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Question

Find the value of k for which the given system of equations has no solution.$$kx + 3y = k - 3; 12x + ky = k$$


A
k=5
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B
k=2
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C
k=3
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D
None of these
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Solution

The correct option is C None of these
A system of equations, $$a_1x+b_1y+c_1=0$$ and $$a_2x+b_2y+c_2=0$$ has no solution for the following condition:
$$\displaystyle \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$$

The equations in the given question are:
$$kx+3y-k+3=0$$ and $$a_{1}=k,b_{1}=3,c_{1}=3$$
$$12x+ky-k=0$$ and $$a_{2}=12,b_{2}=k,c_{2}=-k$$
Substituting the above values in the condition mentioned above for no solution:
$$\dfrac{a_{1}}{b_{1}}=\dfrac{a_{2}}{b_{2}}$$
$$\dfrac{k}{12}=\dfrac{3}{k}$$

$$k^{2}=36$$
$$\Rightarrow k=6$$

Mathematics

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