question_answer

The system of equations

\[px+(p+1)y+(p-1)z=0,\]

\[(p+1)x+py+(p+2)z=0,\]

\[(p-1)x+(p+2)y+pz=0\]

has a non-trivial solution for

A)  exactly two real values of p                        

B)  no real value of p

C)  exactly one real value of p       

D)  infinitely many real values of p

Correct Answer: C

Solution :

     [c]   The system of equations has non-trivial solution. \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\Delta =0\] Applying \[{{R}_{2}}\to {{R}_{2}}-{{R}_{1}}\] and \[{{R}_{3}}\to {{R}_{3}}-{{R}_{1}},\] we get Applying \[{{R}_{3}}\to {{R}_{3}}+{{R}_{2}},\]we get \[\Rightarrow \,\,\,\,\,\,\,\,\,\,4(-p-(p+1))=0\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,2p+1=0\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,p=-1/2\] Therefore, exactly one real value of p exists.