<p>Match the Columns: Given determinant conditions, match the relationships.</p><p>(a) \(\begin{vmatrix} 1 & 2a & a \\ 1 & 3b & b \\ 1 & 4c & c \end{vmatrix} = 0\)</p>
Step-by-Step Solution
Key Concept: Determinants equal to zero encode linear relationships between variables; expanding and simplifying reveals the hidden algebraic connections.
<p><strong>Step 1:</strong> Expand the determinant in part (a):</p><p>\(\begin{vmatrix} 1 & 2a & a \\ 1 & 3b & b \\ 1 & 4c & c \end{vmatrix} = -bc - 2a(c-b) + a(4c-3b) = -bc + 2ac - ab = 0\)</p><p><strong>Step 2:</strong> This gives the relation: \(\frac{2}{b} = \frac{1}{a} + \frac{1}{c}\)</p><p><strong>Step 3:</strong> For part (b): \(\begin{vmatrix} 2a & 3a & 1 \\ 3b & 2b & 1 \\ c & c & 1 \end{vmatrix} = 0\) yields \(bc + ac - 5ab = 0\)</p><p><strong>Step 4:</strong> For part (c): \(\begin{vmatrix} a & 2 & 1 \\ b & 3 & 1 \\ c & 4 & 1 \end{vmatrix} = 0\) establishes a collinearity condition.</p><p>∴ Answer is a-r, b-s, c-p, d-q.</p>
Correct Answer: a-r, b-s, c-p, d-q