Question

Let $A=\begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix},$ where $i^{2}=-1.$ Let $I$ denotes the identity matrix of order $2,$ then $I+A+A^{2}+A^{3}\ldots \ldots .A^{110}$ is equal to

• A. $\begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix}$
• B. $\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$
• C. $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$
• D. $\begin{bmatrix} -1 & 0 \\ 0 & 0 \end{bmatrix}$

Answer 1

Answer: (A)

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