Hello, Cthul!
I don't understand your title.
What doesn't work?
Find the inverse of: .$\displaystyle \begin{bmatrix}\text{-}2&\text{-}7 \\ 1&4\end{bmatrix}$
. . $\displaystyle (A)\;\begin{bmatrix}4&7\\ \text{-}1&\text{-}2\end{bmatrix}\qquad (B)\;\begin{bmatrix}\text{-}2&1\\7&\text{-}4\end{bmatrix} \qquad (C)\;\begin{bmatrix}\text{-}4&\text{-}7 \\ 1 & 2 \end{bmatrix}$ . . $\displaystyle (D)\;\begin{bmatrix}1 & \text{-}7 \\ 2&\text{-}4\end{bmatrix}\qquad (E)\;\begin{bmatrix}\text{-}1 & 2 \\ 4&\text{-}7\end{bmatrix} $
Answer (C) works . . . check it out!
It makes $\displaystyle \begin{bmatrix}1&0 \\ 0&1\end{bmatrix}$
I tried finding an inverse.
I got:
$\displaystyle \frac{1}{15}\begin{bmatrix}4&1 \\ \text{-}7&\text{-}2\end{bmatrix}$$\displaystyle \begin{bmatrix}1&0 \\ 0&1\end{bmatrix}=A$
How'd you find it works for C?
EDIT: I see what I did wrong now, nevermind. I know how to do it now.