设 $A$ 为 2 阶可逆矩阵, $A^{-1}=\left(\begin{array}{ll}a_{11} & a_{12} \\ a_{21} & a_{22}\end{array}\right)$. 将 $A$ 第一行的 2 倍加到第二行上, 得到矩阵 $B$, 则 $B^{-1}=$
A
$\left(\begin{array}{ll}a_{11}-\frac{1}{2} a_{12} & a_{12} \\ a_{21}-\frac{1}{2} a_{22} & a_{22}\end{array}\right)$.
B
$\left(\begin{array}{ll}a_{11} & a_{12}+\frac{1}{2} a_{11} \\ a_{21} & a_{22}+\frac{1}{2} a_{21}\end{array}\right)$.
C
$\left(\begin{array}{ll}a_{11}-2 a_{12} & a_{12} \\ a_{21}-2 a_{22} & a_{22}\end{array}\right)$.
D
$\left(\begin{array}{ll}a_{11}+2 a_{12} & a_{12} \\ a_{21}+2 a_{22} & a_{22}\end{array}\right)$.
E
F