If (h,k,1) is the solution of 2x+3y+z=9,x+2y+3z=6 and 3x+y+2z=8 then h+k+l=

2x + 3y-5z = 7, x + y + z = 6,3x-4y + 2z = 1, then x =

For what value of k, the system of linear equations x-3y+kz=0, 2x+y+3z=-1, 3x-2y+2z=1 has a unique solution? a) K=1 b)k!=1 c) K!=-1 d) K=0

3x+4y+7z=4, 2x+y+ 3z= 3, x + 2y +3z =8.

Let the system of linear equations 2x+3y-z=0,2x+ky-3z=0 and 2x-y+z=0 have non trivial solution then (x)/(y)+(y)/(z)+(z)/(x)+k will be

2x + 3y + z = 11, x + 2y + z = 8, x-y + z = 2

If the system of equations, 2x + 3y-z = 0, x + ky -2z = 0 " and " 2x-y+z = 0 has a non-trivial solution (x, y, z), then (x)/(y) + (y)/(z) + (z)/(x) + k is equal to

Solve 2x-y+3z=9,x+y+z=6 and x-y+z=2 by matrix method.

Add 2x - 3y + z , 5y - x + 7z and 3x - y - 6z.

The solution of the system of equations is x-y+2z=1,2y-3z=1 and 3x-2y+4y=2 is