$K_{1} < K_{2}$ be the values of K for which the planes $K x + 4 y + z = 0, 4 x + K y + 2 z = 0 a n d 2 x + 2 y + z = 0$ intersects in a straight line. Let $P = [\begin{matrix} K_{2} & K_{1} \\ K_{1} & K_{2} \end{matrix}],$ then $P^{- 1} =$

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$\frac{1}{7} [\begin{matrix} 4 & - 3 \\ - 3 & 4 \end{matrix}]$

$\frac{1}{5} [\begin{matrix} 3 & - 2 \\ 2 & 3 \end{matrix}]$

$\frac{1}{12} [\begin{matrix} 4 & - 2 \\ - 2 & 4 \end{matrix}]$

$\frac{1}{5} [\begin{matrix} 3 & - 2 \\ - 2 & 3 \end{matrix}]$

answer is B.

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Detailed Solution

For three planes to intersect along a line $|\begin{matrix} K & 4 & 1 \\ 4 & K & 2 \\ 2 & 2 & 1 \end{matrix}| = 0 \Rightarrow K (K - 4) + 8 - 2 K = 0$

$\Rightarrow k^{2} - 6 k + 8 = 0 \Rightarrow k = 2, 4$

$∴$ Determinant $= |\begin{matrix} 4 & 2 \\ 2 & 4 \end{matrix}| = 12$

$∴$ $P^{- 1} = \frac{1}{12} [\begin{matrix} 4 & - 2 \\ - 2 & 4 \end{matrix}]$

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