A_n×n,n > 1. If (1, 0, 1, 0, 0, ..., 0) ∈ Rⁿ belongs to the null space of A then

The correct option is:

|A| = 0

Let us understand why this option is correct by breaking down the definition of the null space of a matrix and its relation to the determinant and invertibility of the matrix.

First, recall the definition of the null space of a matrix. For an $n\times n$ matrix $A$, the null space (or kernel) of $A$, denoted by $N\left(A\right)$, is the set of all vectors $x\in {ℝ}^n$ such that:
$Ax=0$
where $0$ is the zero vector in ${ℝ}^n$.

In this problem, we are given that a specific vector $v=(1,0,1,0,0,\mathrm{...},0)\in {ℝ}^n$ belongs to the null space of $A$.
Since $v$ has non-zero components (specifically, the first and third components are 1), it is a non-zero vector:
$v\ne 0$

Because $v\in N\left(A\right)$ and $v\ne 0$, the matrix equation $Ax=0$ has a non-trivial (non-zero) solution.
According to the fundamental properties of square matrices, the homogeneous system of linear equations $Ax=0$ has a non-trivial solution if and only if the matrix $A$ is singular (not invertible).

A square matrix $A$ is singular if and only if its determinant is equal to zero:
$\left|A\right|=0$

Therefore, since there exists a non-zero vector in the null space of $A$, the determinant of $A$ must be zero, confirming that $\left|A\right|=0$ is the correct option.