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Q.

The equations $x + y + z = 1, 2 x + y - z = 0, x + y + k z = 2$ has unique solutions then

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a

$k = 0$

b

$k = 1$

c

$k \neq 1$

d

$k \neq 0$

answer is C.

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Detailed Solution

$[A B] = [\begin{matrix} 1 & 1 & 1 \\ 2 & 1 & - 1 \\ 1 & 1 & K \end{matrix} \begin{matrix} 1 \\ 0 \\ 2 \end{matrix}] \begin{matrix} R_{2} - 2 R_{1} \\ R_{3} - R_{1} \end{matrix} ~ [\begin{matrix} 1 & 1 & 1 \\ 0 & - 1 & - 3 \\ 0 & 0 & K - 1 \end{matrix} \begin{matrix} 1 \\ - 2 \\ 1 \end{matrix}]$

Since, $k - 1 \neq 0$

= $k \neq 1$

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The equations x+y+z=1,2x+y−z=0,x+y+kz=2 has unique solutions then