Question

If \( p \neq a \), \( q \neq b \), \( r \neq c \), and the system of equations \[ px + ay + az = 0 \] \[ bx + qy + bz = 0 \] \[ cx + cy + rz = 0 \] has a non-trivial solution, then the value of \[ \frac{p}{p - a} + \frac{q}{q - b} + \frac{r}{r - c} \] is:

• A. 1
• B. 2
• C. \( \frac{1}{2} \)
• D. 0

Hint:

When solving systems of equations with non-trivial solutions, compute the determinant of the coefficient matrix. For the solution to exist, the determinant must be zero, which gives the necessary relationship between the variables.

Answer 1

The Correct Option is B

Consider a system of equations that has a non-trivial solution, implying that the determinant of the system's matrix must be zero. Given the matrix \[ \Delta = \begin{vmatrix} p & a & a \\ b & q & b \\ c & c & r \end{vmatrix} = 0 \] we apply column operations \(C_2 \rightarrow C_2 - C_1\) and \(C_3 \rightarrow C_3 - C_1\) to simplify the determinant: \[ \Delta = \begin{vmatrix} p & a-p & a-p \\ b & q-b & b \\ c & 0 & r-c \end{vmatrix} = 0 \] Expanding along \(C_3\), we calculate: \[ \Delta = (a-p)\begin{vmatrix} b & q-b \\ c & 0 \end{vmatrix} - (r-c)\begin{vmatrix} p & a-p \\ b & q-b \end{vmatrix} = 0 \] \[ = (a-p)(b \cdot 0 - c(q-b)) + (r-c)(p(q-b) - b(a-p)) = 0 \] \[ = -(a-p)c(q-b) + (r-c)(pq - pb - ab + bp) = 0 \] \[ = (a-p)c(q-b) - (r-c)(pq - ab) = 0 \] \[ = (pq - ab)(r-c) - (q-b)(a-p)c = 0 \] Dividing by \((pq-ab)(r-c)\) and \((q-b)\), we simplify: \[ \frac{c}{r-c} + \frac{p-a}{q-b} + \frac{b}{r-c} = 0 \] \[ \frac{p-a}{q-b} - \frac{q-r}{q-b} = 2 \] This results in the relation: \[ \frac{p-r}{q-b} = 2 \] confirming that for a non-trivial solution to exist, specific relations between the parameters must hold. Thus, the correct answer is Option B.