Question

If $A=\left[\begin{array}{ccc}1& 2& 1\\ 0& 1& -1\\ 3& -1& 1\end{array}\right]$, then $A^3-3A^2-A+9I$ is equal to

• A. Zero matrix
• B. $I$
• C. $A$
• D. $A^2$

Answer 1

The correct option is A Zero matrix

$A=\left[\begin{array}{ccc}1& 2& 1\\ 0& 1& -1\\ 3& -1& 1\end{array}\right]$
$\Rightarrow A^2=\left[\begin{array}{ccc}1& 2& 1\\ 0& 1& -1\\ 3& -1& 1\end{array}\right]\left[\begin{array}{ccc}1& 2& 1\\ 0& 1& -1\\ 3& -1& 1\end{array}\right]=\left[\begin{array}{ccc}4& 3& 0\\ -3& 2& -2\\ 6& 4& 5\end{array}\right]$
$\Rightarrow A^3=A^2.A=\left[\begin{array}{ccc}4& 3& 0\\ -3& 2& -2\\ 6& 4& 5\end{array}\right]\left[\begin{array}{ccc}1& 2& 1\\ 0& 1& -1\\ 3& -1& 1\end{array}\right]=\left[\begin{array}{ccc}4& 11& 1\\ -9& -2& -7\\ 21& 11& 7\end{array}\right]$
$A^3-3A^2-A+9I$$=\left[\begin{array}{ccc}4& 11& 1\\ -9& -2& -7\\ 21& 11& 7\end{array}\right]-3\left[\begin{array}{ccc}4& 3& 0\\ -3& 2& -2\\ 6& 4& 5\end{array}\right]-\left[\begin{array}{ccc}1& 2& 1\\ 0& 1& -1\\ 3& -1& 1\end{array}\right]+9\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]$
$\therefore A^3-3A^2-A+9I=O$
Hence, option A.