Answer

Solution

1. 1

   Since $A$ is singular, $\det A = 0$, so $\operatorname{rank}(A) < 5$. By the rank--nullity theorem,

2. 2

   $\dim \ker A = 5 - \operatorname{rank}(A) \geq 1$. The null space therefore contains a nontrivial

3. 3

   subspace, giving infinitely many solutions in $\mathbb R^5$.

Answer structure / marking notes

Content note