The answer is “no”. Let \(B = \begin{bmatrix} a & b \\ c & d\end{bmatrix}\). Then \(B \begin{bmatrix} 1 & -2 \\ -2 & 4\end{bmatrix} = I_2\) implies that \[\begin{bmatrix}a-2b & -2a+4b \\ c-2d & -2c+4d\end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}.\] Comparing the entries in the first row of both sides, we see that \begin{eqnarray} a - 2b & = & 1 \\ -2a + 4b & = & 0 \\ \end{eqnarray} Adding two times the first equation to the second equation gives \(0 = 2\), which is impossible.