Question

Let the Eigen vector of the matrix $\left[\begin{array}{cc}1& 2\\ 0& 2\end{array}\right]$ be 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 the value of (a + b)?

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

Answer 1

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

A = $\left[\begin{array}{cc}1& 2\\ 0& 2\end{array}\right]$ $\because$ A is UTM so ${\lambda }_1$ = 1 , ${\lambda }_2$ = 2 Now let Eigen vectors are $X_1$ = $\left[\begin{array}{c}1\\ a\end{array}\right]$ & $X_2$ = $$\left[\begin{array}{c}1\\ b\end{array}\right]$$

According to definition, we have,

$A{X}_1={\lambda }_1{X}_1$ & $A{X}_2={\lambda }_2{X}_2$

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

$\left[\begin{array}{c}(1+2a)\\ 2a\end{array}\right]=\left[\begin{array}{c}1\\ a\end{array}\right]$ & $\left[\begin{array}{c}(1+2b)\\ 2b\end{array}\right]=\left[\begin{array}{c}2\\ 2b\end{array}\right]$

$\Rightarrow$ a = 0 & b = $\frac{1}{2}$