Question

For what value of $k$ will the system of linear equations has infinite number of solution.
$kx+4y=k-4$
$16x+ky=k$

Answer 1

For $\infty$lymany solutions

$kx+4y=k-4$

$16x+ky=k$

$\left|\begin{array}{cc}k& 4\\ 16& k\end{array}\right|=0$

$\Rightarrow k^2-4\times 16=0$

$k^2=64$

$\Rightarrow k=8,-8$

$\therefore$ for $k=8$

$\Rightarrow$ yes $\infty$ly many solution but for $k=-8$, it is absurd.

$\therefore k=81$ correct answer.