Upper Triangular Block Matrix Determinant by induction

Set $$ X=\begin{bmatrix}A & C\\0 & B\end{bmatrix} $$ If $A$ is not invertible, then its columns are linearly dependent, hence the first $m$ columns of $X$ are linearly dependent and also $X$ is not invertible. In this case the relation $\det X=\det A\det B$ is true. So we can assume $A$ is invertible; if Gaussian elimination on $A$ requires row switches, then collect all row switches in a permutation matrix $P$, so elimination on $PA$ can be done without row switches and $PA=LU$ where $L$ is lower triangular and $U$ is upper unitriangular. Consider the matrix $$P'=\begin{bmatrix}P & 0 \\ 0 & I_n\end{bmatrix}$$ so $$ P'X= \begin{bmatrix}P & 0 \\ 0 & I_n\end{bmatrix} \begin{bmatrix}A & C\\0 & B\end{bmatrix}= \begin{bmatrix}PA&PC\\0&B\end{bmatrix}= \begin{bmatrix}LU&PC\\0&B\end{bmatrix}= \begin{bmatrix}L & 0 \\ 0 & I_n\end{bmatrix} \begin{bmatrix}U & L^{-1}PC\\0 & B\end{bmatrix} $$ Now, as $$ \begin{bmatrix}L & 0 \\ 0 & I_n\end{bmatrix} $$ is lower triangular, we clearly have that its determinant is equal to $\det L$. Since $U$ is upper unitriangular, with $m$ times repeated Laplace development we get that $$ \det\begin{bmatrix}U & L^{-1}PC\\0 & B\end{bmatrix}=\det B $$ Therefore $$ \det P'X=\det P'\det X=\det L\det B $$ On the other hand, $P'$ is a permutation matrix generated by as many row swaps as $P$, so $\det P=\det P'$. Also $\det A=\det L\det U=\det L$.

Hence $\det X=\det A\det B$.

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We can also do it by induction. For $i=1,2,\dots,m$, denote by $A_i$ the matrix obtained from $A$ by removing the first column and the $i$-th row; $C_i$ is the matrix obtained from $C$ by removing the $i$-th row.

The base case of induction is for $m=1$, which is obvious. Suppose $m>1$ and develop $\det X$ along the first column: \begin{multline} \det X= (-1)^{1+1}a_{11}\det\begin{bmatrix} A_1 & C_1 \\ 0 & B\end{bmatrix}+ (-1)^{2+1}a_{21}\det\begin{bmatrix} A_2 & C_2 \\ 0 & B\end{bmatrix}\\ +\dots+ (-1)^{m+1}a_{m1}\det\begin{bmatrix} A_m & C_m \\ 0 & B\end{bmatrix} \end{multline} By induction hypothesis, we have, for $i=1,2,\dots,m$, $$ \det\begin{bmatrix} A_i & C_i \\ 0 & B\end{bmatrix}=\det A_i\det B $$ so \begin{align} \det X&= (-1)^{1+1}a_{11}\det A_1\det B+ (-1)^{2+1}a_{21}\det A_2\det B\\ &\qquad\qquad+\dots+ (-1)^{m+1}a_{m1}\det A_m\det B\\ &= \bigl((-1)^{1+1}a_{11}\det A_1+ (-1)^{2+1}a_{21}\det A_2+\dots+ (-1)^{m+1}a_{m1}\det A_m\bigr)\det B\\ &=\det A\det B \end{align}

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