Question

If $A=\left[\begin{array}{ccc}1& 1& 0\\ 1& 2& 1\\ 2& 1& 0\end{array}\right]$ then find $A^3-3A^2-I.$

Answer 1

Given $A=\left[\begin{array}{ccc}1& 1& 0\\ 1& 2& 1\\ 2& 1& 0\end{array}\right]$

$A^2=A\times A=\left[\begin{array}{ccc}1& 1& 0\\ 1& 2& 1\\ 2& 1& 0\end{array}\right]\left[\begin{array}{ccc}1& 1& 0\\ 1& 2& 1\\ 2& 1& 0\end{array}\right]=\left[\begin{array}{ccc}1+1& 1+2& 1\\ 1+2+2& 1+4+1& 2\\ 2+1& 2+2& 1\end{array}\right]=\left[\begin{array}{ccc}2& 3& 1\\ 5& 6& 2\\ 3& 4& 1\end{array}\right]$

$A^3=A\times A\times A=A^2\times A=\left[\begin{array}{ccc}2& 3& 1\\ 5& 6& 2\\ 3& 4& 1\end{array}\right]\left[\begin{array}{ccc}1& 1& 0\\ 1& 2& 1\\ 2& 1& 0\end{array}\right]=\left[\begin{array}{ccc}2+3+2& 2+6+1& 3\\ 5+6+4& 5+12+2& 6\\ 3+4+2& 3+8+1& 4\end{array}\right]=\left[\begin{array}{ccc}7& 9& 3\\ 15& 19& 6\\ 9& 12& 4\end{array}\right]$

Using the above matrix in the given matrix equation

$A^3-3A^2-I=\left[\begin{array}{ccc}7& 9& 3\\ 15& 19& 6\\ 9& 12& 4\end{array}\right]-3\left[\begin{array}{ccc}2& 3& 1\\ 5& 6& 2\\ 3& 4& 1\end{array}\right]-\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

$=\left[\begin{array}{ccc}7-6-1& 9-9& 3-3\\ 15-15& 19-18-1& 0\\ 9-9& 12-12& 4-3-1\end{array}\right]=\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]=0$ (solution)