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,
$\left|\begin{array}{ccc}t& (t+1)& (t-1)\\ (t+1)& t& (t+2)\\ (t-1)& (t+2)& 1\end{array}\right|=0$
On applying $C_2\to C_2-C_1$
and $C_3\to C_3-C_2,$
we get $\Rightarrow$ $\left|\begin{array}{ccc}t& 1& -2\\ t+1& -1& 2\\ t-1& 3& -2\end{array}\right|=0$
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.