# Prove about determinant of a block matrix.

Let $X \in Mat_p(\mathbb{K})$, $C \in Mat_q(\mathbb{K})$, $D\in Mat_{p,q}(\mathbb{K})$ where $p+q=n$.

Consider the function

$\alpha: \mathbb{K}^p\times...\times\mathbb{K}^p\rightarrow\mathbb{K}$

Defined by

$\alpha(X_1,...,X_p)=det\begin{bmatrix} X_{p\times p} & D_{p\times q} \\ 0_{q\times p } & C_{q\times q} \end{bmatrix}$

where $X\in Mat_p(\mathbb{K})$ in columns $X_1,...,X_p$

Prove:

$\alpha (X_1,...,X_p)=det(X).det(\begin{bmatrix} I_{p\times p} & D_{p\times q} \\ 0_{q\times p } & C_{q\times q}\end{bmatrix}$

My Work: (Edited with the hint of Jepsilon)

Suppose $X$ is invertible, then

$\begin{pmatrix} X & D\\ 0 & C \end{pmatrix}=\begin{pmatrix} X & 0\\ 0 & I \end{pmatrix}\begin{pmatrix} I & X^{-1}D\\ 0 & C \end{pmatrix}$

For other way, we have:

$\alpha(X_1,...,X_p)=det\begin{pmatrix} X & D\\ 0 & C \end{pmatrix}=det\begin{pmatrix} X & 0\\ 0 & I \end{pmatrix}det\begin{pmatrix} I & X^{-1}D\\ 0 & C \end{pmatrix}=XC$

Here i'm stuck. Can someone help me?

Note: I was thinking the case when $X$ is non-invertible. I think in that case $\alpha(X_1,...,X_p)$ is $0$. But i don't sure of that. Can someone help me with this other case?

$$\begin{pmatrix} X & D\\ 0 & C \end{pmatrix}=\begin{pmatrix} X & 0\\ 0 & I \end{pmatrix}\begin{pmatrix} I & X^{-1}D\\ 0 & C \end{pmatrix}$$
• Okay, then if i apply determinant i go to obtain this, no? $det(\begin{pmatrix} X & D\\ 0 & C \end{pmatrix}=det\begin{pmatrix} X & 0\\ 0 & I \end{pmatrix}det\begin{pmatrix} I & X^{-1}D\\ 0 & C \end{pmatrix}=XC$ no?, but that is my confusion. then how can i get obtain$det(X).det(\begin{bmatrix} I_{p\times p} & D_{p\times q} \\ 0_{q\times p } & C_{q\times q}\end{bmatrix}$ Thanks for help me! I appreciate that! Commented Jan 25, 2018 at 22:09