Question

$A=\left[\begin{array}{ccc}1& 2& 2\\ 2& 1& 2\\ 2& 2& 1\end{array}\right],\text{then}{A}^3-4A^2-6A=$

• A. $0$
• B. $A$
• C. $-A$
• D. $I$

Answer 1

The correct option is C $-A$

$A^2=\left[\begin{array}{ccc}1& 2& 2\\ 2& 1& 2\\ 2& 2& 1\end{array}\right]\left[\begin{array}{ccc}1& 2& 2\\ 2& 1& 2\\ 2& 2& 1\end{array}\right]=\left[\begin{array}{ccc}9& 8& 8\\ 8& 9& 8\\ 8& 8& 9\end{array}\right]$
$A^3=A^2\times A=\left[\begin{array}{ccc}9& 8& 8\\ 8& 9& 8\\ 8& 8& 9\end{array}\right]\left[\begin{array}{ccc}1& 2& 2\\ 2& 1& 2\\ 2& 2& 1\end{array}\right]=\left[\begin{array}{ccc}41& 42& 42\\ 42& 41& 42\\ 42& 42& 41\end{array}\right]$
$A^3-4A^2-6A=\left[\begin{array}{ccc}41& 42& 42\\ 42& 41& 42\\ 42& 42& 41\end{array}\right]-\left[\begin{array}{ccc}36& 32& 32\\ 32& 36& 32\\ 32& 32& 36\end{array}\right]\left[\begin{array}{ccc}6& 12& 12\\ 12& 6& 12\\ 12& 12& 6\end{array}\right]$
$=\left[\begin{array}{ccc}-1& -2& -2\\ -2& -1& -2\\ -2& -2& -1\end{array}\right]=-A$