Question

For what value of k the following pair of linear equations Kx- 3y = 1 -12x+ ky = 2 Has 1) a unique solution 2) infinitely many solutions 3) no solution

Answer 1

kx-3y=1
-12x+ky=2
1)unique soln.
   condition
a1/a2≠b1/b2
∵k/-12≠3/k
  k*k≠-12*3
  k²≠-36
  k≠+-6

2)infinitely many soln.
  condition
a1/a2=b1/b2=c1/c2

k/-12=3/k=1/2
now,k*k=-12*3
       k²=-36
       k²=+-6←
Now, 3*2=k
         6=k←
from these two, +6 is common
so k=6

3)No soln.
condition
a1/a2=b1/b2≠c1/c2

k/-12=3/k≠1/2
Now, k*k=-12*3
         k²=-36
         k=+-6←

Now, 3*2≠k
         6≠k