Let A \in \R^{N \times N} represent a square matrix,
\begin{align*}
A^{-1} A_m &= e_m \\
A^{-1} \begin{pmatrix} A_1 & A_2 & \dots & A_M \end{pmatrix} &= \begin{pmatrix} e_1 & e_2 & \dots & e_M \end{pmatrix}
\end{align*}
\begin{align*}
\end{align*}
\begin{align*}
A^{-1} \left( \sum_{n=1}^N x_n A_n \right) = x ,\quad x \in \R^{N}
\end{align*}
This is because
\begin{align*}
A x = \sum_{n=1}^N x_n A_n \\
A^{-1} A x = A^{-1} \left( \sum_{n=1}^N x_n A_n \right) \\
x = A^{-1} \left( \sum_{n=1}^N x_n A_n \right) \\
\end{align*}
If A^{-1} exists, then any y \in \R^N can be expressed as a linear combination y = x_1 A_1 + \dots + x_N A_N = A x.
The weights x = \set{x_n}_{n=1}^N can be obtained using A^{-1} y.
This is because
\begin{align*}
y = A x ,\\
A^{-1} y = A^{-1} A x ,\\
A^{-1} y = x . \\
\end{align*}
Discussion