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

kx + 3y = k - 3,12x + ky = k.

Given: kx + 3y = k - 3 – eq 1

12x + ky = k – eq 2

Here,

a₁ = k, b₁ = 3, c₁ = - (k - 3)

a₂ = 12, b₂ = k, c₂ = - k

Here,

Given that system of equations has no solution

∴ = ≠

= ≠

Here,

=

k×k = 3×12

k² = √36

K = ±6 eq 3

Also,

≠

3× - k - (k - 3)× k

- 3k ≠ - k² + 3k

K² - 3k - 3k≠ 0

K²6k≠ 0

K(k - 6) ≠0

K ≠ 0 and k ≠ 6 eq 4

From eq 3 and eq 4 we can conclude

K = 6

∴ k = - 6