Let $X$ be a smooth hypersurface in Projective space $\mathbb{P}^n$ of degree $ d$ defined by the equation $f=0$. Given that we have a vector bundle $E$ of rank $r\geq1$ on $X$ such that we have the following exact sequence on $\mathbb{P}^n$:
$$0\rightarrow O(-1)^{rd}\rightarrow O^{rd}\rightarrow E\rightarrow 0.$$ My question is as follows. What is the morphism from $O(-1)^{rd}\rightarrow O^{rd}$? A paper indicated that it is given by a $rd\times rd$ matrix of linear forms. Why is this? I am not able to see it. If this is so, can we say where in $\mathbb{P}^n$ the determinant of that matrix vanishes?
I do not understand your confusion about the size of the matrix. If you fix a basis (to avoid confusion), the map is given as $\oplus_{i=1}^m O(-1)e_i\to \oplus_{i=1}^m Ov_i$, where $e_i, v_i$ are just place holders. A map from $O(-1)e_i\to Ov_j$ is given by a linear form (possibly zero) $l_{ij}$. Thus, you get $m^2$ linear forms $l_{ij}$ given by an $m\times m$ matrix.
For the latter part, notice that determinant of this matrix is a homogeneous polynomial $F$ of degree $rd$. At any point $f\neq 0$, $E=0$ and thus the matrix must be non-singular there. This says, the only prime factor of $F$ is $f$ and by degree considerations, $F$ is $f^r$ up to a non-zero constant.