Question

If $A=\left[\begin{array}{cc}1& 3\\ 3& 1\end{array}\right]$, show that $A^2-2A$ is a scalar matrix.

Answer 1

Given matrix

$A=\left[\begin{array}{cc}1& 3\\ 3& 1\end{array}\right]$

$A^2=\left[\begin{array}{cc}1& 3\\ 3& 1\end{array}\right]\left[\begin{array}{cc}1& 3\\ 3& 1\end{array}\right]$

$=\left[\begin{array}{cc}1\times 1+3\times 3& 1\times 3+3\times 1\\ 3\times 1+1\times 3& 3\times 3+1\times 1\end{array}\right]=\left[\begin{array}{cc}1+9& 3+3\\ 3+3& 1+9\end{array}\right]$

$=\left[\begin{array}{cc}10& 6\\ 6& 10\end{array}\right]$

$2A=2\left[\begin{array}{cc}1& 3\\ 3& 1\end{array}\right]=\left[\begin{array}{cc}2& 6\\ 6& 2\end{array}\right]$

$\therefore A^2-2A=\left[\begin{array}{cc}10& 6\\ 6& 10\end{array}\right]-\left[\begin{array}{cc}2& 6\\ 6& 2\end{array}\right]=\left[\begin{array}{cc}10-2& 6-6\\ 6-6& 10-2\end{array}\right]$

$=\left[\begin{array}{cc}8& 0\\ 0& 8\end{array}\right]=8\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=8I$

$\therefore A^2-2A$ is scalar matrix