Question

If $A=\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]$ and $B=A^2+A^4+A^6+A^8+A^{10},$ then the value of $det\left(B\right)$ is equal to

Answer 1

Given: $A=\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]$
$\Rightarrow A^2=\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]=\left[\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right]$
$\Rightarrow A^4=\left[\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right]\left[\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$
$\Rightarrow A^4=I_2$ (identity matrix of order 2)
$\Rightarrow A^6=A^4\cdot A^2=A^2$
$\Rightarrow A^8=A^4\cdot A^4=I_2$
$\Rightarrow A^{10}=A^8\cdot A^2=A^2$
$\therefore B=A^2+I_2+A^2+I_2+A^2$
$\Rightarrow B=3A^2+2I_2$
$\Rightarrow B=\left[\begin{array}{cc}-3& 0\\ 0& -3\end{array}\right]+\left[\begin{array}{cc}2& 0\\ 0& 2\end{array}\right]$
$\Rightarrow B=\left[\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right]$
Hence, the value of $det\left(A\right)$ is $1.$