For what value of k will the following

The given system of equations is

3x + y = 1

(2k − 1)x + (k − 1)y = 2k + 1

Here, a1 = 3, b1 = 1, c1 = 1

a2 = 2k − 1, b2 = k − 1, c2 = 2k + 1

The given system of linear equations has no solution.

$\therefore \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$

$\Rightarrow \frac{3}{2 k-1}=\frac{1}{k-1} \neq \frac{1}{2 k+1}$

$\Rightarrow \frac{3}{2 k-1}=\frac{1}{k-1}$ and $\frac{1}{k-1} \neq \frac{1}{2 k+1}$

Now,

$\frac{3}{2 k-1}=\frac{1}{k-1}$

$\Rightarrow 3 k-3=2 k-1$

$\Rightarrow 3 k-2 k=-1+3$

$\Rightarrow k=2$

When k = 2,

$\frac{1}{k-1}=\frac{1}{2-1}=1$ and $\frac{1}{2 k+1}=\frac{1}{2 \times 2+1}=\frac{1}{5}$

Thus, for $k=2, \frac{1}{k-1} \neq \frac{1}{2 k+1}$

Hence, the given system of linear equations will have no solution when k = 2.