Question

The number of distinct real values of $\lambda$, for which the determinant $\begin{vmatrix}-\lambda^2 & 1 & 1 \\ 1 & -\lambda^2 & 1 \\ 1 & 1 & -\lambda^2\end{vmatrix}$ vanishes, is

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (C)

$R_1 \rightarrow R_1+R_2+R_3$
$\left(2-\lambda^2\right)\begin{vmatrix}1 & 1 & 1 \\ 1 & -\lambda^2 & 1 \\ 1 & 1 & -\lambda^2\end{vmatrix}=0$
$C _1 \rightarrow C _1- C _2$ and $C _2 \rightarrow C _2- C _3$
$\left(2-\lambda^2\right)\begin{vmatrix}0 & 0 & 1 \\ 1+\lambda^2 & -\lambda^2-1 & 1 \\ 0 & 1+\lambda^2 & -\lambda^2\end{vmatrix}=0 \Rightarrow\left(2-\lambda^2\right)\left[1+\lambda^2\right]^2=0$
$\therefore \lambda^2=2 \Rightarrow \lambda= \pm \sqrt{2} \Rightarrow$ two values of $\lambda$