Question

Let the homogeneous system of linear equations $px + y + z = 0, x + qy + z = 0,$ and $x + y + rz = 0,$ where $p ,q ,r \neq 1$, have a non-zero solution, then the value of $\frac{1}{1-p}+\frac{1}{1-q}+\frac{1}{1-r}$ is

• A. -1
• B. 1
• C. 2
• D. 0

Answer 1

Answer: (B)

Given equations are $px + y + z = 0, x + qy + z = 0, x + y + rz = 0$
Since the system have a non-zero solution, then
$\begin{vmatrix} p & 1 & 1 \\[0.3em] 1& q & 1 \\[0.3em] 1 & 1 &r \end{vmatrix}$ =0
Applying $C_2 \rightarrow C_2 -C_1$
and $C_3 \rightarrow C_3-C_2$
$\Rightarrow \begin{vmatrix} p & 1-p & 0 \\[0.3em] 1& q-1 & 1-q \\[0.3em] 1 & 1 &r-1 \end{vmatrix}$ =0
$\Rightarrow (1 - P) (1 - q) (1 - r) \begin{vmatrix} \frac{ p}{1-p} & 1 & 0\\[0.3em] \frac{ 1}{1-q}& -1 & 1 \\[0.3em] \frac{1}{1-r} & 0 &-1 \end{vmatrix}$ =0
$\Rightarrow (1-p)(1-q)(1-r)
\left[\frac{p}{1-p}(1)-1\left(-\frac{1}{1-q}-\frac{1}{1-r}\right)\right]=0$
Since, $p, q, r \neq$ 1
$=\frac{p}{1-p}+\frac{1}{1-q}+\frac{1}{1-r}=0$
$\Rightarrow \frac{1}{1-p}-1+\frac{1}{1-q}+\frac{1}{1-r}=0$
$\Rightarrow \frac{1}{1-p}+\frac{1}{1-q}+\frac{1}{1-r}=1$