Question

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

Answer 1

We have $A=\left|\begin{array}{ccc}1& 0& 2\\ 0& 2& 1\\ 2& 0& 3\end{array}\right|$

$A^2=A.A=\left|\begin{array}{ccc}1& 0& 2\\ 0& 2& 1\\ 2& 0& 3\end{array}\right|\left|\begin{array}{ccc}1& 0& 2\\ 0& 2& 1\\ 2& 0& 3\end{array}\right|$

$=\left|\begin{array}{ccc}1+0+4& 0+0+0& 2+0+6\\ 0+0+2& 0+4+0& 0+2+3\\ 2+0+6& 0+0+0& 4+0+9\end{array}\right|$

$=\left|\begin{array}{ccc}5& 0& 8\\ 2& 4& 5\\ 8& 0& 13\end{array}\right|$

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

$=\left|\begin{array}{ccc}5+0+16& 0+0+0& 10+0+24\\ 2+0+10& 0+8+0& 4+4+15\\ 8+0+26& 0+0+0& 16+0+39\end{array}\right|$

$=\left|\begin{array}{ccc}21& 0& 34\\ 12& 8& 23\\ 34& 0& 55\end{array}\right|$

Now, $A^3-6A^2+7A+k{l}_3=O$
$=\left|\begin{array}{ccc}21& 0& 34\\ 12& 8& 23\\ 34& 0& 55\end{array}\right|-\left|\begin{array}{ccc}30& 0& 48\\ 12& 24& 30\\ 48& 0& 78\end{array}\right|+\left|\begin{array}{ccc}7& 0& 14\\ 0& 14& 7\\ 14& 0& 21\end{array}\right|+\left|\begin{array}{ccc}k& 0& 0\\ 0& k& 0\\ 0& 0& k\end{array}\right|$
$=\left|\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right|$
$\Rightarrow 21-30+7+k=0$

so the value of $k=2$