We assume a singular $2×2$ matrix $A=\begin{pmatrix} a & b \\ c & d \\ \end{pmatrix}.$ Then $ad=bc$ and $A^2=(a+d)A$ and $A^n=(a+d)^{n-1}A.$ So
$$B=\sum_{n = 0}^{+\infty} \frac {A^{2n}}{(2n)!} = I+ \sum_{n = 1}^{+\infty} \frac {(a+d)^{2n-1}}{(2n)!}A$$
By the ratio test, the series $\sum_{n = 1}^{+\infty} \frac {(a+d)^{2n-1}}{(2n)!}$ converges and we have:
$$\sum_{n = 1}^{+\infty} \frac {(a+d)^{2n-1}}{(2n)!} = \frac{\cosh (a+d) -1}{a+d}$$
Could $B$ be a zero matrix?
$B=0\implies 0=\det A=(\frac {a+d}{1-\cosh (a+d)})^2\implies a+d=0$, a contradiction.