Let A be a $$2 \times 2$$ matrix with real entries. Let I be the $$2 \times 2$$ identity matrix. Denote by $$tr(A)$$, the sum of diagonal entries of $$A$$. Assume that $$A^2 = I$$. Statement-1: If $$A \ne I$$ and $$A \ne -I$$, then $$\det A = -1$$. Statement-2: If $$A \ne I$$ and $$A \ne -I$$, then $$tr(A) \ne 0$$.

1. Analyze the matrix equation

We are given that $$A^2 = I$$ for a $$2 \times 2$$ matrix with real entries. Taking the determinant on both sides:

$$\det(A^2) = \det(I)$$

$$(\det A)^2 = 1 \implies \det A = 1 \text{ or } \det A = -1$$

2. Apply the characteristic equation

Every square matrix satisfies its own characteristic equation. For a $$2 \times 2$$ matrix $$A$$, this is given by:

$$A^2 - \text{tr}(A) \cdot A + (\det A) \cdot I = O$$

We can substitute our given condition $$A^2 = I$$ into this equation:

$$I - \text{tr}(A) \cdot A + (\det A) \cdot I = O$$

$$\text{tr}(A) \cdot A = (1 + \det A) \cdot I$$

3. Evaluate Statement-1

Let us analyze what happens if $$\det A = 1$$. Substituting $$\det A = 1$$ into our relation:

$$\text{tr}(A) \cdot A = (1 + 1) \cdot I$$

$$\text{tr}(A) \cdot A = 2I$$

If $$\text{tr}(A) = 0$$, then $$O = 2I$$, which is impossible since the identity matrix is not a zero matrix. Therefore, $$\text{tr}(A) \neq 0$$. We can now isolate $$A$$:

$$A = \frac{2}{\text{tr}(A)} \cdot I$$

Squaring both sides gives:

$$A^2 = \frac{4}{(\text{tr}(A))^2} \cdot I^2 \implies I = \frac{4}{(\text{tr}(A))^2} \cdot I$$

$$(\text{tr}(A))^2 = 4 \implies \text{tr}(A) = 2 \text{ or } \text{tr}(A) = -2$$

If $$\text{tr}(A) = 2$$, then $$A = \frac{2}{2}I = I$$.

If $$\text{tr}(A) = -2$$, then $$A = \frac{2}{-2}I = -I$$.

Thus, if $$\det A = 1$$, then $$A$$ must be either $$I$$ or $$-I$$. By contraposition, if $$A \neq I$$ and $$A \neq -I$$, then $$\det A$$ cannot be $$1$$, meaning it must be $$-1$$.

Statement-1 is True.

4. Evaluate Statement-2

Let us analyze what happens if $$\det A = -1$$. Substituting $$\det A = -1$$ into our relation:

$$\text{tr}(A) \cdot A = (1 - 1) \cdot I$$

$$\text{tr}(A) \cdot A = O$$

Since $$A \neq O$$ (because $$A^2 = I$$), this equation implies that $$\text{tr}(A) = 0$$.

We can confirm this by constructing a counterexample where $$A \neq I$$, $$A \neq -I$$, but $$\text{tr}(A) = 0$$:

$$A = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$$

Let us check its properties:

$$A^2 = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I$$

$$\text{tr}(A) = 0 + 0 = 0$$

Here, $$A \neq I$$ and $$A \neq -I$$, yet $$\text{tr}(A) = 0$$. Therefore, the claim that $$\text{tr}(A) \neq 0$$ is false.

Statement-2 is False.

Final Answer

Statement-1 is True and Statement-2 is False.