Question

For what value of k doesthe linear equation Kx+y=k^2 ,x+ky =1 have no solution

Answer 1

Given:

kx +y = k²

x +ky = 1

• Condition : For No solution

To find:

• The value of k

Solution:

Comparing both the equations with ax +by +c = 0

From kx +y = k²

==> kx +y -k² = 0

We get

$a_1 = k \\ </p><p>b_1 = 1 \\ </p><p>c_1 = -(k {}^{2} )$

From x +ky = 1

==>x +ky -1 = 0

We get

$a_1 = 1 \\ </p><p>b_1 = k \\ </p><p>c_1 = -1$

Condition for no solution

$\boxed{\frac{ a_1 }{a _2} = \frac{b _1}{b _ 2} \ne \frac{c _1}{c_2} }$

Therefore, we get

$\frac{ a_1 }{a _2} = \frac{b _1}{b _ 2}$

$\implies \frac{k}{1} = \frac{1}{k} \\ \\ \implies \: {{k}^{2} } = 1 \\ \\ \implies \red{ \boxed{k = 1}}$

Hence, k = 1 is the Required Value