Proof explanation about matrix for block. Prove: 
$$\det\left[\begin{array}[cc]\\A&C\\
0&B\end{array}\right]=\det(A)\det(B)$$
Proof:
$$A = Q_A R_A, \quad B = Q_B R_B$$
be QR decompositions of $A$ and $B$. Then
\begin{align*}
\det \begin{bmatrix} A & C \\ 0 & B \end{bmatrix} &= \det \begin{bmatrix} Q_A R_A & Q_A Q_A^T C \\ 0 & Q_B R_B \end{bmatrix} = \det \left( \begin{bmatrix} Q_A \\ & Q_B \end{bmatrix} \begin{bmatrix} R_A & Q_A^T C \\ 0 & R_B \end{bmatrix} \right) \\
&= \det \begin{bmatrix} Q_A \\ & Q_B \end{bmatrix} \det \begin{bmatrix} R_A & Q_A^T C \\ 0 & R_B \end{bmatrix} = \det Q \det R,
\end{align*}
where
$$Q := \begin{bmatrix} Q_A \\ & Q_B \end{bmatrix}, \quad R := \begin{bmatrix} R_A & Q_A^T C \\ 0 & R_B \end{bmatrix}.$$
Notice that $R$ is (upper) triangular, so its determinant is equal to the product of its diagonal elements, so
$$\det R = \det \begin{bmatrix} R_A & 0 \\ 0 & R_B \end{bmatrix}.$$
Combining what we have,
\begin{align*}
\det \begin{bmatrix} A & C \\ 0 & B \end{bmatrix} &= \det Q \det R = \det \begin{bmatrix} Q_A \\ & Q_B \end{bmatrix} \det \begin{bmatrix} R_A \\ & R_B \end{bmatrix} \\
&= \det Q_A \det Q_B \det R_A \det R_B = \det (Q_AR_A) \det (Q_B R_B) \\
&= \det A \det B.
\end{align*}

I don't understand this line:$$\det \begin{bmatrix} Q_A \\ & Q_B \end{bmatrix} \det \begin{bmatrix} R_A \\ & R_B \end{bmatrix} \\
= \det Q_A \det Q_B \det R_A \det R_B$$

Can explain me that detail?
 A: The effort of proving 
$$
\det \begin{bmatrix} Q_A \\ & Q_B \end{bmatrix} 
= \det Q_A \det Q_B
$$
is basically the same as the effort to show $$
\det \begin{bmatrix} A&C \\0 & B \end{bmatrix} 
= \det A \det B.
$$
You can either work slowly (and painfully) by working on calculating the determinant by definition, or you can do the following. Use Schur Decomposition (or the Jordan Form) to write $A=SJS^{-1}$, $B=TKT^{-1}$, with $J,K$ upper triangular. Then 
\begin{align}
\begin{bmatrix} A&C\\ 0&B\end{bmatrix} &=\begin{bmatrix} SJS^{-1}&C\\ 0&TKT^{-1}\end{bmatrix} 
=\begin{bmatrix} S&0\\ 0&T\end{bmatrix} \begin{bmatrix} J&S^{-1}CT\\ 0&K\end{bmatrix} \begin{bmatrix} S^{-1}&0\\ 0&T^{-1}\end{bmatrix} \\ \ \\
&=\begin{bmatrix} S&0\\ 0&T\end{bmatrix} \begin{bmatrix} J&S^{-1}CT\\ 0&K\end{bmatrix} \begin{bmatrix} S&0\\ 0&T\end{bmatrix} ^{-1}.
\end{align}
So 
$$
\det\begin{bmatrix} A&C\\ 0&B\end{bmatrix}=\det \begin{bmatrix} J&S^{-1}CT\\ 0&K\end{bmatrix} 
=\det J\,\det K=\det A\,\det B.
$$
The fact the determinant of the block matrix with $J,K$ in the diagonal is $\det J\,\det K$ follows easily from the fact that the matrix is upper triangular, and its diagonal is made up of the diagonals of $J$ and $K$. 
A: Determinants multiply.
With $A$ square $m$ by $m,$ and $B$ square $n$ by $n,$ also $C$ $m$ by $n,$
$$
\left(
\begin{array}{c|c}
I_m & 0 \\ \hline
0 & B
\end{array}
\right)
\left(
\begin{array}{c|c}
A & C \\ \hline
0 & I_n
\end{array}
\right) =
\left(
\begin{array}{c|c}
A & C \\ \hline
0 & B
\end{array}
\right)
$$
