Question

If $A=\left[\begin{array}{cc}-1& 0\\ 0& 2\end{array}\right]$, then $A^3-A^2=$

• A. $2A$
• B. $2I$
• C. $A$
• D. $I$

Answer 1

The correct option is A $2A$

$A^2=A\times A=\left[\begin{array}{cc}-1& 0\\ 0& 2\end{array}\right]\left[\begin{array}{cc}-1& 0\\ 0& 2\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 4\end{array}\right]$
$A^3=A^2\times A=\left[\begin{array}{cc}1& 0\\ 0& 4\end{array}\right]\left[\begin{array}{cc}-1& 0\\ 0& 2\end{array}\right]=\left[\begin{array}{cc}-1& 0\\ 0& 8\end{array}\right]$
$A^3-A^2=\left[\begin{array}{cc}-1& 0\\ 0& 8\end{array}\right]-\left[\begin{array}{cc}1& 0\\ 0& 4\end{array}\right]=\left[\begin{array}{cc}-2& 0\\ 0& 4\end{array}\right]$
$=2\left[\begin{array}{cc}-1& 0\\ 0& 2\end{array}\right]=2A$