Question

The system of homogeneous equations $ tx+(t+1)y+(t-1)z=0, $ $ (t+1)x+ty+(t+2)z=0 $ $ (t-1)x+(t+2)y+tz=0, $ has non-trival solutions for

• A. exactly three real values oft
• B. exactly two real values oft
• C. exactly one real value oft
• D. infinite number of values oft

Answer 1

Answer: (C)

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.