Use the given information to find A. $$   (-2A^{-1}-I) ^T = \begin{bmatrix}
        1 & -2 \\
        -4 & 7 \\
        \end{bmatrix}
$$
I did it like this:
$$   -2A^{-1}-I = \begin{bmatrix}
        1 & -4 \\
        -2 & 7 \\
        \end{bmatrix}
$$
=     $\dfrac{1}{(1)(7)-(-4)(-2)}$
Did I start the problem correctly? Any help is very much appreciated!
 A: First find $A^{-1}$ from the given equality
$$A^{-1}=-\frac12\left(I+\begin{bmatrix}
        1 & -2 \\
        -4 & 7 \\
        \end{bmatrix}^T\right)=\begin{bmatrix}
        -1 & 2 \\
        1 & -4 \\
        \end{bmatrix}$$
and then invert this matrix to find
$$A=\frac12\begin{bmatrix}
        -4 & -2 \\
        -1& -1 \\
        \end{bmatrix}$$
A: It looks like you started off right insofar as you took the transpose of each side as as step towards isolating $A$.  But the it looks as if you immediately tried to invert
$\begin{bmatrix} 1 & -4 \\ -2 & 7 \end{bmatrix},$ judging from the fact that the somewhat cryptic piece of an equation
$= - \dfrac{1}{(1)(7) - (-4)(-2)} \tag{1}$
which occurs next contains the determinant of
$-2A^{-1}-I = \begin{bmatrix} 1 & -4 \\ -2 & 7 \end{bmatrix} \tag{2}$
in the denominator.  I recommend the following:  from (2), add $I$ to both sides, yielding
$-2A^{-1} = \begin{bmatrix} 2 & -4 \\ -2 & 8 \end{bmatrix}, \tag{3}$
and then divide by $-2$:
$A^{-1} = \begin{bmatrix} -1 & 2 \\ 1 & -4 \end{bmatrix}; \tag{4}$
now just compute the inverse, using the fact that $(A^{-1})^{-1} = A$ and $\det A^{-1} = 2$; then the result is
$A = \begin{bmatrix} -2 & -1 \\ -\frac{1}{2} & -\frac{1}{2} \end{bmatrix}, \tag{5}$
as may easily be checked by calculating $AA^{-1}$; I leave this to you!
Hope this helps!  Cheerio,
and as always,
Fiat Lux!!!
