### 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: Let's start by expanding the left-hand side of the equation using the distributive property and the fact that $\mathbf{I}$ is the identity matrix.
\[(x \mathbf{I} + y \mathbf{A})^2 = (x \mathbf{I} + y \mathbf{A})(x \mathbf{I} + y \mathbf{A}) = x^2 \mathbf{I} + xy \mathbf{A} + xy \mathbf{A} + y^2 \mathbf{A}^2\]
Next, let's simplify this expression by using the fact that $\mathbf{A}^2 = -\mathbf{I}$, which we can verify by multiplying $\mathbf{A}$ by itself.
\[x^2 \mathbf{I} + xy \mathbf{A} + xy \mathbf{A} + y^2 \mathbf{A}^2 = x^2 \mathbf{I} + 2xy \mathbf{A} - y^2 \mathbf{I}\]
Now, let's set this equal to $\mathbf{A}$ and compare the coefficients of $\mathbf{I}$ and $\mathbf{A}$ on both sides of the equation.
We get a system of two equations in two variables, $x$ and $y$.
\[x^2 \mathbf{I} + 2xy \mathbf{A} - y^2 \mathbf{I} = \mathbf{A} \implies \begin{cases} x^2 - y^2 = 0 \\ 2xy = 1 \end{cases}\]
Let's solve this system by substituting $y = \frac{1}{2x}$ from the second equation into the first equation and simplifying.
\[x^2 - y^2 = 0 \implies x^2 - \frac{1}{4x^2} = 0 \implies 4x^4 - 1 = 0 \implies (2x^2 - 1)(2x^2 + 1) = 0\]
Since we are looking for positive real numbers $x$ and $y$, we can ignore the factor $2x^2 + 1$, which is always positive, and focus on the factor $2x^2 - 1$, which is zero when $x = \pm \frac{1}{\sqrt{2}}$.
We choose the positive root, $x = \frac{1}{\sqrt{2}}$, and plug it back into the second equation to find $y = \frac{1}{2x} = \frac{\sqrt{2}}{2}$.
Therefore, the ordered pair $(x,y)$ that satisfies the given equation is $(\frac{1}{\sqrt{2}}, \frac{\sqrt{2}}{2})$.