The system of equations `2x-ay = 4`, `5x-ky = 5` is in consistant with no unique solution. If k and a are natural numbers, then the minimum possible value of `(a + k)` is equal to

If the system of equations 2x+ay+6z=8, x+2y+z=5, 2x+ay+3z=4 has a unique solution then 'a' cannot be equal to :

If the system of equations 2x+ay+6z=8, x+2y+z=5, 2x+ay+3z=4 has a unique solution then 'a' cannot be equal to :

The system of equation kx + y + z = 1, x + ky + z = k and x + y + kz = k^(2) has no solution if k equals :

For what value of k does the system of equations " " x + 2y = 3, 5x + ky + 7 = 0 have (i) a unique solution, (ii) no solution ? Also, show that there is no value of k for which the given system equations has infinitely many solutions.

Value of k for which system of equations kx + 2y = 5, 3x + y =1 has unique solution is

The system of equations kx + y + z =1, x + ky + z = k and x + y + zk = k^(2) has no solution if k is equal to :

The value of k for which the system of equations 2x - 3y =1 and kx + 5y = 7 have a unique solution, is

The system of equation kx+y+z=1, x +ky+z=k and x+y+kz=k^(2) has no solution if k equals