The problem asks how many $2 \times 2$ matrices $A$ with real entries satisfy $A^2 = I_2$, where $I_2$ is the $2 \times 2$ identity matrix. Here's the analysis:

Key Conditions:

A matrix $A$ satisfies $A^2 = I_2$ if $A$ is:

1. Idempotent with respect to the identity matrix: $A^2 = I_2$ means $A$ must be its own inverse ($A^{-1} = A$).
2. The eigenvalues of $A$ must satisfy the characteristic equation $\lambda^2 = 1$, which gives $\lambda = 1$ or $\lambda = -1$.

Types of $A$:

From these properties, $A$ can belong to these cases:

1. The Identity Matrix $I_2$: Trivially satisfies $A^2 = I_2$.
2. The Negative Identity Matrix $-I_2$: Trivially satisfies $A^2 = I_2$.
3. Diagonalizable Matrices with Eigenvalues $1$ and $-1$: These matrices have the form: $P \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} P^{-1},$ where $P$ is any invertible matrix. This leads to infinitely many such matrices.

Conclusion:

The number of $2 \times 2$ matrices $A$ with $A^2 = I_2$ is infinitely many. The correct answer is: D. Infinitely many.

Would you like a detailed explanation of any step or example matrices? Here are some questions to explore further:

1. What does it mean for a matrix to be diagonalizable?
2. How can we find $P$ for diagonalizing a matrix?
3. What are examples of matrices satisfying $A^2 = I_2$?
4. How does the determinant of $A$ relate to this problem?
5. What happens when $A$ is not diagonalizable?

Tip: When working with matrices, always verify their eigenvalues and diagonalizability for conditions like $A^2 = I_2$.