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Given the sys...

Question

Given the system of equation $p x + y + z = 1$ , $x + p y + z = p$ , $x + y + p z = p^{2}$ , then for what value of $p$ does this system have no solution -

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Solution

The correct option is A $- 2$
$p x + y + z = 1$
$x + p y + z = p$
$x + y + p z = p^{2}$
you can use Row Echelon from
But here a trick is easy to see
put $p = - 2$
$- 2 x + y + z = 1$
$x - 2 y + z = - 2$
$x + y - 2 z = 4$ Add
$\Rightarrow 0 = 3$ Not possible
$∴$ For $P = - 2 :$ No solution

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