Let A be a `2 xx 2` matrix with non-zero entries and let A^2=I, where i is a `2 xx 2` identity matrix, Tr(A) i= sum of diagonal elements of A and `|A|` = determinant of matrix A. Statement 1:Tr(A)=0 Statement 2:`|A|`=1

Options

A. Statement 1 is false, statement 2 is true.
B. Statement 1 is true, statement 2 is true, statement 2 is a correct explanation for statement 1.
C. Statement 1 is true, statement 2 is true, statement 2 is a correct explanation for statement 1.
D. Statement 1 is true, statement 2 is false.

Explanation

To solve the problem, we need to analyze the properties of the matrix \( A \) given that \( A^2 = I \), where \( I \) is the \( 2 \times 2 \) identity matrix.

Let \( A \) be represented as:
\[ A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \]

### Step 1: Calculate \( A^2 \)
We compute \( A^2 \):
\[ A^2 = A \cdot A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \cdot \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} a^2 + bc & ab + bd \\ ac + dc & bc + d^2 \end{pmatrix} \]

### Step 2: Set \( A^2 \) equal to the identity matrix
Since \( A^2 = I \), we have:
\[ \begin{pmatrix} a^2 + bc & ab + bd \\ ac + dc & bc + d^2 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

From this, we can derive the following equations:
1. \( a^2 + bc = 1 \)
2. \( ab + bd = 0 \)
3. \( ac + dc = 0 \)
4. \( bc + d^2 = 1 \)

### Step 3: Analyze the equations
From equations 2 and 3, we can factor them:
- From \( ab + bd = 0 \), we can factor out \( b \): \( b(a + d) = 0 \)
- From \( ac + dc = 0 \), we can factor out \( c \): \( c(a + d) = 0 \)

Since \( A \) has non-zero entries, \( b \neq 0 \) and \( c \neq 0 \). Therefore, we must have:
\[ a + d = 0 \implies d = -a \]

### Step 4: Substitute \( d \) into the equations
Now substituting \( d = -a \) into equations 1 and 4:
1. \( a^2 + bc = 1 \)
2. \( bc + (-a)^2 = 1 \) simplifies to \( bc + a^2 = 1 \)

Both equations are the same, confirming our substitution is consistent.

### Step 5: Calculate the trace and determinant
The trace \( \text{Tr}(A) \) is given by:
\[ \text{Tr}(A) = a + d = a - a = 0 \]

The determinant \( |A| \) is calculated as:
\[ |A| = ad - bc = a(-a) - bc = -a^2 - bc \]
From \( a^2 + bc = 1 \), we can substitute \( bc = 1 - a^2 \):
\[ |A| = -a^2 - (1 - a^2) = -a^2 - 1 + a^2 = -1 \]

### Conclusion
1. **Statement 1**: \( \text{Tr}(A) = 0 \) is **true**.
2. **Statement 2**: \( |A| = 1 \) is **false** (we found \( |A| = -1 \)).