Finding a matrix How might I find matrix $M\in M_2(\mathbb C)$ such that $M^t A =M^{-1}$
where $A=\left[
                           \begin{array}{cc}
                             a & a  \\
                             a & a+1 \\
                           \end{array}
                         \right]\in M_2(\mathbb N)$ without using the brute force method of writing $M=\left[
                           \begin{array}{cc}
                             a & b  \\
                             c & d \\
                           \end{array}
                         \right]$? (I don't quite know how to solve the resulting system even in the brute force case anyway.)
Thanks in advance.

Clearly $a$ is nonzero, or else $M^\top A$ is non-invertible and it cannot be equal to the invertible matrix $M^{-1}$. Now,
$$
\begin{align}
A=\begin{pmatrix}a&a\\a&a+1\end{pmatrix}
&=\begin{pmatrix}a&a\\a&a\end{pmatrix}+\begin{pmatrix}0&0\\0&1\end{pmatrix}\\
&=\begin{pmatrix}\sqrt{a}\\ \sqrt{a}\end{pmatrix}\begin{pmatrix}\sqrt{a}&\sqrt{a}\end{pmatrix}+\begin{pmatrix}0\\1\end{pmatrix}\begin{pmatrix}0&1\end{pmatrix}\\
&=\begin{pmatrix}\sqrt{a}&0\\ \sqrt{a}&1\end{pmatrix}
\begin{pmatrix}\sqrt{a}&\sqrt{a}\\0&1\end{pmatrix} = (M^{-1})^\top M^{-1}.
\end{align}
$$
So you may set $M=\begin{pmatrix}\sqrt{a}&\sqrt{a}\\0&1\end{pmatrix}^{-1}=\begin{pmatrix}\frac{1}{\sqrt{a}}&-1\\0&1\end{pmatrix}$.
