Question 20

Let $v = (0, 1, 0, \dots, 0)^T \neq \mathbf{0}$. Since $Mv = \mathbf{0}$:

1. The nullity of $M$ is at least $1$, so $M$ is not invertible and $\det(M) = 0$ (Option (B) is always true and Option (A) is false).
2. Because the null space of $M$ is a vector subspace over $\mathbb{R}$, for every non-zero scalar $c \in \mathbb{R}$ (for instance, $c = 1$ and $c = 2$), the vector $cv \neq \mathbf{0}$ also lies in the null space of $M$. Thus there are infinitely many (and in particular at least two) non-zero vectors in the null space of $M$ (Option (D) is always true).
3. The rank of $M$ can be any integer from $0$ (if $M = 0$) to $n - 1$, so Option (C) is not always true.