Can $A^3=0$ imply $|I+A|=0$? Suppose $A$ is a non-zero matrix such that $A^3=0$. Prove the following assertions or provide counter examples:- 
$(1) A^2$ is a zero matrix
$(2) A+A^2$ can have zero trace
$(3) A-A^2$ can have zero trace
$(4) I+A$ is  singular.
My Attempt:- I know if $A^3=0$ then  $A^2=0$ can be true (though not always). I have no idea whether $tr(A+A^2)=0$ or $tr(A-A^2)=0$ is possible or not if $A^3=0$. But when I looked closely at $|I+A|$ then I found that
$$|I+A|=0$$
For $2\times2$ matrix, we have
$$\Rightarrow |A|+tr(A)+1=0 $$
$$\Rightarrow \lambda_1\lambda_2+\lambda_1+\lambda_2+1=0 $$ 
where $\lambda_1$ and $\lambda_2$ are the two eigenvalues of $A$
$$\Rightarrow \lambda_1(\lambda_2+1)+1.(\lambda_2+1)=0 $$
$$\Rightarrow (\lambda_2+1)(\lambda_1+1)=0 $$
$$\Rightarrow \lambda_2=-1, \lambda_1=-1  \tag1$$
But we have $$A^3=0$$
$$\Rightarrow |A^3|=0$$
$$\Rightarrow |A|^3=0$$
$$\Rightarrow |A|=0$$
So,
either $\lambda_1=0$ or $\lambda_2=0$ (or both may be zero) which contradicts with equation $(1)$. So, $I+A$ is non singular.
Am I Correct ?
 A: Notice that $$(I+A)(I-A+A^2)=I,$$ hence $I+A$ is nonsingular. 
For $(2), (3)$, try the matrix $\begin{bmatrix} 0 & 1 \\ 0 & 0\end{bmatrix}$
A: You're on the right track.
Your first sentence about option (1) is correct, it can be ruled out. 
Your thoughts about $2\times2$ matrices and option (4) are also correct. 
However, here we might deal with bigger matrices. Especially because for a $2\times2$ matrix $A$, we do have $A^3=0 \implies A^2=0$.
It's generally a good practice to think about examples. A typical example for $A^3=0$ is
$$\pmatrix{0&1&0\\0&0&1\\0&0&0}$$
This example rules out options (2) and (3), so we're indeed only left with (4).
For this, can you find an inverse for $I+A$, knowing $A^3=0$?
A: $(a)$ can be true, set $A = \pmatrix{ 0 & 1 \\ 0 & 0}$.
$A^3 = 0$ implies that $\sigma(A) = \{0\}$, because the minimal polynomial of $A$ divides $x^3$.
This in turn means that $\sigma(A^k) = \{0\}$ so $\operatorname{Tr} A^k = 0$ for all $k \in \mathbb{N}$. Therefore $(b)$ and $(c)$ are always true:
$$\operatorname{Tr}(A \pm A^2) = \operatorname{Tr} A \pm \operatorname{Tr} A^2 = 0$$
Also, $\sigma(I + A) = \{1\}$ so $\det(I + A) = 1 \ne 0$, so $(d)$ is certainly false.
