Identity Matrix - Question

Does there exist a

2 \times 2

matrix

B

such that

B [\begin{matrix} 1 & - 2 \\ - 2 & 4 \end{matrix}] = I_{2}

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