For what values of `p`, the equations :`x+y+z=1 , x+2y+4z=p , x+4y+10z=p^2` have a solution ? Solve them completely in each case.

The values of x, y, z for the equations x-y+z=1, 2x-y=1, 3x+3y-4z=2 are

A solution set of the equations x+2y+z=1 , x+3y+4z=k , x+5y+10z=k^(2) is

the sum of values of p for which the equations x+y+z=1x+2y +4z=p and x+4y +10z = p^(2) have a solution is "____"

The values of x, y, z for the equations 5x-y+4z=5, 2x+3y+5z=2, 5x-2y+6z=1 are

The value of alpha for which the system of equations x+y+z=1, x+2y+4z=alpha, x+4y+10z=alpha^2 has no solution is (A) -1 (B) 0 (C) 3 (D) 2

The system of equation -2x + y + z = 1, x – 2y + z = –2, x + y + lambdaz = 4 will have no solution if

Find inverse of matrix,solve the equation x-y+z=4,2x+y-3z=0x+y+z=2

Solve x+y+z=1 , 2x+2y+2z=2 , 3x+3y+3z=4