Question

The system of equations
$kx + (k + 1)y + (k - 1) z = 0$
$(k + 1)x + ky + (k + 2) z = 0$
$(k - 1)x + (k + 2)y + kz = 0$
has a non-trivial solution for

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

Answer 1

Answer: (C)

To have a non-trivial solution, we must have
$\begin{vmatrix}k & k+1 & k-1 \\ k+1 & k & k+2 \\ k-1 & k+2 & k\end{vmatrix}=0 \Rightarrow 2 k+1=0 \Rightarrow k=\frac{-1}{2}$