### 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 positive real numbers $x$ and $y$ such that the matrix $(x \mathbf{I} + y \mathbf{A})^2$ equals $\mathbf{A}$.
I can start by expanding the square and simplifying the matrix expression.
I have $(x \mathbf{I} + y \mathbf{A})^2 = (x \mathbf{I})^2 + 2x y \mathbf{I} \mathbf{A} + (y \mathbf{A})^2 = x^2 \mathbf{I} + 2xy \mathbf{A} + y^2 \mathbf{A}^2.$
I can use the fact that $\mathbf{I}$ is the identity matrix and $\mathbf{A}^2 = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} = -\mathbf{I}$ to further simplify this expression.
I get $(x \mathbf{I} + y \mathbf{A})^2 = (x^2 - y^2) \mathbf{I} + 2xy \mathbf{A}.$
Now I want this to equal $\mathbf{A}$, so I need to compare the corresponding entries of the two matrices.
I have $x^2 - y^2 = 0$ and $2xy = 1$.
The first equation implies that $x = \pm y$, and the second equation implies that $x$ and $y$ are both nonzero.
Since I want $x$ and $y$ to be positive, I can choose $x = y$ and ignore the negative possibility.
Then I can plug in $x = y$ into the second equation and get $2x^2 = 1$, or $x^2 = \frac{1}{2}$.
Taking the positive square root, I get $x = \frac{\sqrt{2}}{2}$.
This also implies that $y = \frac{\sqrt{2}}{2}$, since $x = y$.
Therefore, the ordered pair $(x,y)$ that satisfies the given equation is $\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$.