Maximum Determinant of Three by Three Matrix with Restricted Entries

To find the maximum possible value of the determinant of a $3 \times 3$ matrix $P$ with entries taken from the set $\{-1, 0, 1\}$, we analyze the upper bound using Hadamard's Inequality and matrix properties.

Step 1: Applying Hadamard's Inequality

Let $R_1, R_2, R_3$ be the row vectors of $P$. Hadamard's inequality states that: $|\det(P)| \le \|R_1\| \|R_2\| \|R_3\|$

Since each entry $a_{ij} \in \{-1, 0, 1\}$, we have $a_{ij}^2 \le 1$. Therefore, the squared norm of each row vector satisfies: $\|R_i\|^2 = \sum_{j=1}^3 a_{ij}^2 \le 1 + 1 + 1 = 3 \implies \|R_i\| \le \sqrt{3}$

Applying this to Hadamard's inequality gives: $|\det(P)| \le \sqrt{3} \times \sqrt{3} \times \sqrt{3} = 3\sqrt{3} \approx 5.196$

Since $P$ is an integer matrix, $\det(P)$ must be an integer. Hence, $|\det(P)| \le 5$.

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Step 2: Refining the Upper Bound

Case 1: All rows have the maximum norm $\|R_i\|^2 = 3$

In this case, all entries of $P$ belong to $\{-1, 1\}$.

The inner product of any two distinct rows $R_i \cdot R_j = c_{ij} = \sum_{k=1}^3 a_{ik} a_{jk}$. Since $a_{ik} a_{jk} \in \{-1, 1\}$, the sum of three such terms must be an odd integer, so $c_{ij} \in \{-3, -1, 1, 3\}$.

• If $c_{ij} = \pm 3$, then $R_i = \pm R_j$, which implies $\det(P) = 0$.
• For a non-zero determinant, we must have $c_{ij} \in \{-1, 1\}$ for all $i \neq j$.

Now consider the product $P P^T$:

$$P P^T = \begin{pmatrix} 3 & c_{12} & c_{13} \\ c_{12} & 3 & c_{23} \\ c_{13} & c_{23} & 3 \end{pmatrix}$$

The determinant of $P P^T$ is: $\det(P P^T) = (\det P)^2 = 3(9 - c_{23}^2) - c_{12}(3 c_{12} - c_{13} c_{23}) + c_{13}(c_{12} c_{23} - 3 c_{13})$

Using $c_{12}^2 = c_{13}^2 = c_{23}^2 = 1$: $(\det P)^2 = 3(8) - (3 - c_{12} c_{13} c_{23}) + (c_{12} c_{13} c_{23} - 3) = 18 + 2 c_{12} c_{13} c_{23}$

Since $c_{12} c_{13} c_{23} \in \{-1, 1\}$:

1. If $c_{12} c_{13} c_{23} = 1$, then $(\det P)^2 = 20$. However, $\sqrt{20}$ is not an integer, so no such matrix exists.
2. If $c_{12} c_{13} c_{23} = -1$, then $(\det P)^2 = 16 \implies |\det P| = 4$.

Case 2: At least one row has $\|R_i\|^2 \le 2$

If at least one row contains a zero entry, its length is at most $\sqrt{2}$. By Hadamard's Inequality: $|\det(P)| \le \sqrt{3} \times \sqrt{3} \times \sqrt{2} = \sqrt{18} \approx 4.24$

Since $\det(P)$ must be an integer, $|\det(P)| \le 4$.

Thus, in all cases, the upper bound for the determinant is $4$.

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Step 3: Constructing an Example with $\det(P) = 4$

Consider the matrix:

$$P = \begin{pmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{pmatrix}$$

Expanding along the first row: $\det(P) = 1((-1)(-1) - (1)(1)) - 1((1)(-1) - (1)(1)) + 1((1)(1) - (-1)(1))$ $\det(P) = 1(1 - 1) - 1(-1 - 1) + 1(1 + 1) = 0 + 2 + 2 = 4$

Conclusion

The maximum possible value of the determinant of $P$ is 4.