A question of non-singularity Let $A$ and $B$ be matrices such that $B^2+ AB + 2I = 0$, where I
denotes the identity matrix. Which of the following matrices must be
nonsingular?
(A) $A + 2I$
(B) $B$
(C) $B + 2I$
(D) $A$
I tried using a few tricks assuming each option to be nonsingular and then coming to the given form but to no avail. Any hint is appreciated.
 A: Since $$\det (B+A)\cdot \det (B) = \det (B^2+AB) = \det (-2I) =-2$$ we see that $\det (B)\ne 0$ so $B$ is invertibile.
A: We have
$B^2 + AB + 2I = 0, \tag 1$
so
$(B + A)B = B^2 + AB = -2I, \tag 2$
or
$\left ( -\dfrac{1}{2}(B + A) \right )B = I, \tag 3$
that is,
$B^{-1} = -\dfrac{1}{2}(B + A), \tag 4$
and the correct answer is (B).
A: I'm assuming you're over the real numbers.
Suppose $B$ is singular and take $v\ne0$ such that $Bv=0$; then
$$
0=(B^2+AB+2I)v=B^2v+ABv+2Iv=2v,
$$
a contradiction. Therefore $B$ is nonsingular.
In order to exclude the other cases, let's see whether $A$, $A+2I$ or $B+2I$ can be the zero matrix.
The matrix $A$ can be the zero matrix, because
$$
\begin{bmatrix}
0 & 2 \\
-1 & 0
\end{bmatrix}^2=-2I
$$
Also $A+2I$ can be the zero matrix: the identity to satisfy would be
$$
B^2-2B+2I=0
$$
and the matrix
\begin{bmatrix}
2 & -1 \\
2 & 0
\end{bmatrix}
satisfies it. I looked for a matrix with trace $2$ and determinant $2$, so Hamilton-Cayley solves the problem.
Can $B+2I=0$?

Just for completeness, the same computations can be performed over any field, with the same result, provided the field doesn't have characteristic $2$. Indeed, the initial contradiction we found is definitely not such when $2=0$.
In this case the identity becomes $B^2+AB=0$ and we see that $B$ can as well be the zero matrix, with $A$ any matrix at all. If $B$ is invertible, then $A=B$.
There is no case distinction for $A+2I=A$ and $B+2I=B$. Thus in the case of characteristic $2$, none of the four conditions is forced.
