<p>If the system of linear equations has a non-zero solution:<br/><span>\[\begin{align} x + 2ay + az &= 0 \\ x + 3by + bz &= 0 \\ x + 4cy + cz &= 0 \end{align}\]</span> then <span>\(a, b, c\)</span>:</p>
Step-by-Step Solution
Key Concept: For a homogeneous system of linear equations to have a non-zero solution, the coefficient matrix must be singular (determinant = 0). We need to find the condition on a, b, c that makes the determinant zero.
<p><strong>Step 1:</strong> Write the coefficient matrix and set its determinant to zero for non-zero solutions to exist.</p><p>The coefficient matrix is:</p><p>$$A = \begin{pmatrix} 1 & 2a & a \\ 1 & 3b & b \\ 1 & 4c & c \end{pmatrix}$$</p><p><strong>Step 2:</strong> Calculate the determinant by expanding along the first column.</p><p>$$\det(A) = 1\begin{vmatrix} 3b & b \\ 4c & c \end{vmatrix} - 1\begin{vmatrix} 2a & a \\ 4c & c \end{vmatrix} + 1\begin{vmatrix} 2a & a \\ 3b & b \end{vmatrix}$$</p><p><strong>Step 3:</strong> Evaluate the 2×2 determinants.</p><p>$$= (3bc - 4bc) - (2ac - 4ac) + (2ab - 3ab)$$</p><p>$$= -bc - (-2ac) + (-ab)$$</p><p>$$= -bc + 2ac - ab$$</p><p><strong>Step 4:</strong> For non-zero solution, set determinant equal to zero.</p><p>$$-bc + 2ac - ab = 0$$</p><p>$$2ac = ab + bc$$</p><p><strong>Step 5:</strong> Divide throughout by $abc$ (assuming $a, b, c \neq 0$).</p><p>$$\frac{2ac}{abc} = \frac{ab}{abc} + \frac{bc}{abc}$$</p><p>$$\frac{2}{b} = \frac{1}{c} + \frac{1}{a}$$</p><p><strong>Step 6:</strong> Rearrange to identify the progression.</p><p>$$\frac{2}{b} = \frac{1}{a} + \frac{1}{c}$$</p><p>This is the condition for $a, b, c$ to be in Harmonic Progression (H.P.), since the reciprocals $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in A.P.</p><p><strong>∴ Answer:</strong> C</p>
Correct Answer: C