<p>If the value of the determinant \(\begin{vmatrix} a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c \end{vmatrix}\) is positive, then \((a, b, c > 0)\)</p>
Step-by-Step Solution
Key Concept: Expand the determinant and recognize that for positive a, b, c, the expression abc + 2 - (ab + bc + ca) can be rewritten as (a-1)(b-1)(c-1) + (a-1)(b-1) + (b-1)(c-1) + (c-1)(a-1), which reveals that the determinant's positivity depends on the product (a-1)(b-1)(c-1) and symmetric terms.
<p><strong>Step 1:</strong> Expand the determinant along the first row:</p><p>Δ = a(bc - 1) - 1(c - 1) + 1(1 - b)</p><p>Δ = abc - a - c + 1 + 1 - b</p><p>Δ = abc - a - b - c + 2</p><p><strong>Step 2:</strong> Rewrite by regrouping:</p><p>Δ = abc - ab - ac - bc + a + b + c - 1 + ab + bc + ca - a - b - c + 3</p><p>Δ = (a - 1)(b - 1)(c - 1) + [ab + bc + ca - a - b - c + 1] + 1</p><p>Or more directly: Δ = (a-1)(b-1)(c-1) + (a-1)(b-1) + (b-1)(c-1) + (c-1)(a-1) + 1</p><p><strong>Step 3:</strong> For Δ > 0 with a, b, c > 0:</p><p>If a, b, c > 1: all terms are positive, so Δ > 0 ✓</p><p>If a, b, c < 1: (a-1)(b-1)(c-1) > 0 and other terms positive, so Δ > 0 ✓</p><p>If values are mixed (some > 1, some < 1): determinant can be negative</p><p><strong>∴ Answer: B (Either all three are greater than 1, or all three are less than 1)</strong></p>
Correct Answer: B