What should be the value(s) of $k$, for which the pair of linear equations $kx+y=k^2$ and $x+ky=0$ have infinitely many solutions

Question

What should be the value(s) of $k$, for which the pair of linear equations $kx+y=k^2$ and $x+ky=0$ have infinitely many solutions.

For, the pair of linear equations, to have infinitely many solutions, we have the following condition,

$$(k/1)=(1/k)=((k^2)/0)$$

Does a solution exist?

Answer 1

Put $x=-ky$ in the first equation to get $-k^{2}y+y=k^{2}$. This (and the given system) has a unique solution if $1-k^{2} \neq 0$ and the system has no solution when $1-k^{2}=0$ or $k=\pm 1$. So it never has infinitely many solutions.

Answer 2

$$kx+y=k^2$$ $$x+ky=0$$ $\implies$ $$k(-ky)+y=k^2\implies y(1-k^2)=k^2\implies y=\frac{k^2}{1-k^2}$$ Then from this $$x=-ky=-k\left(\frac{k^2}{1-k^2}\right)\implies x=-\frac{k^3}{1-k^2}$$ If $k=\pm1$ then there is no solution. Otherwise, there is a unique solution for each $k\neq\pm1.$