Question

For the matrix $P=\left[\begin{array}{ccc}3& -2& 2\\ 0& -2& 1\\ 0& 0& 1\end{array}\right]$ one of the eige nvalues is equal to $-2$. Which of the followign is an eigen vector?

• A. $\left[\begin{array}{c}3\\ -2\\ 1\end{array}\right]$
• B. $\left[\begin{array}{c}3\\ 2\\ -1\end{array}\right]$
• C. $\left[\begin{array}{c}1\\ -2\\ 3\end{array}\right]$
• D. $\left[\begin{array}{c}2\\ 5\\ 0\end{array}\right]$

Answer 1

The correct option is D $\left[\begin{array}{c}2\\ 5\\ 0\end{array}\right]$

If ${\lambda }_i$ is the eigen value of matrix and $X_i$ the corresponding eigen vector then,
$A\left[X_i\right]={\lambda }_i\left[X_i\right]$
$\Rightarrow (A-\lambda I)X=0\lambda (A+2I)X=0$
$\left[\begin{array}{ccc}5& -2& 2\\ 0& 0& 1\\ 0& 0& 3\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$
$R_1\to R_3-3R_2$
$\left[\begin{array}{ccc}5& -2& 2\\ 0& 0& 1\\ 0& 0& 0\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$
$\Rightarrow 5x-2y+2z=0$ & $z=0$
so, $x=\frac{2}{5}y$ & $z=0$
Let $y=k,x=\frac{2}{5}k,z=0$
So, $X=\left[\begin{array}{c}\frac{2}{5}k\\ k\\ 0\end{array}\right]=\left[\begin{array}{c}2\\ 5\\ 0\end{array}\right]$
$\therefore$ EIgen vector corresponding to eigen value $-2$ is $[250{]}^T$