Let `A=[a_("ij")]` be a matrix of order 2 where `a_("ij") in {-1, 0, 1}` and adj. `A=-A`. If det. `(A)=-1`, then the number of such matrices is ______ .

Explanation

To solve the problem, we need to find the number of 2x2 matrices \( A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) where \( a, b, c, d \in \{-1, 0, 1\} \), given the conditions that \( \text{adj}(A) = -A \) and \( \det(A) = -1 \).

### Step-by-step Solution:

1. **Understanding the adjugate and determinant relationship**:
The adjugate of a 2x2 matrix \( A \) is given by:
\[ \text{adj}(A) = \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \]
The condition \( \text{adj}(A) = -A \) implies:
\[ \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} = \begin{pmatrix} -a & -b \\ -c & -d \end{pmatrix} \]
This leads to the equations:
- \( d = -a \)
- \( -b = -b \) (which is always true)
- \( -c = a \)
- \( a = -d \)

From \( d = -a \) and \( a = -d \), we can conclude that \( d = -a \).

2. **Finding the determinant**:
The determinant of matrix \( A \) is given by:
\[ \det(A) = ad - bc \]
Substituting \( d = -a \):
\[ \det(A) = a(-a) - bc = -a^2 - bc \]
We know \( \det(A) = -1 \), so:
\[ -a^2 - bc = -1 \implies a^2 + bc = 1 \]

3. **Analyzing possible values for \( a \)**:
Since \( a \) can take values from \{-1, 0, 1\}, we analyze each case:
- **Case 1**: \( a = 1 \)
\[ 1^2 + bc = 1 \implies 1 + bc = 1 \implies bc = 0 \]
Possible pairs for \( (b, c) \) are \( (0, 0), (1, 0), (0, -1), (-1, 0) \). Thus, \( d = -1 \).
- Matrices:
- \( \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \)
- \( \begin{pmatrix} 1 & 1 \\ 0 & -1 \end{pmatrix} \)
- \( \begin{pmatrix} 1 & -1 \\ 0 & -1 \end{pmatrix} \)
- \( \begin{pmatrix} 1 & 0 \\ -1 & -1 \end{pmatrix} \)
- \( \begin{pmatrix} 1 & 0 \\ 1 & -1 \end{pmatrix} \)

- **Case 2**: \( a = 0 \)
\[ 0^2 + bc = 1 \implies bc = 1 \]
Possible pairs for \( (b, c) \) are \( (1, 1), (-1, -1) \). Thus, \( d = 0 \).
- Matrices:
- \( \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \)
- \( \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix} \)

- **Case 3**: \( a = -1 \)
\[ (-1)^2 + bc = 1 \implies 1 + bc = 1 \implies bc = 0 \]
Possible pairs for \( (b, c) \) are \( (0, 0), (1, 0), (0, -1), (-1, 0) \). Thus, \( d = 1 \).
- Matrices:
- \( \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} \)
- \( \begin{pmatrix} -1 & 1 \\ 0 & 1 \end{pmatrix} \)
- \( \begin{pmatrix} -1 & -1 \\ 0 & 1 \end{pmatrix} \)
- \( \begin{pmatrix} -1 & 0 \\ -1 & 1 \end{pmatrix} \)
- \( \begin{pmatrix} -1 & 0 \\ 1 & 1 \end{pmatrix} \)

4. **Counting the matrices**:
- From Case 1: 5 matrices
- From Case 2: 2 matrices
- From Case 3: 5 matrices

Total number of matrices = \( 5 + 2 + 5 = 12 \).

### Final Answer:
The number of such matrices is **12**.