Finding the non singular matrix. $(1)$  Suppose $A$ and $B$ satisfy the relation $B^2+AB+2I=0$. Are the following four matrices necessarily non singular?  
(a)$A$   ( b)$B$        (c) $A+2I$             (d) $B+2I$
My attempt:-                      $$B^2+AB+2I=0$$
$$\Rightarrow B^2+AB= -2I$$
$$\Rightarrow B(A+B)= -2I$$
$$\Rightarrow |B(A+B)|= |-2I|$$
$$\Rightarrow |B||A+B|= (-2)^3|I|=-8$$
So, $$|B|\not= 0$$
Reqd. Answer is $B$.
Am I Correct?
 A: Yes.
If $B$ were singular there would be some nonzero vector $v$ such that $Bv=0$, but then
$$(B^2+AB+2I)v=0$$
$$\implies 2Iv=0$$
$$\implies v=0$$
which contradicts the assumption that $v$ is nonzero.
A: You could choose $A=0$, $B= \sqrt{2}i I$, so $A$ could be singular.
You could choose $A=-2I$, $B=(1+i)I$, so $A+2I$ could be singular.
You could choose $B=-2I$, $A=3I$, so $B+2I$ could be singular.
A: You are correct that $B$ has to be invertible. Let's construct an example which shows the matrices $A$, $A + 2I$ and $B + 2I$ can be singular.
$B$ is invertible so we can express $A$ from $B^2 + AB + 2I = 0$ as $$A = B + 2B^{-1}$$
Using the Spectral theorem, we conclude that the eigenvalues of $A$ are of the form $x + \frac2x$, where $x \in \sigma(B)$.
$A$ is singular if and only if $0\in \sigma(A)$ so $$0  = x + \frac2x \iff x = \pm i\sqrt{2}$$
Meaning $\pm i\sqrt{2} \in \sigma (B)$.
On the other hand, $A + 2I$ is singular if and only if $-2 \in \sigma(A)$, which happens when
$$-2 = x + \frac2x \iff x = -1 \pm i$$ 
So it has to be $-1 \pm i \in \sigma(B)$.
Now, setting $B = \begin{pmatrix} \sqrt{2}i & 0 & 0\\ 0 & -2 & 0\\0 & 0 & -1 + i \end{pmatrix}$ yields $$A = B + 2B^{-1} = \begin{pmatrix} 0 & 0 & 0\\ 0 & -3 & 0\\0 & 0 & -2 \end{pmatrix}$$
so $A, A + 2I$ and $B + 2I$ are all singular.
Therefore, $(b)$ is the only correct option.
Interestingly, the above discussion also shows that  $A, A + 2I$ and $B + 2I$ cannot all simultaneously be singular if the matrices are $2 \times 2$ since then $\sigma(B)$ can have at most two elements.  
