Find the value of k for which each of the following systems of linear equations has an infinite number of solutions:

(k – 3)x + 3y = k,

kx + ky = 12.

Given: (k – 3)x + 3y = k – eq 1

kx + ky = 12 – eq 2

Here,

a₁ = (k - 3), b₁ = 3, c₁ = - k

a₂ = k , b₂ = k, c₂ = - 12

Given that system of equations has infinitely many solution

∴ = =

= =

Here,

=

3×( - 12) = - k×(k)

- 36 = - k²

K² = 36

k = √36

k = ±6

k = 6 and k = - 6 – eq 3

Also,

=

K(k - 3) = 3k

K² - 3k = 3k

K² - 6k = 0

K(k - 6) = 0

K = 0 and k = 6 – eq 4

From – eq 3 and – eq 4

k = 6

∴ k = 6