For the system of equations `x+y+z=4, y+2z=5 and x+y+pz=q` to have no solution (A) `p=1 and q=4` (B) `p=1 and q!=4` (C) `p!=1 and q=4` (D) `p!=1 and q!=4`

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 "____"

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.

For what values of p and q are the system of equations : 2x+5y+pz=q;x+2y+3z=14x+y+z=6 are consistent?

Let the homogeneous system of linear equations px+y+z=0,x+qy+z=0, and x+y+rz=0 ,where p,q,r!=1, have anon-zero solution,then the value of (1)/(1-p)+(1)/(1-q)+(1)/(1-r) is

SECTION MULTIPLE CORRECT CHOICE TYPE Example 41 Let a, b, c , p, q be five different non-zero real numbers and x, y, z be three numbers satisfying the system of equations a a-p aq then (a) x + y + z = a + b + c-p-q abc (b) x = pq 0 in (c) y = (a-p)(h-p)(c-p) p(p-q) (a-q)(b-q)(c-q) aq-p) (d) z = b.