<p>If the system of linear equations<br>\(x + 2ay + az = 0\)<br>\(x + 3by + bz = 0\)<br>\(x + 4cy + cz = 0\)<br>has a non-zero solution, then \(a, b, c\)</p>
Step-by-Step Solution
Key Concept: For a homogeneous system to have non-trivial solutions, the coefficient matrix determinant must equal zero. This creates a constraint relationship among the parameters a, b, c that can be extracted by expanding the determinant and factoring.
<p><strong>Step 1:</strong> For non-trivial solution, set det(A) = 0 where coefficient matrix is:</p><p>$$\begin{vmatrix} 1 & 2a & a \\ 1 & 3b & b \\ 1 & 4c & c \end{vmatrix} = 0$$</p><p><strong>Step 2:</strong> Expand along first column:</p><p>$$1(3bc - 4bc) - 1(2ac - 4ac) + 1(2ab - 3ab) = 0$$</p><p>$$-bc + 2ac - ab = 0$$</p><p><strong>Step 3:</strong> Rearrange and factor:</p><p>$$ab + bc = 2ac$$</p><p>Divide by abc (assuming non-zero):</p><p>$$\frac{1}{c} + \frac{1}{a} = \frac{2}{b}$$</p><p><strong>Step 4:</strong> This means <strong>a, b, c are in Harmonic Progression (H.P.)</strong> or equivalently, $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in A.P.</p><p>∴ Answer: C (a, b, c are in H.P.)</p>
Correct Answer: C