Find the value of k for which each of the following systems of equations has no solution:

kx + 3y = 3,12x + ky = 6.

Given: kx + 3y = 3 – eq 1

12x + ky = 6 – eq 2

Here,

a₁ = k, b₁ = 3, c₁ = - 3

a₂ = 12, b₂ = k, c₂ = - 6

Here,

Given that system of equations has no solution

∴ = ≠

= ≠

Here,

=

k×k = 3×12

k² = √36

K = ±6 eq 3

Also,

≠

3× - 6 ≠ - 3× k

- 18 ≠ - 3k

3k≠18

K≠6 eq 4

From eq 3 and eq 4 we can conclude

K = 6

∴ k = - 6