Science:Math Exam Resources/Courses/MATH152/April 2016/Question B 03 (a)/Solution 1

We plug $A=\left[\begin{array}{ccc}2& 2& -1\\ 0& 0& 1\\ 0& -4& 5\end{array}\right]$ and $\lambda =1$ into $(A-\lambda I)v=0$ to get

$(A-\lambda I)v=\left[\begin{array}{ccc}2-1& 2& -1\\ 0& 0-1& 1\\ 0& -4& 5-1\end{array}\right]v=0.$

Let now $v=\left[\begin{array}{c}a\\ b\\ c\end{array}\right]$. Then, we have

$\left[\begin{array}{ccc}1& 2& -1\\ 0& -1& 1\\ 0& -4& 4\end{array}\right]\left[\begin{array}{c}a\\ b\\ c\end{array}\right]=\left[\begin{array}{c}a+2b-c\\ -b+c\\ -4b+4c\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right].$

This means that $b=c$ and $a=-b.$ So, we choose $b=1$ to get

$v=\left[\begin{array}{c}-1\\ 1\\ 1\end{array}\right].$