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 (Test3) we get an outcome like this:
$$
\begin{bmatrix}
1.000000 & 0.000000 \\
1.000000 & 1.000000
\end{bmatrix} = \\
\begin{bmatrix}
0.894427 & -0.447214 \\
0.447214 & 0.894427
\end{bmatrix}
\begin{bmatrix}
1.341641 & 0.447214 \\
0.447214 & 0.894427
\end{bmatrix}
$$
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:
$$
\begin{bmatrix}
1 & 0 \\
1 & 1
\end{bmatrix} =
\begin{bmatrix}
2/\sqrt{5} & -1/\sqrt{5} \\
1/\sqrt{5} & 2/\sqrt{5}
\end{bmatrix}
\begin{bmatrix}
3/\sqrt{5} & 1/\sqrt{5} \\
1/\sqrt{5} & 2/\sqrt{5}
\end{bmatrix}
$$
Finding the left polar decomposition is left as an exercise to the reader.
EDIT.
However, the square root of a real-valued general symmetric positive definite $2 \times 2$ matrix can be calculated exactly:
$$
\begin{bmatrix} a & b \\ b & c \end{bmatrix}^2 = \begin{bmatrix} A & B \\ B & C \end{bmatrix} \quad \Longrightarrow \\
\begin{bmatrix} a & b \\ b & c \end{bmatrix}\begin{bmatrix} a & b \\ b & c \end{bmatrix} =
\begin{bmatrix} a^2+b^2 & b(a+c) \\ b(a+c) & b^2+c^2 \end{bmatrix} = \begin{bmatrix} A & B \\ B & C \end{bmatrix}
$$
It follows that:
$$
\begin{cases} a^2+b^2 = A \\ b(a+c) = B \\ b^2+c^2 = C \end{cases} \quad \Longrightarrow \quad
\begin{cases} a^2 = A - b^2 \\ b^2(a^2+2ac+c^2) = B^2 \\ c^2 = C - b^2 \end{cases}
$$
Using the determinants gives an additional equation:
$$
(ac-b^2)^2 = AC-B^2 \quad \Longrightarrow \quad ac = \sqrt{AC-B^2}+b^2
$$
Making smart combinations already gives the end-result:
$$
b^2(a^2+2ac+c^2) = B^2 \quad \Longrightarrow \\ b^2([A - b^2]+2[\sqrt{AC-B^2}+b^2]+[C - b^2]) = B^2
\quad \Longrightarrow \\ b^2(A+2\sqrt{AC-B^2}+C) = B^2 \quad \Longrightarrow \\ b = \frac{B}{\sqrt{A+2\sqrt{AC-B^2}+C}}
$$
Together with
$$
a = \sqrt{A-b^2} \quad ; \quad c = \sqrt{C-b^2}
$$
So here comes a much simpler solution. First form the transpose times the original matrix:
$$
\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}^T\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} =
\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} =
\begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} A & B \\ B & C \end{bmatrix} =
\begin{bmatrix} a & b \\ b & c \end{bmatrix}^2
$$
With the above formulas:
$$
b = \frac{B}{\sqrt{A+2\sqrt{AC-B^2}+C}} = \frac{1}{\sqrt{2 + 2\sqrt{1} + 1}} = 1/\sqrt{5} \\
a = \sqrt{A-b^2} = \sqrt{2 - 1/5} = 3/\sqrt{5} \\
c = \sqrt{C-b^2} = \sqrt{1 - 1/5} = 2/\sqrt{5}
$$
So we have, indeed:
$$
\sqrt{\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}^T\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}} =
\begin{bmatrix} 3/\sqrt{5} & 1/\sqrt{5} \\ 1/\sqrt{5} & 2/\sqrt{5} \end{bmatrix}
$$
At last, the orthogonal matrix is found with:
$$
\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}
\begin{bmatrix} 3/\sqrt{5} & 1/\sqrt{5} \\ 1/\sqrt{5} & 2/\sqrt{5} \end{bmatrix}^{-1} =
\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}
\begin{bmatrix} 2/\sqrt{5} & -1/\sqrt{5} \\ -1/\sqrt{5} & 3/\sqrt{5} \end{bmatrix} / 1 =
\begin{bmatrix} 2/\sqrt{5} & -1/\sqrt{5} \\ 1/\sqrt{5} & 2/\sqrt{5} \end{bmatrix}
$$
Which leads to the same conclusion:
$$
\begin{bmatrix}
1 & 0 \\
1 & 1
\end{bmatrix} =
\begin{bmatrix}
2/\sqrt{5} & -1/\sqrt{5} \\
1/\sqrt{5} & 2/\sqrt{5}
\end{bmatrix}
\begin{bmatrix}
3/\sqrt{5} & 1/\sqrt{5} \\
1/\sqrt{5} & 2/\sqrt{5}
\end{bmatrix}
$$
