<p>The least value of the product <em>xyz</em> for which the determinant \(\begin{vmatrix} x & 1 & 1 \\ 1 & y & 1 \\ 1 & 1 & z \end{vmatrix}\) is non-negative, is</p>
Step-by-Step Solution
Key Concept: Expand the determinant and use AM-GM inequality on the constraint xyz ≥ constraint_value, recognizing that equality in AM-GM gives the minimum value when the determinant equals zero (boundary condition).
<p><strong>Step 1:</strong> Expand the determinant along the first row:</p><p>Δ = x(yz - 1) - 1(z - 1) + 1(1 - y)</p><p>Δ = xyz - x - z + 1 + 1 - y</p><p>Δ = xyz - x - y - z + 2</p><p><strong>Step 2:</strong> For the determinant to be non-negative: xyz - x - y - z + 2 ≥ 0</p><p>Therefore: xyz ≥ x + y + z - 2</p><p><strong>Step 3:</strong> For the minimum value of xyz, we set the determinant equal to zero (boundary condition):</p><p>xyz = x + y + z - 2</p><p><strong>Step 4:</strong> By symmetry, the minimum occurs when x = y = z. Let x = y = z = a:</p><p>a³ = 3a - 2</p><p>a³ - 3a + 2 = 0</p><p>(a - 1)²(a + 2) = 0</p><p>So a = 1 or a = -2</p><p><strong>Step 5:</strong> Check values:</p><p>• When a = 1: xyz = 1³ = 1</p><p>• When a = -2: xyz = (-2)³ = -8</p><p><strong>Step 6:</strong> Verify using AM-GM on the constraint xyz ≥ x + y + z - 2. At x = y = z = 1: LHS = 1, RHS = 1 ✓. For x = y = z = -2: LHS = -8, RHS = -8 ✓</p><p>Since we need the least value and -8 < 1, the minimum value is <strong>-8</strong>.</p><p>∴ Answer: C (or -8, depending on options)</p>
Correct Answer: C