Question

If $A=\left|\begin{array}{cc}1& -1\\ -1& 1\end{array}\right|$, then show that $A^2=2A\&A^3=4A$

Answer 1

Given that:

$A=\left|\begin{array}{cc}1& -1\\ -1& 1\end{array}\right|$

Now,

$A^2=A\times A$

$A\times A=\left|\begin{array}{cc}1& -1\\ -1& 1\end{array}\right|\times \left|\begin{array}{cc}1& -1\\ -1& 1\end{array}\right|$

$A\times A=\left|\begin{array}{cc}1\times 1+(-1)\times (-1)& -1\times 1+1\times (-1)\\ -1\times 1+1\times (-1)& 1\times 1+(-1)\times (-1)\end{array}\right|$

$A^2=\left|\begin{array}{cc}2& -2\\ -2& 2\end{array}\right|$

$A^2=2\left|\begin{array}{cc}1& -1\\ -1& 1\end{array}\right|$

$A^2=2A$

Hence proved.

Now,

$A^3=A^2\times A$

$=2A\times A$ (Since we proved $A^2=2A$)

$=2A^2$

$=2\times 2A$

$A^3=4A$

Hence proved.