设 $\mathrm{A}$ 为 2 阶可逆矩阵, 且 $(2 A)^{-1}=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]$, 则 $\mathrm{A}=$
$\text{A.}$ $\frac{1}{2}\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]^{-1}$
$\text{B.}$ $2\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]^{-1}$
$\text{C.}$ $\frac{1}{2}\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]$
$\text{D.}$ $2\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]$