Question

Find the value(s) of k for which the pair of linear equations $kx + y = {k^2}$ and
$x + ky = 1$ have infinitely many solutions.

Answer 1

Hint: One should have knowledge about the relation between the coefficients of the two equations when they have infinitely many solutions. If the equations ${a_1}x + {b_1}y + {c_1} = 0$ and ${a_2}x + {b_2}y + {c_2} = 0$ have infinitely many solutions then-
$\dfrac{{\mathrm a}_1}{{\mathrm a}_2}=\dfrac{{\mathrm b}_1}{{\mathrm b}_2}=\dfrac{{\mathrm c}_1}{{\mathrm c}_2}$

Complete step by step answer:
It is given that-
$kx + y = {k^2}$
$kx + y - {k^2} = 0$ ….(1)
x + ky = 1
x + ky - 1 = 0 ….(2)
It is given that equations (1) and (2) have infinitely many solutions. Hence, the ratios of their coefficients are in proportion. Hence-
$\dfrac{\mathrm k}1=\dfrac1{\mathrm k}=\dfrac{-\mathrm k^2}{-1}\\\mathrm{So},\;\\\dfrac{\mathrm k}1=\dfrac1{\mathrm k}\;....\left(3\right)\\\dfrac1{\mathrm k}=\dfrac{\mathrm k^2}1\;....\left(4\right)$
On solving equation (3)-
$\mathrm k=\dfrac1{\mathrm k}\\\mathrm k^2=1\\\mathrm k=\pm1\;\\$

On solving equation (4)-
$\dfrac1{\mathrm k}=\mathrm k^2\\\mathrm k^3=1\\\mathrm k=1$
In these two solutions, k = 1 satisfies both equations but k = -1 satisfies only one equation.

Therefore, k = 1 is the required answer.

Note: The most common mistake in this question is that students write k = -1 in the answer as well, which is wrong. k = -1 satisfies only one equation, but we need those values which satisfy both.