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Matching the following
Column - I
Column - II
I.
The number of values of k if the system of equations $(k + 1) x + 8y = 4k$ and $kx + (k + 3) y = 3k - 1$ has infinite solutions
P.
$x = 2, y = 2, z = 2$
II.
Solutions of $2x - y + 3z = 8$
$- x + 2y + z = 4$
$3x + y - 4z = 0$
Q.
Inconsistent
III.
The system $2x + 6y = - 11$
$6x + 20 y - 6z = - 3$
$6y - 18z = - 1$
R
1

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$I - R, II - Q, III - P$

$I - P, II - Q, III - R$

$I - Q, II - R, III - P$

$I - R, II - P, III - Q$

answer is C.

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Detailed Solution

Condition: $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$

(1) $\frac{k + 1}{k} = \frac{8}{k + 3} = \frac{4 k}{3 k - 1};$

$case (1) : \frac{k + 1}{k} = \frac{8}{k + 3}$

$(k + 1) (k + 3) - 8 k = 0; k^{2} - 4 k + 3 = 0$

$k = 1, k = 3$

$C a s e (2) : \frac{8 k}{k + 3} = \frac{4 k}{3 k - 1}; 8 k (3 k - 1) = 4 k (k + 3)$

$\Rightarrow 24 k^{2} - 8 k - 4 k^{2} - 12 k = 0; 20 k^{2} - 20 k = 0$

$20 k (k - 1) = 0 \Rightarrow k = 0, k = 1$

$Case (3): \frac{k + 1}{k} = \frac{4 k}{3 k - 1}; 3 k^{2} + 2 k - 1 - 4 k^{2} = 0$

$k^{2} - 2 k + 1 = 0$

${(k - 1)}^{2} = 0 ∴ k = 1$

Hence $k = 1$

$\begin{array}{l} (2) 2 x - y + 3 z = 8 \\ - x + 2 y + z = 4 \\ 3 x + 4 y - 4 z = 0 \end{array}$

Verify $x = 2, y = 2, z = 2$

(3) $A D = [\begin{array}{l} \begin{matrix} 2 & 6 & 0 & - 11 \end{matrix} \\ \begin{matrix} 6 & 20 & - 6 & - 3 \end{matrix} \\ \begin{matrix} 0 & 6 & - 18 & - 1 \end{matrix} \end{array}]$

$\to_{R_{2} - 3 R_{1}}^{R_{2}} [\begin{array}{l} \begin{matrix} 2 & 6 & 0 & - 11 \end{matrix} \\ \begin{matrix} 0 & 2 & - 6 & 30 \end{matrix} \\ \begin{matrix} 0 & 6 & - 18 & - 1 \end{matrix} \end{array}]$

$\to_{R_{3} - 3 R_{2}}^{R_{3}} [\begin{array}{l} \begin{matrix} 2 & 6 & 0 & - 11 \end{matrix} \\ \begin{matrix} 0 & 2 & - 6 & 30 \end{matrix} \\ \begin{matrix} 0 & 0 & 0 & - 91 \end{matrix} \end{array}]$

Rank $AD = 3$

Rank $A = 2$ $2 \neq 3$

$∴$ The system is inconsistent

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