Find the value of $k$ , so that the following system of equations has no solution :
$3 x + y = 1; (2 k - 1) x + (k - 1) y = (2 k - 1)$ .

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Solution

Given that,
$\begin{matrix} 3 x + y = 1 (2 k - 1) x + (k - 1) y = (2 k - 1) (A : B) = [\begin{matrix} 31 : 1 (2 k - 1) (k - 1) : (2 k - 1) \end{matrix}] = [\begin{matrix} 1 \frac{1}{3} : \frac{1}{3} (2 k - 1) (k - 1) : (2 k - 1) \end{matrix}] = ⎡ ⎢ ⎣ \begin{matrix} 1 \frac{1}{3} : \frac{1}{3} 0 (- \frac{k}{3} + \frac{2}{3}) : \frac{- 2}{3} (2 k - 1) \end{matrix} ⎤ ⎥ ⎦ \end{matrix}$
It $k (A) \neq k (A : B)$ then it has no solution .
So,
$\begin{matrix} \frac{- k}{3} + \frac{2}{3} = 0 k = 2 \end{matrix}$

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