What is the volume of a pyramid in $\mathbb{R^4}$ with five corners at $\langle0,0,0,0\rangle$ and the rows of $I$
My attempted solution:
For a triangle in $\mathbb{R^2}$, the following formula gives the volume (area in $\mathbb{R^2}$):
$$A = \frac{1}{2}\cdot\begin{vmatrix}x_1 & y_1 & 1\\ x_2 & y_2 & 1\\ x_3 & y_3 & 1\end{vmatrix}$$
where $\langle x_i, y_i \rangle$ is a vertex of the triangle. Now if you take a close look, you'll notice $A \in \mathbb{R^3}$, while the our triangle is in $\mathbb{R^2}$, this is because our triangle exists in a plane $z = 1$, hence the $1's$ in column three of $A$.
My Understanding of this formula: We need $A$ to be in $\mathbb{R^3}$ as we have three vertices of a triangle in two-dimensions, and we can't take the determinant of a $2 \times 3$ matrix, so instead we take $\mathbb{R^2}$ as a plane in $\mathbb{R^3}$ and calculate the determinant from a $3 \times 3$ matrix.
Now I tried to generalize this formula for $A \in \mathbb{R^5}$, with our pyramid in $\mathbb{R^4}$, by my (possibly wrong) interpretation of the above formula like so:
$$A = \frac{1}{2}\cdot\begin{vmatrix}x_1 & y_1 & z_1 & \alpha_1& 1\\ x_2 & y_2 & z_2 & \alpha_2& 1\\ x_3 & y_3 & z_3 & \alpha_3& 1\\ x_4 & y_4 & z_4 & \alpha_4& 1\\ x_5 & y_5 & z_5 & \alpha_5& 1\\ \end{vmatrix}$$
where $\langle x_i, y_i, z_i, \alpha_{i} \rangle$ is a vertex of the pyramid in $\mathbb{R^4}$.
Plugging in numbers we get
$$A = \frac{1}{2}\cdot\begin{vmatrix}0 & 0 & 0 & 0& 1\\ 1 & 0 & 0 & 0& 1\\ 0 & 1 & 0 & 0& 1\\ 0 & 0 & 1 & 0& 1\\ 0 & 0 & 0 & 1& 1\\ \end{vmatrix} = \frac{1}{2}$$
which is wrong as the correct answer is $\frac{1}{24}$.
What have I done wrong? Is my interpretation of the original formula wrong? If so what is the correct interpretation? How can you correctly calculate the volume of a pyramid in $\mathbb{R^4}$?