Question

Find the values of k for which the pair of linear equations Kx+y=k^{​​​​​​2} and x+ky=1 have infinitely many solutions.

Answer 1

Given $kx+y=k^2$ --- (1)
$x+ky=1$ ---- (2)
Both the equations are in the form of:

$a_{1}x+b_{1}y+c_1=0and{a}_{2}x+b_{2}y+c_2=0$.
In the equation (1), we have
$a_1$ = k, $b_1$ = 1,$c_1$ = $-k^2$.
In the equation (2),
$a_2$ = 1,$b_2$ = k,$c_2$ = -1.
Given that the equation has infinitely many solutions.

So, $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$
$k=\frac{1}{k}=\frac{-k^2}{-1}$

$k^2=1$ or $k=k^2$

$\Rightarrow k=1.$