Question

${\left(\begin{array}{cc}2& -1\\ 3& -2\end{array}\right)}^n=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$ if $n$ is

• A. odd
• B. any natural number
• C. even
• D. not possible

Answer 1

The correct option is C even

${\left[\begin{array}{cc}2& -1\\ 3& -2\end{array}\right]}^n=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$
$A^n=I$
$A^2=\left[\begin{array}{cc}2& -1\\ 3& -2\end{array}\right]\left[\begin{array}{cc}2& -1\\ 3& -2\end{array}\right]=\left[\begin{array}{cc}4-3& -2+2\\ 6-6& -3+4\end{array}\right]$
$=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$
$\therefore \left({\left|A^2\right|}^{n/2}\right)=I$
$\therefore \left({\left|I\right|}^{n/2}\right)=I$
$\therefore$ n is even.