Question

The eigen vectors of the matrix $\left[\begin{array}{cc}1& 2\\ 0& 2\end{array}\right]$ are written in the form $\left[\begin{array}{c}1\\ a\end{array}\right]$ and $\left[\begin{array}{c}1\\ b\end{array}\right]$
What is a + b ?

• A. 0
• B. $\frac{1}{2}$
• C. 1
• D. 2

Answer 1

The correct option is B $\frac{1}{2}$

Given matrix,
$A=\left[\begin{array}{cc}1& 2\\ 0& 2\end{array}\right]$
The characteristic equation is $|A-\lambda I|=0$
$\left|\begin{array}{cc}1-\lambda & 2\\ 0& 2-\lambda \end{array}\right|=0$
$(\lambda -1)(\lambda -2)=0$
$\therefore \lambda =1,2$
Thus the eigen values are 1, 2.
It x, y be the components of an eigen vector corresponding to the eigen value 1, then $(A-\lambda I)X=O$
$\left[\begin{array}{cc}1-\lambda & 2\\ 0& 2-\lambda \end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=0$
corresponding to $\lambda$ =1, we have
$\left[\begin{array}{cc}0& 2\\ 0& 1\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]$
$y=0$
Let $x=k,k\in R$
So, $X_1=\left[\begin{array}{c}k\\ 0\end{array}\right]\sim \left[\begin{array}{c}1\\ 0\end{array}\right]=\left[\begin{array}{c}1\\ a\end{array}\right]$ (given)
$\therefore \left[\begin{array}{c}1\\ 0\end{array}\right]$ eigen vector for eigen value $\lambda$ = 1
Corresponding to $\lambda =2,wehave\left[\begin{array}{cc}-1& 2\\ 0& 0\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]$
$-x+2y=0$
$x=2y$
Now, let y = k, x = 2k then

$X_2=\left[\begin{array}{c}2k\\ k\end{array}\right]\sim \left[\begin{array}{c}2\\ 1\end{array}\right]\sim \left[\begin{array}{c}1\\ \frac{1}{2}\end{array}\right]=\left[\begin{array}{c}1\\ b\end{array}\right]$ (given)
So, $a+b=0+1/2=\frac{1}{2}$
Alternative solution :
$|A-\lambda I|=\left|\begin{array}{cc}1-\lambda & 2\\ 0& 2-\lambda \end{array}\right|=0$
$(1-\lambda )(2-\lambda )=0$
$\Rightarrow \lambda =1,2$
${\lambda }_1=1,X_1=\left[\begin{array}{c}1\\ a\end{array}\right]$
${\lambda }_2=1,X_2=\left[\begin{array}{c}1\\ b\end{array}\right]$
$AX=\lambda X$
$\Rightarrow \left[\begin{array}{cc}1& 2\\ 0& 2\end{array}\right]\left[\begin{array}{c}1\\ a\end{array}\right]=1\left[\begin{array}{c}1\\ a\end{array}\right]$
$\Rightarrow 1+2a=1$
$\Rightarrow a=0$

$\left[\begin{array}{cc}1& 2\\ 0& 2\end{array}\right]\left[\begin{array}{c}1\\ b\end{array}\right]=2\left[\begin{array}{c}1\\ b\end{array}\right]$
$\Rightarrow 1+2b=2$
$\Rightarrow b=\frac{1}{2}$
$a+b=0+\frac{1}{2}=\frac{1}{2}$