Question

Let $A=\left[\begin{array}{cc}0& i\\ i& 0\end{array}\right],$ where $i^2=-1.$ Let $I$ denotes the identity matrix of order $2,$ then $I+A+A^2+A^3\dots \dots .A^{110}$ is equal to

• A. $\left[\begin{array}{cc}0& i\\ i& 0\end{array}\right]$
• B. $\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$
• C. $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$
• D. $\left[\begin{array}{cc}-1& 0\\ 0& 0\end{array}\right]$

Answer 1

Answer: (A)

$A^2=\left[\begin{array}{cc}0& i\\ i& 0\end{array}\right]\left[\begin{array}{cc}0& i\\ i& 0\end{array}\right]=\left[\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right]=-I$
$A^3=A^{2}A=-I\cdot A=-A$
$A^4={\left(A^2\right)}^2=I$
$A^5=A,A^6=-I,A^7=-A,A^8=I\dots .$ So on
$I+A+A^2+A^3=I+A-I-A=0$
$A^4+A^5+A^6+A^7=0\dots \dots \dots \dots A^{104}+A^{105}+A^{106}+A^{107}=0$
$A^{108}+A^{109}+A^{110}=I+A-I=A$
$\Rightarrow I+A+A^2+.\dots \dots .+A^{110}=A$