MDoubt No 000018 - GATE

GATE [EE, 2007, 2 marks]
Question : The linear operation $L(x)$ is defined by the cross product $L(x)=b\times X$, where $b=\begin{bmatrix} 0 & 1 & 0 \end{bmatrix}^T$ and $X=\begin{bmatrix} x_1 & x_2 & x_3 \end{bmatrix}^T$ are three dimensional vectors. The $3\times 3$ matrix $M$ of this operation sstisfies $$L(x)=M\begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}$$ Then the eigen values of $M$ are

(a) $0,+1,-1$
(b) $1,-1,1$
(c) $i,-i,1$
(d) $i,-i,0$

Solution : As $L(x)=M\begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}$ and $L(x)=b\times X$
$\therefore M\begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}=b\times X \quad\quad\ldots(1)$
Since $M$ is $3\times 3$ matrix,
let $M=\begin{bmatrix} a_1 & a_2 & a_3\\ a_4 & a_5 & a_6\\ a_7 & a_8 & a_9 \end{bmatrix}$ and
$b\times X=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 0 & 1 & 0\\ x_1 & x_2 & x_3 \end{vmatrix}$
$=x_3\hat{i}+0\hat{j}-x_1\hat{k}$
$\begin{bmatrix} x_3 & 0 & -x_1\\ \end{bmatrix}^T$
Equation $(1)$ becomes
$\begin{bmatrix} a_1 & a_2 & a_3\\ a_4 & a_5 & a_6\\ a_7 & a_8 & a_9 \end{bmatrix}\begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}=\begin{bmatrix} x_3\\ 0\\ -x_1 \end{bmatrix}$
On comparing
$\begin{bmatrix} 0 & 0 & 1\\ 0 & 0 & 0\\ -1 & 0 & 0 \end{bmatrix}\begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}=\begin{bmatrix} x_3\\ 0\\ -x_1 \end{bmatrix}$
So, $M=\begin{bmatrix} 0 & 0 & 1\\ 0 & 0 & 0\\ -1 & 0 & 0 \end{bmatrix}$
For eigen vales of $M$
$|M-\lambda I|=0$
$\begin{vmatrix} -\lambda & 0 & 1\\ 0 & -\lambda & 0\\ -1 & 0 & -\lambda \end{vmatrix}=0$
$-\lambda(\lambda^2-0)+1(0-\lambda)=0$
$\lambda^3+\lambda=0$
$\lambda(\lambda^2+1)=0$
$\lambda=0,\pm i$
Eigen values are $i, -i, 0$
CORRECT ANSWER : D