$ and $A_{22}$ commute, there are probably conditions to make this formal blocked determinant thing work.

share|improve this question

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.