The statement reads as follows "If is invertible, there are no special solutions in the null space."
According to me, its actually true because if the matrix is invertible, it means that it has full rank, and its null space contains only the zero vector. However, the questions asks you to explain why the statement is false.
Is there a way for this to be false? Or, is the question wrong and the statement is true?
Assuming that "special solution" means "nonzero solution", then the statement is true. The proof is straightforward:
Consider $\mathbf{x}$ in the null space of $A$, so $A\mathbf{x} = \mathbf{0}$. Since $A$ is invertible, we can apply $A^{-1}$ to the equation: \begin{align} && A\mathbf{x} &= \mathbf{0} \\ \implies && A^{-1}A\mathbf{x} &= A^{-1}\mathbf{0} \\ \implies && \mathbf{x} &= \mathbf{0} \\ \end{align}