Let n > 1. Consider an n×n matrix M with its elements from R. Let the vector (0, 1, 0,0,...,0) ∈ Rⁿ be in the null space of M. Which of the following options is/are always correct?

The correct option is: Determinant of M is 0.

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

By definition, the null space (or kernel) of an $n\times n$ matrix $M$ consists of all vectors $v\in {ℝ}^n$ such that:
$Mv=0$
where $0$ is the zero vector in ${ℝ}^n$.

We are given that the vector $v=(0,1,0,0,...,0)$ is in the null space of $M$. Notice that this vector is non-zero because its second component is $1$.

The existence of a non-zero vector in the null space of a square matrix has major implications:
1. It implies that the linear transformation represented by the matrix $M$ is not injective (not one-to-one), since both the non-zero vector $v$ and the zero vector map to the zero vector.
2. Consequently, the matrix $M$ is not invertible (singular).
3. A fundamental theorem in linear algebra states that a square matrix is invertible if and only if its determinant is non-zero. Since $M$ is not invertible, its determinant must be zero.

Thus, the determinant of $M$ is always 0.