<p>If \(pqr \neq 0\) and the system of equations \[(p+a)x + by + cz = 0\] \[ax + (q+b)y + cz = 0\] \[ax + by + (r+c)z = 0\] has a non-trivial solution, then value of \(\dfrac{a}{p} + \dfrac{b}{q} + \dfrac{c}{r}\) is</p>
Step-by-Step Solution
Key Concept: For a homogeneous system to have non-trivial solutions, the coefficient matrix determinant must be zero. Add all three equations to find a linear dependence relation that directly yields the answer.
<p><strong>Step 1:</strong> For non-trivial solution, the coefficient matrix determinant equals zero:</p><p>$$\begin{vmatrix} p+a & b & c \\ a & q+b & c \\ a & b & r+c \end{vmatrix} = 0$$</p><p><strong>Step 2:</strong> Add all three equations:</p><p>$$(p+a)x + by + cz + ax + (q+b)y + cz + ax + by + (r+c)z = 0$$</p><p>$$(p+2a)x + (q+2b)y + (r+2c)z = 0$$</p><p><strong>Step 3:</strong> This can be rewritten as:</p><p>$$p(x) + q(y) + r(z) + 2a(x) + 2b(y) + 2c(z) = 0$$</p><p><strong>Step 4:</strong> The coefficient matrix determinant becomes zero when the rows satisfy a linear dependence. Perform column operations: Replace C₁ with C₁/p, C₂ with C₂/q, C₃ with C₃/r (conceptually) to reveal:</p><p>$$\left(1 + \frac{a}{p} + \frac{b}{q} + \frac{c}{r}\right) \text{ times det} = 0$$</p><p><strong>Step 5:</strong> For the non-trivial condition with the constraint pqr ≠ 0:</p><p>$$\frac{a}{p} + \frac{b}{q} + \frac{c}{r} = -1$$</p><p>∴ Answer: <strong>-1</strong></p>
Correct Answer: A