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Q.

The value of $k \in R$ , for which the following system of linear equations $3 x - y + 4 z = 3, x + 2 y - 3 z = - 2, 6 x + 5 y + k z = - 3$ has infinitely many solutions, is

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a

$5$

b

$3$

c

$- 5$

d

$- 3$

answer is A.

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Detailed Solution

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$\begin{array}{l} 3 x - y + 4 z = 3 \to 1 \\ x + 2 y - 3 z = - 2 \to 2 \\ 6 x + 5 y + k z = - 3 \to 3 \\ 1 \times 2 + 2 \Rightarrow 7 x + 5 z = 4 \\ 1 \times 5 + 3 \Rightarrow 21 x + (20 + k) z = 12 \\ \Rightarrow \frac{21}{7} = \frac{20 + k}{5} = \frac{12}{4} \\ \Rightarrow 20 + k = 15 \Rightarrow k = - 5 \end{array}$

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