To find the volume of the region that is bordered by 4 points in 3D space To find the volume of the region that in the points $A(x_1,y_1,z_1),B(x_2,y_2,z_2),C(x_3,y_3,z_3),D(x_0,y_0,z_0)$. 
Let's define a 4X4 matrix to determine plane equation  that are on $A,B,C$
$$\mathbf{M}=\begin{bmatrix}x & y & z& 1\\x_1 & y_1 & z_1& 1\\x_2 & y_2 & z_2& 1\\x_3 & y_3 & z_3& 1\end{bmatrix}$$
We know that plane equation can be found by $$det(\mathbf{M})=0$$ ,
Let's put $D(x_0,y_0,z_0)$ to plane equation and to get :
$$\mathbf{M_{3}}=\begin{bmatrix}x_0 & y_0 & z_0& 1\\x_1 & y_1 & z_1& 1\\x_2 & y_2 & z_2& 1\\x_3 & y_3 & z_3& 1\end{bmatrix}$$
We know if the point $D$ is on the plane (A,B,C) , $det(\mathbf{M_3})=0$. The point will satisfy the plane equation but if it is not on the plane we will have a value.
Is it true that volume $V$ of the region that is bordered by 4 points can written as
$V=\frac{1}{3} |det(\mathbf{M_3})|$ ?
I wrote it as intuition without proof. Can you help me to prove the lemma ? (or disprove)
And more generally ; Can the region value that is bordered by ($n+1$) points in the n-dimensional space be calculated by
$V_n=\frac{1}{n} |det(\mathbf{M_{n}})|$ ?
Please note that if $n=2$ ,
 The formula satisfies to find the area of triangle in 2D space
$$\mathbf{M_{2}}=\begin{bmatrix}x_0 & y_0 & 1\\x_1 & y_1 & 1\\x_2 & y_2 & 1\end{bmatrix}$$
$V_2=A=\frac{1}{2} |det(\mathbf{M_{2}})|$ 
Thanks for answers and comments
EDIT: I have found the paper in web while googling. You can see the related part in page 154 about N-dimensional volume calculation. As @Christian Blatter  pointed, the numarical factor must be $1\over n!$  not $1\over n$
 A: Given $n+1$ points ${\bf a}_k$ $(0\leq k\leq n)$ in ${\mathbb R}^n$ one considers their convex hull $S$ which is a (maybe degenerate) $n$-dimensional simplex. This simplex is spanned by the $n$ vectors ${\bf a}_k-{\bf a}_0$ $(1\leq k\leq n)$, and its volume is equal to the volume of the parallelotope $P$ spanned by these vectors, divided by $n!$ (see below). The volume of $P$ is, up to sign,  equal to  the determinant of the matrix having the coordinates of the spanning vectors ${\bf a}_k-{\bf a}_0$ in its columns. So
$${\rm vol}(S)={1\over n!}\>\biggl|\det\bigl[{\bf a}_1-{\bf a}_0,\ {\bf a}_2-{\bf a}_0,\ \ldots,\ {\bf a}_n-{\bf a}_0\bigr]\biggr|\ .\tag{1}$$
Consider now your matrices $M_2$ and $M_3$. If you subtract the first row from all the other rows you obtain  matrices $\hat M_2$ and $\hat M_3$ whose last column is zero, apart from the element in the upper right corner, which remains $1$. It's easy to see that the determinants of $\hat M_2$, $\hat M_3$, and therefore the determinants of $M_2$ and $M_3$, coincide, up to sign,  with the determinant appearing in $(1)$. But your conjectured numerical factor is wrong.
I cannot see how the equations of the  hyperplanes bounding $S$ should play a rôle here.
A proof of $(1)$ rests on induction and an application of Fubini's theorem. For arbitrary $n\geq1$ consider the special simplex  with vertices ${\bf 0}$ and $h{\bf e}_k$ $(1\leq k\leq n)$:
$$S_n(h):=\{(x_1,x_2,\ldots,x_n)\>|\>x_k\geq0 \ (1\leq k\leq n), \ x_1+x_2+\ldots+x_n\leq h\}\ .$$
It is geometrically obvious that ${\rm vol}_n\bigl(S_n(h)\bigr)=h^n\>{\rm vol}_n\bigl(S_n(1)\bigr)$.
We now proceed as follows:
$$\eqalign{{\rm vol}_{n+1}\bigl(S_{n+1}(1)\bigr)&=\int_{S_{n+1}(1)} 1\ {\rm d}(x_1,\ldots,x_n)\cr&=\int_0^1\left(\int_{S_n(1-x_{n+1})}1\ {\rm d}(x_1,\ldots,x_n)\right)\>dx_{n+1}\cr &=\int_0^1{\rm vol}_n\bigl(S_n(1-x_{n+1})\bigr)\>dx_{n+1}\cr
&={\rm vol}_n\bigl(S_n(1)\bigr)\int_0^1(1-x_{n+1})^n\>dx_{n+1}\cr
&={1\over n+1}{\rm vol}_n\bigl(S_n(1)\bigr)\ .\cr}$$
