Question

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

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

• A. k = 6
• B. k = -6
• C. k = -3
• D. k = 3

Answer 1

The correct option is B k = -6

Equations are written as

kx + 3y – 3 = 0 .....(1)

12x + ky – 6 = 0 ....... (2)

$a_1=k,b_1=3,c_1=-3$

$a_2=12,b_2=k,c_2=-6$

for no solution we must have:

$\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$

Now,

$\frac{k}{12}=\frac{3}{k}\ne \frac{-3}{-6}$

$\frac{k}{12}=\frac{3}{k}$
and
$\frac{3}{k}\ne \frac{1}{2}$

$k^2=36,k\ne 6$

Hence, k = -6

Hence, the given system of equations is has no solution if k =-6.