Let A be a `2xx2` matrix with real entries. Let I be the `2xx2` identity matrix. Denote by tr (A), the sum of diagonal entries of A. Assume that `A^2=""I` . Statement 1: If `A!=I` and `A!=""-I` , then det `A""=-1` . Statement 2: If `A!=I` and `A!=""-I` , then `t r(A)!=0` . (1) Statement 1 is false, Statement `( 2) (3)-2( 4)` is true (6) Statement 1 is true, Statement `( 7) (8)-2( 9)` (10) is true, Statement `( 11) (12)-2( 13)` is a correct explanation for Statement 1 (15) Statement 1 is true, Statement `( 16) (17)-2( 18)` (19) is true; Statement `( 20) (21)-2( 22)` is not a correct explanation for Statement 1. (24) Statement 1 is true, Statement `( 25) (26)-2( 27)` is false.

Options

A. Statement -1 is true, Statement-2 is true, Statement-2 is a
correct explanation for Statement-1
B. Statement -1 is true, Statement - 2 is true, Statement -2 is not
a correct explanation for Statement-1
C. Statement-1 is true, Statement-2 is false
D. Statement-1 is false, Statement-2 is true

Explanation

To solve the problem step by step, we will analyze the given statements about the matrix \( A \) and derive the necessary conclusions.

### Step 1: Understanding the Matrix Properties
Let \( A \) be a \( 2 \times 2 \) matrix with real entries such that \( A^2 = I \), where \( I \) is the \( 2 \times 2 \) identity matrix. This implies that \( A \) is an involutory matrix, meaning that when squared, it returns the identity matrix.

### Step 2: Writing the Matrix
We can represent the matrix \( A \) as:
\[ A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \]
where \( a, b, c, d \) are real numbers.

### Step 3: Finding \( A^2 \)
Calculating \( A^2 \):
\[ A^2 = A \cdot A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} a^2 + bc & ab + bd \\ ac + dc & bc + d^2 \end{pmatrix} \]
Setting this equal to the identity matrix \( I \):
\[ \begin{pmatrix} a^2 + bc & ab + bd \\ ac + dc & bc + d^2 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

### Step 4: Forming Equations
From the equality of matrices, we get the following equations:
1. \( a^2 + bc = 1 \) (Equation 1)
2. \( ab + bd = 0 \) (Equation 2)
3. \( ac + dc = 0 \) (Equation 3)
4. \( bc + d^2 = 1 \) (Equation 4)

### Step 5: Analyzing the Determinant
The determinant of matrix \( A \) is given by:
\[ \text{det}(A) = ad - bc \]
We need to show that if \( A \neq I \) and \( A \neq -I \), then \( \text{det}(A) = -1 \).

### Step 6: Using the Properties of \( A \)
From Equation 1 and Equation 4, we can substitute \( bc \) from one equation into the other:
- From Equation 1: \( bc = 1 - a^2 \)
- From Equation 4: \( bc = 1 - d^2 \)

Setting these equal gives:
\[ 1 - a^2 = 1 - d^2 \implies a^2 = d^2 \]
This implies \( a = d \) or \( a = -d \).

### Step 7: Exploring Cases
1. **Case 1**: If \( a = d \), then from Equation 1:
\[ a^2 + bc = 1 \implies a^2 + bc = 1 \]
And from Equation 2:
\[ ab + bd = 0 \implies a(b + b) = 0 \implies a = 0 \text{ or } b = 0 \]
If \( b = 0 \), then \( A \) is diagonal, leading to \( A = \begin{pmatrix} a & 0 \\ 0 & a \end{pmatrix} \) which contradicts \( A \neq I \) and \( A \neq -I \).

2. **Case 2**: If \( a = -d \), then substituting gives:
\[ a^2 + bc = 1 \text{ and } bc + a^2 = 1 \]
This leads to \( \text{det}(A) = -1 \).

### Step 8: Conclusion for Statement 1
Thus, we conclude that if \( A \neq I \) and \( A \neq -I \), then \( \text{det}(A) = -1 \) holds true.

### Step 9: Analyzing the Trace
The trace \( \text{tr}(A) = a + d \). Since \( a = -d \), we have:
\[ \text{tr}(A) = a + (-a) = 0 \]
Thus, \( \text{tr}(A) \neq 0 \) is false.

### Final Conclusion
- **Statement 1**: True
- **Statement 2**: False