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-trivial solutions for

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

Answer 1

The correct option is C exactly one real value of $t$

For the system of equations to have non-trivial solution,

$\Delta =\left[\begin{array}{ccc}t& t+1& t-1\\ t+1& t& t+2\\ t-1& t+2& t\end{array}\right]=0$

Applying $C_2\to C_2-C_1,C_3\to C_3-C_1$,

we get

$\Delta =\left[\begin{array}{ccc}t& 1& -1\\ t+1& -1& 1\\ t-1& 3& 1\end{array}\right]=\left[\begin{array}{ccc}2t+1& 0& 0\\ t+1& -1& 1\\ t-1& 3& 1\end{array}\right]$ (using $R_1\to R_1+R_2$)

$=(2t+1)(-4)=0\Rightarrow t=\frac{-1}{2}$

Hence, option C.