If $A = \left[ {\begin{array}{*{20}{c}}1&3\\2&1\end{array}} \right]$, then determinant of ${A^2} – 2A$ is

If $A = \left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\a&b&{ – 1}\end{array}} \right]$, then  ${A^2} = $

Let $A = \left[ {\begin{array}{*{20}{c}}1&2\\3&4\end{array}} \right]$ and $B = \left[ {\begin{array}{*{20}{c}}a&0\\0&b\end{array}} \right]\;,a,b \in N$ then

If $A = [a\,\,b],B = [ – b – a]$ and $C = \left[ \begin{array}{l}\,\,\,\,a\\ – a\end{array} \right]$, then the correct statement is

Let $A$ denote the matrix $\left[\begin{array}{ll}0 & i \\ i & 0\end{array}\right]$, where $i^2=-1$, and let $I$ denote the identity matrix $\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$. Then, $I+A+A^2+\ldots+A^{2010}$ is