JEE Advanced 2018 · Paper 2, Mathematics question 8

Upper bound (Hadamard). $|\det P|$ is at most the product of the row lengths.

• If any entry is 0, that row has length at most $\sqrt2$, so $|\det P|\le\sqrt2\cdot\sqrt3\cdot\sqrt3\approx4.24$. Hence $\det P\le4$.
• If all entries are $\pm1$, subtract the first row from the others. Their entries become $0$ or $\pm2$, so the determinant is a multiple of 4. Also $|\det P|\le(\sqrt3)^3\approx5.2$, so $\det P\le4$.

Attained.
$$\det\begin{pmatrix}1&1&1\\1&-1&1\\1&1&-1\end{pmatrix}=1(1-1)-1(-1-1)+1(1+1)=4.$$

Answer: 4