Let P be a matrix of order 3 x 3 such that all the entries in P are from the set $\{-1, 0, 1\}.$ Then, the maximum possible value of the determinant of P is ____.

Let P be a matrix of order 3 × 3 such that all the entries in P are from the set {-1, 0, 1}. Then, the maximum possible value of the determinant of P is ____. $P = (p_{ij})_{3 \times 3}, \quad p_{ij} \in \{-1, 0, 1\}$ $\text{Find } \max (\det P)$ $\det P = \vec{r_1}\cdot(\vec{r_2}\times\vec{r_3})$ $|\det P| \le \|\vec{r_1}\|\,\|\vec{r_2}\|\,\|\vec{r_3}\| \le (\sqrt{3})^3 = 3\sqrt{3} \approx 5.196$ $\Rightarrow |\det P| \le 5 \quad (\det P \text{ is an integer})$ $\text{If some entry is } 0: \ \|\vec{r_i}\| \le \sqrt{2}$ $|\det P| \le \sqrt{2}\cdot 3 = 3\sqrt{2} \approx 4.24 \Rightarrow |\det P| \le 4$ $\text{If no entry is } 0: P \text{ has all entries } \pm 1$ $\det(\pm 1 \text{ matrix of order } 3) \text{ is divisible by } 4$ $\Rightarrow |\det P| \le 4$ $\text{So in every case } |\det P| \le 4$ $P = \begin{pmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{pmatrix}$ $\det P = 1(0) - 1(-2) + 1(2) = 4$ $\therefore \max (\det P) = 4$