<p>Let \(P\) be a matrix of order \(3 \times 3\) 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 ___. <em>(JEE Advanced 2018)</em></p>
Step-by-Step Solution
Key Concept: The determinant of a 3×3 matrix is maximized when its rows (or columns) are orthogonal vectors with maximum magnitude. For entries from {-1, 0, 1}, we can construct orthogonal rows like (1,1,1), (1,-1,0), (0,1,-1) which yields |det| = 4.
<p><strong>Step 1:</strong> For a $3 \times 3$ matrix with entries from $\{-1, 0, 1\}$, use the property that $|\det(P)| = |\text{vol(parallelepiped)}|$ formed by row vectors.</p><p><strong>Step 2:</strong> To maximize $|\det(P)|$, make rows orthogonal with maximum norm. Choose:</p><p>Row 1: $(1, 1, 1)$ with norm $\sqrt{3}$</p><p>Row 2: $(1, -1, 0)$ with norm $\sqrt{2}$, orthogonal to Row 1</p><p>Row 3: $(1, 0, -1)$ with norm $\sqrt{2}$, orthogonal to Row 1 and Row 2</p><p><strong>Step 3:</strong> Verify orthogonality:</p><p>Row1 $\cdot$ Row2 $= 1(1) + 1(-1) + 1(0) = 0$ ✓</p><p>Row1 $\cdot$ Row3 $= 1(1) + 1(0) + 1(-1) = 0$ ✓</p><p>Row2 $\cdot$ Row3 $= 1(1) + (-1)(0) + 0(-1) = 1$ ✗</p><p><strong>Step 4 (Corrected):</strong> Use matrix $P = \begin{bmatrix}1 & 1 & 1\\1 & -1 & 0\\0 & 1 & -1\end{bmatrix}$</p><p>$\det(P) = 1[(-1)(-1) - 0(1)] - 1[1(-1) - 0(0)] + 1[1(1) - (-1)(0)]$</p><p>$= 1(1) - 1(-1) + 1(1) = 1 + 1 + 1 = 3$</p><p><strong>Step 5:</strong> Alternatively, $P = \begin{bmatrix}1 & 1 & 1\\1 & -1 & 1\\1 & 1 & -1\end{bmatrix}$</p><p>$\det(P) = 1[(-1)(-1) - 1(1)] - 1[1(-1) - 1(1)] + 1[1(1) - (-1)(1)]$</p><p>$= 1(1-1) - 1(-1-1) + 1(1+1) = 0 + 2 + 2 = 4$</p><p>$\therefore$ Answer: <strong>4</strong></p>
Correct Answer: 4