Question

Let the homogeneous system of linear equations $px+y+z=0,x+qy+z=0,$ and $x+y+rz=0,$ where $p,q,r\ne 1$, 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
$\left|\begin{array}{ccc}p& 1& 1\\ 1& q& 1\\ 1& 1& r\end{array}\right|$ =0
Applying $C_2\to C_2-C_1$
and $C_3\to C_3-C_2$
$\Rightarrow \left|\begin{array}{ccc}p& 1-p& 0\\ 1& q-1& 1-q\\ 1& 1& r-1\end{array}\right|$ =0
$\Rightarrow (1-P)(1-q)(1-r)\left|\begin{array}{ccc}\frac{p}{1-p}& 1& 0\\ \frac{1}{1-q}& -1& 1\\ \frac{1}{1-r}& 0& -1\end{array}\right|$ =0
$\Rightarrow (1-p)(1-q)(1-r)<br/>\left[\frac{p}{1-p}(1)-1\left(-\frac{1}{1-q}-\frac{1}{1-r}\right)\right]=0$
Since, $p,q,r\ne$ 1
$=\frac{p}{1-p}+\frac{1}{1-q}+\frac{1}{1-r}=0$
$\Rightarrow \frac{1}{1-p}-1+\frac{1}{1-q}+\frac{1}{1-r}=0$
$\Rightarrow \frac{1}{1-p}+\frac{1}{1-q}+\frac{1}{1-r}=1$