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Let $A$ be an $n\times n$ matrix over $K$, a commutative ring with identity. Suppose $A$ has the block form $$A=\begin{bmatrix} A_1& 0 & \cdots & 0 \\ 0 & A_2 & \cdots & 0 \\ \vdots & \vdots & & \vdots \\ 0& 0 & \cdots & A_k \end{bmatrix}$$ where $A_i$ is an $r_i\times r_i$ matrix. Prove $\text{det}(A)= \text{det}(A_1)\dotsb \text{det}(A_k)$.

My attempt: We’ll use following lemma, suppose we have an $n\times n$ matrix of the block form $\begin{bmatrix} A &B\\ 0&C\\ \end{bmatrix}$ where $A$ is an $r\times r$ matrix, $C$ is an $s\times s$ matrix, $B$ is $r\times s$, and $0$ denotes the $s\times r$ zero matrix. Then $\text{det} \begin{bmatrix} A &B\\ 0&C\\ \end{bmatrix}=\text{det}(A)\cdot \text{det}(C)$. So $\text{det}(A)=\text{det}(A_1)\cdot \text{det} \begin{bmatrix} A_2 & \cdots & 0 \\ \vdots&& \vdots \\ 0& \cdots & A_k \end{bmatrix}$. By mathematical induction, $\text{det}(A)=\text{det}(A_1)\dotsb \text{det}(A_k)$. Is my proof correct?

Que: let $i,j\in J_n$ such that $i\neq j$. Then $A_{ij}=0$ is in $K$ or $M_{r_i\times r_i}(K)$?

user264745
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    Your attempt is correct, provided you're happy with a proof of that lemma. I don't know what your 'que' is asking – FShrike Oct 24 '22 at 16:34
  • @FShrike let say $r_2=2$. So $A_2\in M_{2\times 2}(K)$. Then $A_{21}$, $(2,1)$ entry of $A$, is zero. I’m asking that zero denotes zero element of $K$ or zero element of $M_{2\times 2}(K)$. – user264745 Oct 24 '22 at 16:58

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