Q27 of 44 Page 3

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

kx + 3y = k - 3,12x + ky = k. (CBSE 2008)

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

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