Let the homogeneous system of linear equations px+y+z=0, x+qy+z=0 and x+y+rz=0, where p,q,r≠1 , have a non-zero solution, then the value of 11−p+11−q+11−r is

4 as non zero solution ⇒A=0p111q111r=0Put p=2, q=321113111r=023r−1−1r−1+11−3=06r−2−r+1−2=05r=3r=3/5∴11−p+11−q+11−r=11−2+11−3+11−3/5=−1+1−2+52=-1+2=1

Let the homogeneous system of linear equations px+y+z=0, x+qy+z=0 and x+y+rz=0, where p,q,r≠1 , have a non-zero solution, then the value of 11−p+11−q+11−r is

4 as non zero solution ⇒A=0p111q111r=0Put p=2, q=321113111r=023r−1−1r−1+11−3=06r−2−r+1−2=05r=3r=3/5∴11−p+11−q+11−r=11−2+11−3+11−3/5=−1+1−2+52=-1+2=1

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Let the homogeneous system of linear equations $p x + y + z = 0, x + q y + z = 0$ and $x + y + r z = 0,$ 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

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4 as non zero solution $\Rightarrow |A| = 0$

$|\begin{matrix} p & 1 & 1 \\ 1 & q & 1 \\ 1 & 1 & r \end{matrix}| = 0$

Put p=2, q=3

$|\begin{matrix} 2 & 1 & 1 \\ 1 & 3 & 1 \\ 1 & 1 & r \end{matrix}| = 0$

$2 (3 r - 1) - 1 (r - 1) + 1 (1 - 3) = 0$

$6 r - 2 - r + 1 - 2 = 0$

$5 r = 3$

$r = 3 / 5$

$∴ \frac{1}{1 - p} + \frac{1}{1 - q} + \frac{1}{1 - r} = \frac{1}{1 - 2} + \frac{1}{1 - 3} + \frac{1}{1 - 3 / 5}$

$= - 1 + \frac{1}{- 2} + \frac{5}{2}$

=-1+2

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Let the homogeneous system of linear equations px+y+z=0, x+qy+z=0 and x+y+rz=0, where p,q,r≠1 , have a non-zero solution, then the value of 11−p+11−q+11−r is

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Let the homogeneous system of linear equations $p x + y + z = 0, x + q y + z = 0$ and $x + y + r z = 0,$ 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