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.
 A: We can rewrite your equation as
$$MM^T = A^{-1},$$
where
$$A^{-1} = \left[\begin{array}{cc} 1 + \frac{1}{a} & -1 \\ -1 & 1\end{array}\right]$$
is real, symmetric positive-definite. The Cholesky decomposition of $A^{-1}$ thus supplies such an $M$:
$$M = \left[\begin{array}{cc}\sqrt{\frac{a+1}{a}} & 0\\-\sqrt{\frac{a}{a+1}} & \sqrt{\frac{1}{a+1}}\end{array}\right].$$
A: You can also rephrase the problem as find a matrix $M$ such that $M^TAM = I$. It is relatively simple once you see that eliminating the matrix entries are easy to handle with $A$.
$$
\pmatrix{1&0\\-1&1}
\pmatrix{a & a  \\a & a+1}\pmatrix{1&-1\\0&1} = \pmatrix{a&0\\0&1}$$
Seeing that we have an extra $a$ term left we remove it by an extra step (assuming $a\neq 0$)
$$
\pmatrix{\frac{1}{\sqrt{a}}&0\\0&1}
\pmatrix{a &0  \\0 & 1}
\pmatrix{\frac{1}{\sqrt{a}}&0\\0&1}= \pmatrix{1&0\\0&1}
$$
Combining the two steps into one, gives 
$$
\underbrace{\pmatrix{\frac{1}{\sqrt{a}}&0\\0&1}\pmatrix{1&0\\-1&1}}_{M^T}
\pmatrix{a & a  \\a & a+1}
\underbrace{\pmatrix{1&-1\\0&1}\pmatrix{\frac{1}{\sqrt{a}}&0\\0&1}}_M = \pmatrix{1&0\\0&1}
$$
A: 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 $M=\begin{pmatrix}\sqrt{a}&\sqrt{a}\\0&1\end{pmatrix}^{-1}=\begin{pmatrix}\frac{1}{\sqrt{a}}&-1\\0&1\end{pmatrix}$.
