Given $\mathbf{A} = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix},$ there exist positive real numbers $x$ and $y$ such that
\[(x \mathbf{I} + y \mathbf{A})^2 = \mathbf{A}.\]Enter the ordered pair $(x,y).$
Question:
Given $\mathbf{A} = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix},$ there exist positive real numbers $x$ and $y$ such that
\[(x \mathbf{I} + y \mathbf{A})^2 = \mathbf{A}.\]Enter the ordered pair $(x,y).$
Answer:
Let's think step by step.

user
\left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right)
Given $\mathbf{A} = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix},$ there exist positive real numbers $x$ and $y$ such that
\[(x \mathbf{I} + y \mathbf{A})^2 = \mathbf{A}.\]Enter the ordered pair $(x,y).$
train