Find the maximum value of the determinant of an arbitrary `3xx3` matrix A, each of whose entries `a_(ij) in {-1,1}`.

Answer

To find the maximum value of the determinant of a `3x3` matrix \( A \) where each entry \( a_{ij} \) can take values from \{-1, 1\}, we can follow these steps:

### Step 1: Define the Matrix
Let the matrix \( A \) be represented as:
\[ A = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix} \]
where \( a_{ij} \in \{-1, 1\} \).

### Step 2: Calculate the Determinant
The determinant of a `3x3` matrix is calculated using the formula:
\[ \text{det}(A) = a_{11}(a_{22}a_{33} - a_{23}a_{32}) - a_{12}(a_{21}a_{33} - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31}) \]

### Step 3: Substitute Values
To find the maximum value, we can try different combinations of \( a_{ij} \) values. We will consider the cases where the entries are either 1 or -1.

### Step 4: Test Different Combinations
Let's try the following combinations of the matrix entries:

1. \( A = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{pmatrix} \)
\[ \text{det}(A) = 0 \]

2. \( A = \begin{pmatrix} 1 & 1 & 1 \\ 1 & -1 & -1 \\ 1 & -1 & 1 \end{pmatrix} \)
\[ \text{det}(A) = 1(1 \cdot 1 - (-1)(-1)) - 1(1 \cdot 1 - (-1)(1)) + 1(1 \cdot (-1) - (-1)(1)) = 1(1 - 1) - 1(1 + 1) + 1(-1 + 1) = 0 - 2 + 0 = -2 \]

3. \( A = \begin{pmatrix} 1 & 1 & 1 \\ -1 & -1 & 1 \\ -1 & 1 & -1 \end{pmatrix} \)
\[ \text{det}(A) = 1((-1)(-1) - (1)(1)) - 1((-1)(-1) - (1)(-1)) + 1((-1)(1) - (-1)(-1)) = 1(1 - 1) - 1(1 + 1) + 1(-1 - 1) = 0 - 2 - 2 = -4 \]

4. \( A = \begin{pmatrix} 1 & 1 & -1 \\ 1 & -1 & 1 \\ -1 & 1 & 1 \end{pmatrix} \)
\[ \text{det}(A) = 1((-1)(1) - (1)(1)) - 1(1(1) - (-1)(-1)) + (-1)(1(1) - (1)(-1)) = 1(-1 - 1) - 1(1 - 1) - 1(1 + 1) = -2 - 0 - 2 = -4 \]

5. \( A = \begin{pmatrix} 1 & 1 & 1 \\ 1 & -1 & -1 \\ -1 & -1 & 1 \end{pmatrix} \)
\[ \text{det}(A) = 1((-1)(1) - (-1)(-1)) - 1(1(1) - (-1)(-1)) + 1(1(-1) - (-1)(1)) = 1(-1 - 1) - 1(1 - 1) + 1(-1 + 1) = -2 - 0 + 0 = -2 \]

After testing various combinations, we find that the maximum determinant occurs with the following matrix:
\[ A = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 1 & -1 \\ 1 & -1 & 1 \end{pmatrix} \]
Calculating the determinant gives:
\[ \text{det}(A) = 4 \]

### Conclusion
The maximum value of the determinant of a `3x3` matrix \( A \) with entries from \{-1, 1\} is \( 4 \).