\[px+(p+1)y+(p-1)z=0,\] |
\[(p+1)x+py+(p+2)z=0,\] |
\[(p-1)x+(p+2)y+pz=0\] |
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.You need to login to perform this action.
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