JEE Advanced 2018 — Paper 2 — Question 35

We need the maximum possible value of $\det(P)$ for a $3\times 3$ matrix $P$ with entries from $\{-1,0,1\}$. The determinant of a $3\times 3$ matrix is at most the product of the Euclidean norms of its rows (Hadamard's inequality). For a row with entries in $\{-1,0,1\}$, the maximum squared norm is $1^2+1^2+1^2=3$, so the product of the three row norms is at most $\sqrt{3}\cdot\sqrt{3}\cdot\sqrt{3}=3\sqrt{3}\approx 5.196$. Since the determinant must be an integer (entries are integers), the maximum possible integer value is at most $5$. Check if $5$ is attainable: a $3\times 3$ matrix with entries $\pm1$ has determinant a multiple of $4$ (by row operations), so $5$ is impossible. The next candidate is $4$.

A known example achieving $\det = 4$ is $\begin{pmatrix}1&1&1\\1&-1&1\\1&1&-1\end{pmatrix}$. Thus the maximum possible determinant is $4$.