1. Key Concepts and Formulas

This problem revolves around the properties of a $2 \times 2$ matrix $A$ that satisfies the condition $A^2 = I$, where $I$ is the $2 \times 2$ identity matrix. Such a matrix is known as an involutory matrix. To analyze its determinant ($\det(A)$) and trace ($\text{tr}(A)$), we will primarily use the Cayley-Hamilton Theorem.

• Cayley-Hamilton Theorem: Every square matrix satisfies its own characteristic equation. For a $2 \times 2$ matrix $A$, its characteristic equation is given by: $\det(A - \lambda I) = 0 \implies \lambda^2 - \text{tr}(A)\lambda + \det(A) = 0$ According to the theorem, the matrix $A$ itself satisfies this equation: $A^2 - \text{tr}(A)A + \det(A)I = O$ where $O$ is the $2 \times 2$ zero matrix.

2. Derivation from the Given Condition

We are given that $A^2 = I$. We can substitute this into the Cayley-Hamilton equation: $I - \text{tr}(A)A + \det(A)I = O$ Why this step? By substituting $A^2=I$, we introduce the given condition into the fundamental relationship provided by Cayley-Hamilton, allowing us to derive specific properties of $A$.

Now, let's rearrange the equation to group the identity matrices: $(1 + \det(A))I - \text{tr}(A)A = O$ This can be rewritten as: $\text{tr}(A)A = (1 + \det(A))I \quad (*)$ Why this step? Rearranging helps isolate the terms involving $A$ and $I$, which will be crucial for our case analysis. This equation establishes a direct relationship between $\text{tr}(A)$, $\det(A)$, $A$, and $I$.

3. Analysis of the Resulting Equation

We need to analyze the implications of equation $(*)$: $\text{tr}(A)A = (1 + \det(A))I$. We consider two main scenarios for matrix $A$:

Scenario 1: $A$ is a scalar multiple of $I$.

• If $A = kI$ for some real scalar $k$.
• Since $A^2 = I$, we have $(kI)^2 = I \implies k^2 I = I \implies k^2 = 1$.
• This implies $k = 1$ or $k = -1$.
• So, $A = I$ or $A = -I$.
• Why this scenario? These are the simplest involutory matrices. However, the statements in the question are conditioned on $A \ne I$ and $A \ne -I$. Therefore, these cases are excluded from the scope of evaluating Statement-1 and Statement-2.

Scenario 2: $A$ is NOT a scalar multiple of $I$.

• If $A$ is not a scalar multiple of $I$, then the matrices $A$ and $I$ are linearly independent in the space of $2 \times 2$ matrices.
• Why linear independence? If $A$ were a scalar multiple of $I$, it would contradict our assumption for this scenario. If $A$ is not $kI$, then $c_1 I + c_2 A = O$ implies $c_1=0$ and $c_2=0$. This property is key to solving equation $(*)$.
• From equation $(*)$, we have $\text{tr}(A)A - (1 + \det(A))I = O$.
• Since $A$ and $I$ are linearly independent, the coefficients of $A$ and $I$ in this linear combination must both be zero.
  • Coefficient of $A$: $\text{tr}(A) = 0$
  • Coefficient of $I$: $-(1 + \det(A)) = 0 \implies 1 + \det(A) = 0 \implies \det(A) = -1$

Conclusion from Analysis: Combining these scenarios, we deduce that for any $2 \times 2$ matrix $A$ with real entries such that $A^2=I$:

• If $A = I$, then $\text{tr}(A) = 2$ and $\det(A) = 1$.
• If $A = -I$, then $\text{tr}(A) = -2$ and $\det(A) = 1$.
• If $A \ne I$ and $A \ne -I$ (which implies $A$ is not a scalar multiple of $I$), then it must be that $\text{tr}(A) = 0$ and $\det(A) = -1$.

Example: Consider $A = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$.

• $A^2 = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I$.
• $A \ne I$ and $A \ne -I$.
• $\text{tr}(A) = 0 + 0 = 0$.
• $\det(A) = (0)(0) - (1)(1) = -1$. This example perfectly fits our derived conclusion for matrices other than $I$ or $-I$.

4. Evaluating Statement-1

Statement-1: If $A \ne I$ and $A \ne -I$, then $\det(A)=-1$.

• Based on our rigorous analysis (Scenario 2), if $A \ne I$ and $A \ne -I$, then $\det(A)$ must be equal to $-1$.
• Therefore, Statement-1 is TRUE.

5. Evaluating Statement-2

Statement-2: If $A \ne I$ and $A \ne -I$, then $\text{tr}(A) \ne 0$.

• Based on our rigorous analysis (Scenario 2), if $A \ne I$ and $A \ne -I$, then $\text{tr}(A)$ must be equal to $0$.
• The statement claims $\text{tr}(A) \ne 0$, which directly contradicts our finding.
• Therefore, Statement-2 is FALSE.

6. Conclusion and Option Selection

• Statement-1 is TRUE.
• Statement-2 is FALSE.

This combination matches option (D).

7. Tips and Common Mistakes

• Don't forget Cayley-Hamilton: For problems involving $A^2$, $\text{tr}(A)$, and $\det(A)$ for $2 \times 2$ matrices, Cayley-Hamilton is often the most direct path.
• Linear Independence: Remember that for $2 \times 2$ matrices, if $A$ is not a scalar multiple of $I$, then $A$ and $I$ are linearly independent. This is a powerful property for solving matrix equations like $c_1 A + c_2 I = O$.
• Test with Examples: After deriving general conclusions, always test them with simple examples (like the one used above) to build confidence and catch potential errors.
• Careful with Conditionals: Pay close attention to the "If... then..." structure of statements. The conditions ($A \ne I$ and $A \ne -I$) are crucial and exclude specific cases.

8. Summary / Key Takeaway

For any $2 \times 2$ matrix $A$ with real entries such that $A^2 = I$:

1. If $A=I$ or $A=-I$, then $\det(A)=1$ and $\text{tr}(A)=\pm 2$.
2. If $A \ne I$ and $A \ne -I$, then $\det(A)=-1$ and $\text{tr}(A)=0$. This comprehensive understanding allows for direct evaluation of statements about such matrices.