TITLE: Upper Triangular Block Matrix Determinant by induction
QUESTION [14 upvotes]: We want to prove that: 
$$\det\begin{pmatrix}A & C \\ 0 & B\\ \end{pmatrix}= \det(A)\operatorname{det}(B),$$
where $A \in M_{m\times m}(R)$, $C \in M_{m\times n}(R)$,$B \in M_{n\times n}(R)$ and $R$ is a commutative ring with $1$.
$\textbf{SOLUTION}:$ 
Let $L = \begin{pmatrix}A & C \\ 0 & B\\ \end{pmatrix},$ then its clear that $L \in M_{(m+n)\times(m+n)}(R)$ and the matrix $L$ can be represented as follows:
$$L = \begin{pmatrix}\begin{bmatrix} 
a_{11} &a_{12}  &\ldots   & a_{1m}  \\ 
a_{21} &a_{22}  &\ldots   & a_{2m}  \\ 
\vdots  & \ddots  & \ddots  &\vdots  \\ 
 a_{11} &a_{12}  &\dots   & a_{mm}  \\
\end{bmatrix}  & \begin{bmatrix}
c_{11} &c_{12}  &\ldots   & c_{1n}  \\ 
c_{21} &c_{22}  &\ldots   & a_{2n}  \\ 
\vdots  & \ddots  & \ddots  &\vdots  \\ 
 c_{m1} &c_{m2}  &\dots   & c_{mn} \\
\end{bmatrix} \\ \\ \begin{bmatrix}
0 &0  &\ldots   & 0  \\ 
0 &0  &\ldots   & 0  \\ 
\vdots  & \ddots  & \ddots  &\vdots  \\ 
 0 &0  &\dots   & 0 \\
\end{bmatrix}& \begin{bmatrix}
b_{11} &b_{12}  &\ldots   & b_{1n}  \\ 
b_{21} &b_{22}  &\ldots   & b_{2n}  \\ 
\vdots  & \ddots  & \ddots  &\vdots  \\ 
 b_{n1} &b_{n2}  &\dots   & b_{nn} \\
\end{bmatrix}\\ \end{pmatrix}$$ Now, for each fixed $i \in \{1,2,\ldots,m,m+1,\ldots,m+n\}$ ($i$ is the row), We have:
$$\det(L) = \sum\limits_{j=1}^m (-1)^{i+j}\alpha_{ij}\det(L_{ij}) + \sum\limits_{j=m+1}^{m+n} (-1)^{i+j}\alpha_{ij}\det(L_{ij})$$
where: 
$$
\ \alpha_{i,j} = \begin{cases} 
      a_{i,j} & i,j = 1,\ldots , m \\
      c_{i,j-m} & i = 1,\ldots , m & j = m+1, \ldots ,m+n \\
      0 & i= m+1, \ldots ,m+n & j = 1,\ldots , m \\
      b_{i-m,j-m} & i = m+1,\ldots ,m+n & j = m+1, \ldots, m+n
   \end{cases}
\\
$$
and
$$
\\ L_{i,j} = \begin{cases} 
      A_{i,j} & i,j = 1,\ldots , m \\
      C_{i,j-m} & i = 1,\ldots , m & j = m+1, \ldots,m+n \\
      0 & i= m+1, \ldots ,m+n & j = 1,\ldots , m \\
      B_{i-m,j-m} & i = m+1,\ldots ,m+n & j = m+1, \ldots ,m+n
   \end{cases}
\\
$$
$\textbf{FIRST CASE}$
Choose $i > m$, then $i \in \{m+1,\ldots, m+n \}$. Now if $n = 0$, the problem is trivial, since $\det(B) = 1$ and therefore:$$\det(L) = \det(A)\det(B) = \det(A)1 = \det(A).$$
Then, lets assume that 
$$\det(L) = \det(A)\det(B),$$
holds for $n = k$.Then choose $i \in \{m+1,\ldots, m+k \}$, then $\forall j < m+1$, I get: $$(-1)^{i+j}\alpha_{ij}\det(L_{ij}) = 0,$$
Hence: 
$$\det(L) = \sum\limits_{j=m+1}^{m+n} (-1)^{i+j}\alpha_{ij}\det(L_{ij})$$
but $i \in \{m+1,m+2, \ldots, m+k\}$ and $j \in \{m+1,m+2, \ldots, m+k\}$ therefore:
$$\det(L) = \sum\limits_{j=m+1}^{m+k} (-1)^{i+j}b_{ij}\det(B_{ij})$$
Notice that the sub-matrix $B_{ij}$ has the form: $\begin{pmatrix}A & C' \\ 0 & B'\\ \end{pmatrix}$. On the other hand, $B'$ have size smaller than $B$, therefore by our induction hypothesis, we have:
$$\det\begin{pmatrix}A & C' \\ 0 & B'\\ \end{pmatrix} = \det(A)\det(B')$$
$
\textbf{SECOND CASE}
$
On the other hand, Choose $i \leq m$, if $m = 0$, then the problem is trivial, since $\det(A) = 1$ and therefore:
$$\det(L) = \det(A)\det(B) = \det(B) = \det(B).$$ Then, lets assume that  $$\det(L) = \det(A)\det(B)$$ holds for $m = k$. Then choose $i \in \{1,\ldots, k \}$, then for all $j>k$, we have: $$(-1)^{i+j}\alpha_{ij}\det(L_{ij}) = 0,$$ therefore: $$\det(L) = \sum\limits_{j=1}^{k} (-1)^{i+j}\alpha_{ij}\det(L_{ij})$$
but $i \in \{1,2, \ldots, k\}$ and $j \in \{1,2, \ldots, k\}$, therefore:
$$\det(L) = \sum\limits_{j=1}^k (-1)^{i+j}a_{ij}\det(A_{ij})$$
Now, the sub-matrix $A_{ij}$ has the form: $\begin{pmatrix}A' & C' \\ 0 & B\\ \end{pmatrix}$. On the other hand, $A'$ has size smaller than $A$, then by the induction hypothesis, we have:
$$\det\begin{pmatrix}A' & C' \\ 0 & B\\ \end{pmatrix} = \det(A')\det(B) $$
I want to know if this is the right proof.

REPLY [7 votes]: If $\det(A)=0,$ then the columns of $A$ are linearly dependent. Then the first $m$ columns of $X$ are linearly dependent.  In that case, $\det(X)=0=\det(A)\,\det(B).$ So, suppose $\det(A)\neq 0.$ Then $A$ is invertible. Notice that 
$$\begin{bmatrix} A & C \\ 0 & B\end{bmatrix} = \begin{bmatrix} A & 0 \\ 0 & I_n\end{bmatrix} \, \begin{bmatrix} I_m & A^{-1}C \\ 0 & B\end{bmatrix}.$$ 
Using the fact that determinant of a product is the product of determinants, we obtain 
$$\det(X)=\det \begin{bmatrix} A & 0 \\ 0 & I_n\end{bmatrix}\, \det \begin{bmatrix} I_m & A^{-1}C \\ 0 & B\end{bmatrix}=\det(A)\,\det(B).$$Arvind