Q. 28

# 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

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