A real $2 \times 2 $ matrix $M$ such that $M^2 = \tiny \begin{pmatrix} -1&0 \\ 0&-1-\epsilon \\ \end{pmatrix}$ , then : 
A real $2 \times 2 $ matrix $M$ such that $$M^2 =  \begin{pmatrix} -1&0 \\ 0&-1-\epsilon  \\   \end{pmatrix}$$
(a) exists for all $\epsilon > 0$.
(b) does not exist for any $\epsilon > 0$.
(c) exists for some $\epsilon > 0$.
(d) None of the above is true.

Attempt: I couldn't think of any theory base to prove / disprove the existence of such a matrix.
Could someone please give me a hint on how to go about this problem?
Thank you very much for your help in this regard.

no matter if $\epsilon \neq 0$ is , there is no real matrix $M.$ 
for $\epsilon = 0, M = \pmatrix{0&b\\-\frac 1b&0}, b\neq 0$
pick a matrix $$M = \pmatrix{a&b\\c&d}, M^2 =\pmatrix{a^2 + bc&(a+d)b\\(a+d)c&bc+d^2}=\pmatrix{-1&0\\0&-1-\epsilon} $$ we have the constraints $$a^2 + bc = -1, bc+d^2 = -1-\epsilon \to a^2 - d^2 = \epsilon \tag 1$$ 
that means if $\epsilon \neq 0,$  then $a+d \neq0\implies b = c = 0.$ from $(1), a^2 = -1, d^2 = -1 - \epsilon$   there are no real solutions as $a^2 = -1.$
if $\epsilon = 0, a^2 - d^2 = 0$ it forces $a = -d, b = c = 0$ but then (1) forces $d^2 = -1$ which cannot be satisfied. the other case is $a = d.$ the constraints are $$a^2 + bc = -1, 2ab = 0 = 2ac, bc + d^2 = -1 $$ two choices: (i) $a = 0$ gives $bc = -1, d = 0$
(ii) $a \neq 0$ which gives $b = 0, c = 0, a^2 = -1$ not possible.
