Find the value of k for which each of the following systems of equations has no solution:
kx+3y=3,12x+ky=6.
Equations are written as
kx + 3y – 3 = 0 .....(1)
12x + ky – 6 = 0 ....... (2)
a1 = k, b1 = 3, c1 = -3
a2 = 12, b2 = k, c2 = -6
for no solution we must have:
a1/a2=b1/b2≠c1/c2Now,
k/12=3/k≠−3/6Hence, k = -6
Hence, the given system of equations is has no solution if k =-6.