determinant of a block-diagonal matrix
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Question : I can do the first part, where $A$ is a $1times1$ matrix. $$Det(C)=Sigma_i (-1)^{i+j} C_{ij}Det(F_{ij})$$ $F_{ij}$ is the co-factor matrix (a matrix that includes all the elements of $C$ except for the ith row and jth column) If we set $j=1$ then the only value for $i$ for which $C_{ij}neq0$ is 1. we also know that $C_{11}=Det(A)$ and therfoer the co-factor matrix is $F_{11}B$. Our expression above becomes $Det(C)=Det(A)det(B)$ as required. The second bit, however, I couldn't do via induction. I did find one method however: $$ C=begin{bmatrix} A & O \ O & I_m end{bmatrix} begin{bmatrix} I_n & O \ O & B end{bmatrix} $$ where $I_n$ is the $ntimes n$ identity matrix. $$ Det(C)=Det (begin{bmatrix} A & O \
