Determine the value of k so that the following linear equations have no solution:
(3k + 1) x + 3y − 2 = 0
(k2 + 1) x + (k − 2) y − 5 = 0
The given system of equations is
(3k + 1) x + 3y − 2 = 0
(k2 + 1) x + (k − 2) y − 5 = 0
Here, a1 = 3k + 1, b1 = 3, c1 = −2
a2 = k2 + 1, b2 = k − 2, c2 = −5
The given system of equations has no solution.
$\therefore \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$
$\Rightarrow \frac{3 k+1}{k^{2}+1}=\frac{3}{k-2} \neq \frac{-2}{-5}$
$\Rightarrow \frac{3 k+1}{k^{2}+1}=\frac{3}{k-2}$ and $\frac{3}{k-2} \neq \frac{2}{5}$
Now,
$\frac{3 k+1}{k^{2}+1}=\frac{3}{k-2}$
$\Rightarrow(3 k+1)(k-2)=3\left(k^{2}+1\right)$
$\Rightarrow 3 k^{2}-5 k-2=3 k^{2}+3$
$\Rightarrow-5 k=5$
$\Rightarrow k=-1$
When k = −1,
$\frac{3}{k-2}=\frac{3}{-1-2}=\frac{3}{-3}=-1$
Thus, for $k=-1, \frac{3}{k-2} \neq \frac{2}{5}$
Hence, the given system of equations has no solution when k = −1.
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