determinant of a block-diagonal matrix

Question:

I can do the first part, where $A$ is a $1\times1$ 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 $n\times n$ identity matrix.

$$Det(C)=Det (\begin{bmatrix} A & O \\ O & I_m \end{bmatrix}) Det( \begin{bmatrix} I_n & O \\ O & B \end{bmatrix})$$

Which is clearly (using similar logic to above) $Det(C)=Det(A)det(B)$

Can anyone say if 1. my proof is valid and 2. show me how to solve this via induction.

Thank you

• I am too lazy to write an answer, but if you apply Laplace formula to the first row of a matrix with $A$ of dimension $n+1$, you get the sum of $n+1$ element containing the determinants of block matrices with block of dimensions $n$ and $m$, where by hypotesis the product rule applies. – enzotib May 24 '17 at 16:16
• We haven't covered the Laplace formula and so, although I can look it up, I don't think that its what the question wants Edit: Laplace is just the the co-factor expansion I did, i didn't know what it was called. Although I thought of that and can't see how. – Toby Peterken May 24 '17 at 16:18