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The system of equations

$\mathrm{-kx+3y-14z=25}\\$

$\mathrm{-15x+4y-kz=3}\\$

$\mathrm{-4x+y+3z=4}\\$

is consistent for all $\mathrm{k}$ in the set
Option: 1
$\mathrm{\mathbb{R}}$

Option: 2
$\mathrm{\mathbb{R}-\left \{ -11,13 \right \}}$

Option: 3
$\mathrm{\mathbb{R}-\left \{ 13 \right \}}$

Option: 4
$\mathrm{\mathbb{R}-\left \{ -11,11 \right \}}$

Answers (1)

best_answer

For cosistance solutions $\mathrm{\Delta \neq0}$

$\mathrm{\Rightarrow\left|\begin{array}{ccc} -k & 3 & -14 \\ -15 & 4 & -k \\ -4 & 1 & 3 \end{array}\right| \neq 0}$

$\mathrm{\Rightarrow-k(12+k)-3(-45-4 k)-14(-15+16) \neq 0}\\$

$\mathrm{\Rightarrow-12 k-k^{2}+135+12 k-14 \neq 0 }\\$

$\mathrm{\Rightarrow k^{2} \neq 121}\\$

$\mathrm{k \neq 11,-11}$

Hence the correct answer is option 4

Posted by

Kuldeep Maurya

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