Discussion
If \(\begin{vmatrix} a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c \end{vmatrix} = 2\), where a, b and c are positive integers, then \(a+b+c\) is equal to
Answer: A
\(a + b + c\) is 6.
Expand the determinant along the first row:
\[ \begin{aligned} \Delta &= a(bc - 1) - (c - 1) + (1 - b) \\ &= abc - a - b - c + 2 \end{aligned} \]
Setting \(\Delta = 2\) gives \(abc = a + b + c\). For positive integers the only solution, up to order, is 1, 2, 3, since \(1 \times 2 \times 3 = 1 + 2 + 3 = 6\). Taking all three equal to 1 gives 1 = 3, and larger values make the product outgrow the sum. So a + b + c = 6.
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