Calculating polar decomposition
A brute force (numerical) method for obtaining the right polar decomposition has been presented in
this answer at MSE.
When applied to the OP's problem (`Test4`) we get an outcome like:
$$
\begin{bmatrix}
2.000000 & -3.000000 \\
1.000000 & 6.000000
\end{bmatrix} = \\
\begin{bmatrix}
0.894427 & -0.447214 \\
0.447214 & 0.894427
\end{bmatrix}
\begin{bmatrix}
2.236068 & 0.000000 \\
0.000000 & 6.708204
\end{bmatrix}
$$
However, when searching for the number $0.447214$ on the internet, we find that it's approximately equal to
What is 1 over square root of 5?.
So let's do an educated guess:
$$
\begin{bmatrix}
2 & -3 \\
1 & 6
\end{bmatrix} =
\begin{bmatrix}
2/\sqrt{5} & -1/\sqrt{5} \\
1/\sqrt{5} & 2/\sqrt{5}
\end{bmatrix}
\begin{bmatrix}
\sqrt{5} & 0 \\
0 & 3\sqrt{5}
\end{bmatrix}
$$
It is noticed that the above question is similar to
another one.
EDIT. But the above is, of course, overkill.
Here comes a much simpler solution. First form the transpose times the original matrix:
$$
\begin{bmatrix} 2 & -3 \\ 1 & 6 \end{bmatrix}^T\begin{bmatrix} 2 & -3 \\ 1 & 6 \end{bmatrix} =
\begin{bmatrix} 2 & 1 \\ -3 & 6 \end{bmatrix} \begin{bmatrix} 2 & -3 \\ 1 & 6 \end{bmatrix} =
\begin{bmatrix} 5 & 0 \\ 0 & 45 \end{bmatrix}
$$
Then make the following Ansatz:
$$
\begin{bmatrix} a & 0 \\ 0 & c \end{bmatrix}^2 =
\begin{bmatrix} a & 0 \\ 0 & c \end{bmatrix}\begin{bmatrix} a & 0 \\ 0 & c \end{bmatrix} =
\begin{bmatrix} a^2 & 0 \\ 0 & c^2 \end{bmatrix} = \begin{bmatrix} 5 & 0 \\ 0 & 45 \end{bmatrix}
$$
So the square root of a diagonal matrix is extremely simple; just take the square roots of the diagonal elements:
$$
\begin{bmatrix} a & 0 \\ 0 & c \end{bmatrix} = \begin{bmatrix} \sqrt{5} & 0 \\ 0 & 3\sqrt{5} \end{bmatrix}
$$
At last, the orthogonal matrix is found with:
$$
\begin{bmatrix} 2 & -3 \\ 1 & 6 \end{bmatrix}\begin{bmatrix} \sqrt{5} & 0 \\ 0 & 3\sqrt{5} \end{bmatrix}^{-1} = \\
\begin{bmatrix} 2 & -3 \\ 1 & 6 \end{bmatrix}\begin{bmatrix} 1/\sqrt{5} & 0 \\ 0 & 1/(3\sqrt{5}) \end{bmatrix} =
\begin{bmatrix} 2/\sqrt{5} & -1/\sqrt{5} \\ 1/\sqrt{5} & 2/\sqrt{5} \end{bmatrix}
$$
Same conclusion:
$$
\begin{bmatrix}
2 & -3 \\
1 & 6
\end{bmatrix} =
\begin{bmatrix}
2/\sqrt{5} & -1/\sqrt{5} \\
1/\sqrt{5} & 2/\sqrt{5}
\end{bmatrix}
\begin{bmatrix}
\sqrt{5} & 0 \\
0 & 3\sqrt{5}
\end{bmatrix}
$$