GATE CS 2026 Set 1 Q10 — Linear Algebra: Matrices, Determinants, System of Linear Equations

We are given that $n > 1$ and the non-zero vector $v = (0, 1, 0, 0, \ldots, 0)^T \in \mathbb{R}^n$ lies in the null space of the $n \times n$ matrix $M$. This means:

Let us analyze each option based on this property:

• Option A & B (Determinant of $M$):
  Since $v$ is the second standard basis vector $e_2$, the product $M e_2$ represents the second column of $M$. Thus, the second column of $M$ is a column of all zeros.

  A matrix with a column of all zeros is singular, meaning its columns are linearly dependent. Therefore, the determinant of $M$ must be $0$.

  • Option A is incorrect.
  • Option B is correct.
• Option C (Rank of $M$):
  The rank of $M$ is not restricted to $1$. For example:
  • If $M$ is the $n \times n$ zero matrix, then $\text{Rank}(M) = 0$.
  • If $M$ is a diagonal matrix with diagonal entries $(1, 0, 1, \ldots, 1)$, then $\text{Rank}(M) = n - 1$. Since $n > 1$, this rank can be greater than $1$.
  Hence, the rank of $M$ is not always $1$.
  Option C is incorrect.
• Option D (Non-zero vectors in the null space):
  The null space of $M$, denoted by $\text{Null}(M)$, is a vector subspace of $\mathbb{R}^n$.

  If a non-zero vector $v \in \text{Null}(M)$, then any scalar multiple $c \cdot v$ (for $c \in \mathbb{R}$) must also belong to $\text{Null}(M)$.

  Since $\mathbb{R}$ contains infinitely many scalars, we can choose $c = 1$ and $c = 2$ to obtain two distinct non-zero vectors in the null space:

  $$v_1 = (0, 1, 0, \ldots, 0)^T \quad \text{and} \quad v_2 = (0, 2, 0, \ldots, 0)^T$$ Thus, there are at least two (in fact, infinitely many) non-zero vectors in the null space of $M$.
  Option D is correct.