Question

Let $A=\left[\begin{array}{ccc}6& 2& 2\\ -2& -3& -1\\ 2& -1& 3\end{array}\right]$ and $f\left(x\right)=|xI-A|$, then $f\left(A\right)$ equals

• A. $0$
• B. $A$
• C. $2A$
• D. $-A$

Answer 1

The correct option is A $0$

$A=\left[\begin{array}{ccc}6& 2& 2\\ -2& -3& -1\\ 2& -1& 3\end{array}\right]$
$|xI-A|=\left[\begin{array}{ccc}x-6& -2& -2\\ 2& x+3& 1\\ -2& 1& x-3\end{array}\right]$
applying $R_3\to R_2+R_3$ and expanding gives
$|xI-A|=\left[\begin{array}{ccc}x-6& -2& -2\\ 2& x+3& 1\\ 0& x+4& x-2\end{array}\right]$
$=x^3-6x^2-10x+36$
$f\left(x\right)=|xI-A|=x^3-6x^2-10x+36$
$f\left(A\right)=A^3-6A^2-10A+36I$
$=\left[\begin{array}{ccc}240& 44& 116\\ -68& -30& -34\\ 140& 14& 78\end{array}\right]-6\left[\begin{array}{ccc}36& 4& 16\\ -8& 6& -4\\ 20& 4& 14\end{array}\right]-10\left[\begin{array}{ccc}6& 2& 2\\ -2& -3& -1\\ 2& -1& 3\end{array}\right]+36\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]$
Hence, option A.