Question

Suppose \( p, q, r \neq 0 \) 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 the value of \[ \frac{a}{p} + \frac{b}{q} + \frac{c}{r} \] is:

• A. \( -1 \)
• B. \( 0 \)
• C. \( 1 \)
• D. \( 2 \)

Hint:

For a homogeneous system \( AX = 0 \) to have a non-trivial solution, the determinant \( |A| \) must be zero.

Answer 1

The Correct Option is A

Step 1: {Construct the determinant of the coefficient matrix} 
\[ |A| = \begin{vmatrix} p+a & b & c \\ a & q+b & c \\ a & b & r+c \end{vmatrix}. \] Since the system has a non-trivial solution, the determinant must be zero: \[ |A| = 0. \] Step 2: {Expand determinant using row operations} 
\[ (p + a)[(q + b)(r + c) - bc] - b[a(r + c) - ca] + c[ab - a(q + b)] = 0. \] \[ (p + a)(qr + qc + br) - b(ar) + c[-aq] = 0. \] Step 3: {Divide by \( pqr \)} 
\[ \frac{pqr}{pqr} + \frac{pqc}{pqr} + \frac{prb}{pqr} + \frac{qra}{pqr} = 0. \] \[ 1 + \frac{c}{r} + \frac{b}{q} + \frac{a}{p} = 0. \] Step 4: {Conclusion} 
a/p + b/q + c/r = -1.