$$A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\space, \space \space \color{green}{tr(A) = a+d}$$$$\begin{pmatrix} a & b \\ c & d \end{pmatrix} \cdot \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} = \begin{pmatrix} 1 & x \\ y & 0 \end{pmatrix}$$$$S_1 = \begin{cases} a + 3b = 1 \\ 2a +4b = x \\ c + 3d = y \\ 2c +4d = 0 \end{cases}$$ $$\begin{pmatrix} a & b \\ c & d \end{pmatrix} \cdot \begin{pmatrix} 2 & 3 \\ 4 & 5 \end{pmatrix} = \begin{pmatrix} x & 3 \\ y+1 & 1 \end{pmatrix}$$$$S_2 = \begin{cases} 2a + 4b = x \\ 3a + 5b = 3 \\ 2c + 4d = y+1 \\ 3c + 5d = 1 \end{cases}$$Dos sistemas $S_1$ e $S_2$, determina-se o $S_3$ :$$S_3 = \begin{cases} a + b = x-1 \\ a + b = 3-x \\ y = -1 \\ c + d = 1 \end{cases} \implies (a,\space b,\space c,\space d) = (1,\space 0, \space 2, \space -1)$$Conhecendo $a$ e $d$, conclui-se que o traço de $A$ é: $\boxed{tr(A) = 0}$ $$\text{Alternativa } \mathbb{(A)}$$