Question

show that the only real number b for which the system x+2y+z=bx;3x+y+2z=by;2x+3y+z=bz has non-zero solution is 6 and solve them, when b=6​

Answer 1

Answer:

last eqn should be 2x+3y+2z= bz

add all the three eqns and take 6 common

6(x+y+z) = b(x+y+z)

6(x+y+z)- b(x+y+z)=0

(x+y +z)(6-b)=0

x+y+z=0

6-b=0

b=6