Step 1: Understanding the Concept:
This problem involves the properties of eigenvalues of a matrix, specifically an involutory matrix (where $A^2 = I$). There are three key matrix theorems that help solve this effortlessly:
1. If $\lambda$ is an eigenvalue of matrix $A$, then $\lambda^k$ is an eigenvalue of $A^k$.
2. The determinant of a matrix is equal to the product of all its eigenvalues: $\det(A) = \lambda_1 \cdot \lambda_2 \cdot \lambda_3$.
3. The trace of a matrix (the sum of its diagonal elements) is equal to the sum of all its eigenvalues: $\text{Tr}(A) = \lambda_1 + \lambda_2 + \lambda_3$.
Step 2: Key Formula or Approach:
Let the three eigenvalues of the $3 \times 3$ matrix $A$ be $\lambda_1$, $\lambda_2$, and $\lambda_3$.
- Given the matrix equation: $A^2 = I$
- Given the determinant property: $\det(A) = -1 \implies \lambda_1 \lambda_2 \lambda_3 = -1$
- Given the sum condition: $\lambda_1 + \lambda_2 + \lambda_3 = 1$
Step 3: Detailed Explanation:
1. Find possible values for the individual eigenvalues: Since $A^2 = I$, the eigenvalues of $A^2$ must match the eigenvalues of the identity matrix $I$. The only eigenvalue of $I$ is 1. Therefore, for any eigenvalue $\lambda$ of $A$:
$$ \lambda^2 = 1 \implies \lambda = \pm 1 $$
This restricts our eigenvalues; each of the three roots $\lambda_1, \lambda_2, \lambda_3$ must be either $+1$ or $-1$.
2. Test the product condition (Determinant):
The product of the eigenvalues must equal $-1$:
$$ \lambda_1 \cdot \lambda_2 \cdot \lambda_3 = -1 $$
For a product of three values (each $\pm 1$) to be negative, there must be either one negative value (roots: $1, 1, -1$) or three negative values (roots: $-1, -1, -1$).
3. Test the sum condition:
Let's check the sum for both remaining possibilities:
- Case 1: If the eigenvalues are $-1, -1, -1$:
$$ \text{Sum} = (-1) + (-1) + (-1) = -3 $$
This contradicts our given condition that the sum must equal $1$.
- Case 2: If the eigenvalues are $1, 1, -1$:
$$ \text{Sum} = 1 + 1 + (-1) = 1 $$
This perfectly satisfies the condition!
Therefore, the unique set of eigenvalues for matrix $A$ is $1, 1, -1$. This matches option (B).
Step 4: Final Answer:
The set of eigenvalues of A is 1, 1, $-$1.