Question

If $A=\left(\begin{array}{c}1\\ 0\\ 2\end{array}\begin{array}{c}0\\ 2\\ 0\end{array}\begin{array}{c}2\\ 1\\ 3\end{array}\right)$ and $A^3-6A^2+7A+k{I}_3=O$ find $k.$

Answer 1

$A=\left(\begin{array}{ccc}1& 0& 2\\ 0& 2& 1\\ 2& 0& 3\end{array}\right)$ , we get
$A^2=\left(\begin{array}{ccc}1& 0& 2\\ 0& 2& 1\\ 2& 0& 3\end{array}\right)\left(\begin{array}{ccc}1& 0& 2\\ 0& 2& 1\\ 2& 0& 3\end{array}\right)=\left(\begin{array}{ccc}5& 0& 8\\ 2& 4& 5\\ 8& 0& 13\end{array}\right)$
$A^3=\left(\begin{array}{ccc}5& 0& 8\\ 2& 4& 5\\ 8& 0& 13\end{array}\right)\left(\begin{array}{ccc}1& 0& 2\\ 0& 2& 1\\ 2& 0& 3\end{array}\right)=\left(\begin{array}{ccc}21& 0& 34\\ 12& 8& 23\\ 34& 0& 55\end{array}\right)$
$A^3-6A^2+7A+k{I}_3=0$, we get $\left(\begin{array}{ccc}21& 0& 34\\ 12& 8& 23\\ 34& 0& 55\end{array}\right)-6\left(\begin{array}{ccc}5& 0& 8\\ 2& 4& 5\\ 8& 0& 13\end{array}\right)+7\left(\begin{array}{ccc}1& 0& 2\\ 0& 2& 1\\ 2& 0& 3\end{array}\right)+k\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)=0$
$\left(\begin{array}{ccc}k-2& 0& 0\\ 0& k-2& 0\\ 0& 0& k-2\end{array}\right)=0$ , which implies $k-2=0$
Therefore the value of $k$ is $2.$