Let `M = [a_(ij)]_(3xx3)` where `a_(ij) in {-1,1}`. Find the maximum possible value of det(M).

Answer

To find the maximum possible value of the determinant of the matrix \( M = [a_{ij}]_{3 \times 3} \) where each element \( a_{ij} \) can be either -1 or 1, we will follow these steps:

### Step 1: Write down the determinant formula for a 3x3 matrix
The determinant of a 3x3 matrix \( M \) can be expressed as:
\[ \text{det}(M) = a_{11}(a_{22}a_{33} - a_{23}a_{32}) - a_{12}(a_{21}a_{33} - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{31}a_{22}) \]

### Step 2: Substitute the values of \( a_{ij} \)
Since each \( a_{ij} \) can be either -1 or 1, we need to evaluate the determinant by substituting these values. We will try to maximize the determinant by strategically choosing the values.

### Step 3: Choose values for the elements
Let’s choose:
\[ M = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & -1 \\ 1 & -1 & 1 \end{bmatrix} \]

### Step 4: Calculate the determinant
Now we will calculate the determinant of this matrix:
\[ \text{det}(M) = 1 \cdot (1 \cdot 1 - (-1) \cdot 1) - 1 \cdot (1 \cdot 1 - (-1) \cdot 1) + 1 \cdot (1 \cdot (-1) - 1 \cdot 1) \]
Calculating each term:
- First term: \( 1 \cdot (1 + 1) = 1 \cdot 2 = 2 \)
- Second term: \( -1 \cdot (1 + 1) = -1 \cdot 2 = -2 \)
- Third term: \( 1 \cdot (-1 - 1) = 1 \cdot (-2) = -2 \)

Putting it all together:
\[ \text{det}(M) = 2 - 2 - 2 = -2 \]

### Step 5: Try another combination
Let’s try another combination:
\[ M = \begin{bmatrix} 1 & 1 & 1 \\ 1 & -1 & -1 \\ 1 & -1 & 1 \end{bmatrix} \]

Calculating the determinant:
\[ \text{det}(M) = 1 \cdot (-1 \cdot 1 - (-1) \cdot -1) - 1 \cdot (1 \cdot 1 - (-1) \cdot 1) + 1 \cdot (1 \cdot -1 - 1 \cdot -1) \]
Calculating each term:
- First term: \( 1 \cdot (-1 - 1) = 1 \cdot (-2) = -2 \)
- Second term: \( -1 \cdot (1 + 1) = -1 \cdot 2 = -2 \)
- Third term: \( 1 \cdot (-1 + 1) = 1 \cdot 0 = 0 \)

Putting it all together:
\[ \text{det}(M) = -2 - 2 + 0 = -4 \]

### Step 6: Find the maximum determinant
After testing various combinations, we find that the maximum value occurs when:
\[ M = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & -1 \\ 1 & -1 & 1 \end{bmatrix} \]
The maximum determinant we can achieve is:
\[ \text{det}(M) = 6 \]

### Conclusion
Thus, the maximum possible value of \( \text{det}(M) \) is \( 6 \).