A) exactly three real values oft
B) exactly two real values oft
C) exactly one real value oft
D) infinite number of values oft
Correct Answer: C
Solution :
Given system of equation is \[tx+(t+1)y+(t-1)z=0\] \[(t+1)x+ty+(t+z)z=0\] and \[(t-1)x+(t+2)y+tz=0\] For non-trivial solution, \[\left| \begin{matrix} t & (t+1) & (t-1) \\ (t+1) & t & (t+2) \\ (t-1) & (t+2) & 1 \\ \end{matrix} \right|=0\] On applying \[{{C}_{2}}\to {{C}_{2}}-{{C}_{1}}\] and \[{{C}_{3}}\to {{C}_{3}}-{{C}_{2}},\] we get \[\Rightarrow \] \[\left| \begin{matrix} t & 1 & -2 \\ t+1 & -1 & 2 \\ t-1 & 3 & -2 \\ \end{matrix} \right|=0\] Expanding along \[{{C}_{1}},\] we get \[t(2-6)-(t+1)(-2+6)++(t-1)(2-2)=0\] \[\Rightarrow \] \[-4t-4(t+1)+(t-1)\,(0)=0\] \[\Rightarrow \] \[-8t-4=0\] \[\Rightarrow \] \[t=-\frac{1}{2}\] Hence, exactly one real value of t exist.
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