Question

Suppose $p, q, r \neq 0$ and system of equation $(p + a)x + by + cz = 0, ax + (q + b)y + cz = 0, ax + by + (r + c)z = 0$, 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

Answer 1

Answer: (A)

-1

Answer 2

The given system of linear equations is:

1. $(p + a)x + by + cz = 0$

2. $ax + (q + b)y + cz = 0$

3. $ax + by + (r + c)z = 0$

This is a homogeneous system of linear equations. For a homogeneous system to have a non-trivial solution, the determinant of the coefficient matrix must be zero. The coefficient matrix is $M = \begin{pmatrix} p+a & b & c \\ a & q+b & c \\ a & b & r+c \end{pmatrix}$. The condition for a non-trivial solution is $\det(M) = 0$.

We calculate the determinant: $\det(M) = (p+a) \begin{vmatrix} q+b & c \\ b & r+c \end{vmatrix} - b \begin{vmatrix} a & c \\ a & r+c \end{vmatrix} + c \begin{vmatrix} a & q+b \\ a & b \end{vmatrix}$

$= (p+a)[(q+b)(r+c) - bc] - b[a(r+c) - ac] + c[ab - a(q+b)]$

$= (p+a)[qr + qc + br + bc - bc] - b[ar + ac - ac] + c[ab - aq - ab]$

$= (p+a)[qr + qc + br] - b[ar] + c[-aq]$

$= p(qr + qc + br) + a(qr + qc + br) - abr - acq$

$= pqr + pqc + pbr + aqr + aqc + abr - abr - acq$

$= pqr + pqc + pbr + aqr$

The condition for a non-trivial solution is $pqr + pqc + pbr + aqr = 0$.

Since $p, q, r \neq 0$, we can divide the entire equation by $pqr$:

$\frac{pqr}{pqr} + \frac{pqc}{pqr} + \frac{pbr}{pqr} + \frac{aqr}{pqr} = 0$

$1 + \frac{c}{r} + \frac{b}{q} + \frac{a}{p} = 0$

Rearranging the terms, we get: $\frac{a}{p} + \frac{b}{q} + \frac{c}{r} = -1$