Question

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

Answer 1

$A=\left(\begin{array}{ccc}1& 0& 2\\ 0& 2& 1\\ 2& 0& 3\end{array}\right)$

$A^2=A\times A=\left(\begin{array}{ccc}1& 0& 2\\ 0& 2& 1\\ 2& 0& 3\end{array}\right)\times \left(\begin{array}{ccc}1& 0& 2\\ 0& 2& 1\\ 2& 0& 3\end{array}\right)=\left(\begin{array}{ccc}5& 0& 8\\ 2& 4& 5\\ 8& 0& 13\end{array}\right)$

$A^3=A^2\times A=\left(\begin{array}{ccc}5& 0& 8\\ 2& 4& 5\\ 8& 0& 13\end{array}\right)\times \left(\begin{array}{ccc}1& 0& 2\\ 0& 2& 1\\ 2& 0& 3\end{array}\right)=\left(\begin{array}{ccc}21& 0& 34\\ 12& 8& 23\\ 34& 0& 55\end{array}\right)$

$\therefore A^3-6A^2+7A+k{I}_3$

$=\left(\begin{array}{ccc}21& 0& 34\\ 12& 8& 23\\ 34& 0& 55\end{array}\right)-6\times \left(\begin{array}{ccc}5& 0& 8\\ 2& 4& 5\\ 8& 0& 13\end{array}\right)+7\times \left(\begin{array}{ccc}1& 0& 2\\ 0& 2& 1\\ 2& 0& 3\end{array}\right)+k\times \left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)=0$

$\Rightarrow k=2$