Q.

If the system of equations x+y+z=5, x+2y+3z=9, x+3y+αz=β has infinitely many solutions, then β-2α equals:

1 21  
2 3  
3 18  
4 5  

Ans.

(2)

x+y+z=5,    x+2y+3z=9,    x+3y+αz=β

D=|11112313α|=0

⇒(2α-9)+(3-α)+(3-2)=0

⇒α=5

Now,  D3=|11512913β|=0

⇒2β-27+9-β+5(3-2)=0

⇒β=13

⇒at α=5, β=13