Question

$IfA=\left[\begin{array}{ccc}1& 1& 3\\ 5& 2& 6\\ -2& -1& -3\end{array}\right],$ then find $A^3$

Answer 1

$A=\left(\begin{array}{ccc}1& 1& 3\\ 5& 2& 6\\ -2& -1& -3\end{array}\right)$

$A^2=\left(\begin{array}{ccc}1& 1& 3\\ 5& 2& 6\\ -2& -1& -3\end{array}\right)\left(\begin{array}{ccc}1& 1& 3\\ 5& 2& 6\\ -2& -1& -3\end{array}\right)$

$=\left(\begin{array}{ccc}1+5-6& 1+2-3& 3+6-9\\ 5+10-12& 5+4-6& 15+12-18\\ -2-5+6& -2-2+3& -6-6+9\end{array}\right)$

$=\left(\begin{array}{ccc}0& 0& 0\\ 3& 3& 9\\ -1& -1& -3\end{array}\right)$

$A^3=A\cdot A^2=\left(\begin{array}{ccc}1& 1& 3\\ 5& 2& 6\\ -2& -1& -3\end{array}\right)\left(\begin{array}{ccc}0& 0& 0\\ 3& 3& 9\\ -1& -1& -3\end{array}\right)$

$=\left(\begin{array}{ccc}3-3& 3-3& 9-9\\ 6-6& 6-6& 18-18\\ -3+3& -3+3& -9+9\end{array}\right)$

$=\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right)$

Hence $A^3=0$.