### Instruction: 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 $(x,y).$.### Response: I need to find two positive real numbers $x$ and $y$ that satisfy the equation $(x \mathbf{I} + y \mathbf{A})^2 = \mathbf{A}$, where $\mathbf{I}$ is the identity matrix and $\mathbf{A}$ is a given matrix.
To simplify the notation, let me write $\mathbf{B} = x \mathbf{I} + y \mathbf{A}$.
Then the equation becomes $\mathbf{B}^2 = \mathbf{A}$.
I can expand $\mathbf{B}^2$ using the distributive property of matrix multiplication: $\mathbf{B}^2 = (x \mathbf{I} + y \mathbf{A})^2 = x^2 \mathbf{I}^2 + 2xy \mathbf{I} \mathbf{A} + y^2 \mathbf{A}^2$.
I can also simplify some terms using the properties of the identity matrix and the given matrix: $\mathbf{I}^2 = \mathbf{I}$, $\mathbf{A}^2 = -\mathbf{I}$, and $\mathbf{I} \mathbf{A} = \mathbf{A} \mathbf{I} = \mathbf{A}$.
Substituting these simplifications into the equation, I get $x^2 \mathbf{I} + 2xy \mathbf{A} - y^2 \mathbf{I} = \mathbf{A}$.
Now, I need to compare the coefficients of the identity matrix and the matrix $\mathbf{A}$ on both sides of the equation.
For the identity matrix, I have $x^2 - y^2 = 0$.
For the matrix $\mathbf{A}$, I have $2xy = 1$.
The first equation implies that either $x = y$ or $x = -y$.
Since I want positive real numbers, I can eliminate the second option.
So $x = y$.
The second equation then becomes $2x^2 = 1$, which implies that $x = \frac{\sqrt{2}}{2}$ and $y = \frac{\sqrt{2}}{2}$.
Therefore, the ordered pair $(x,y)$ that satisfies the equation is $(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})$.