Determine the value of $k$ so that the following linear equations has no solution:
$(3 k + 1) x + 3 y - 2 = 0$ and $(k^{2} + 1) x + (k - 2) y - 5 = 0$

k = - 1

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k = 3

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k = - 7

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k = 8

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Solution

The correct option is A $k = - 1$
Given equations are $(3 k + 1) x + 3 y - 2 = 0$ and $(k^{2} + 1) x + (k - 2) y - 5 = 0$
Here $a_{1} = 3 k + 1, b_{1} = 3$ and $c_{1} = - 2$
$a_{2} = k^{2} + 1, b_{2} = k - 2$ and $c_{2} = - 5$
For no solution, condition is $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$
$\frac{3 k + 1}{k^{2} + 1} = \frac{3}{k - 2}$
$\Rightarrow$ $\frac{3 k + 1}{k^{2} + 1} = \frac{3}{k - 2}$
$\Rightarrow (3 k + 1) (k - 2) = 3 (k^{2} + 1)$
$\Rightarrow$ $3 k^{2} - 5 k - 2 = 3 k^{2} + 3$
$\Rightarrow$ $- 5 k - 2 = 3$
$\Rightarrow - 5 k = 5$
$\Rightarrow k = - 1$
Clearly, $\frac{3}{k - 2} \neq \frac{2}{5}$ for $k = - 1$
Hence, the given system of equations will have no solution for $k = - 1$ .

Suggest Corrections

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