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

Answer

Correct option: B.
$\left[\begin{array}{ll}0 & i \\ i & 0\end{array}\right]$
b
(b)

We have,

$A=\left[\begin{array}{ll}0 & i \\ i & 0\end{array}\right]$

$A^2=\left[\begin{array}{ll}0 & i \\ i & 0\end{array}\right]\left[\begin{array}{ll}0 & i \\ i & 0\end{array}\right]$

$=\left[\begin{array}{cc}i^2 & 0 \\ 0 & i^2\end{array}\right]=\left[\begin{array}{cc}-1 & 0 \\ 0 & -1\end{array}\right]$

$=-\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=-I$

$A^3=A^2 \cdot A=\left[\begin{array}{cc}-1 & 0 \\ 0 & -1\end{array}\right]\left[\begin{array}{ll}0 & i \\ i & 0\end{array}\right]$

$=\left[\begin{array}{cc}0 & -i \\ -i & 0\end{array}\right]$

$=-\left[\begin{array}{ll}0 & i \\ i & 0\end{array}\right]=-A$

and $A^4=A^2, A^2=(-I)(-I)=I=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$

$\therefore I+A+A^2+A^3+A^4+A^5$

$\quad+\ldots+A^{2009}+A^{2009}+A^{2010}$

$=I+A+A^2+A^3+A^4\left[I+A+A^2+A^3\right]$

$+\ldots+A^{2005}\left[I+A+A^2\right]$

$=0+0+\ldots+\left[I+A+A^2\right]$

$=\left[\begin{array}{ll}0 & i \\ i & 0\end{array}\right]$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

The variance of $\alpha$, $\beta$ and $\gamma$ is $9$, then variance of $5$$\alpha$, $5$$\beta$ and $5$$\gamma$ is
If ${a^x} = bc,{b^y} = ca,\,{c^z} = ab,$ then $xyz$=
If $\frac{1}{{p + q}},\;\frac{1}{{r + p}},\;\frac{1}{{q + r}}$ are in $A.P.$, then
$1 + \sum\limits_{r = 0}^{22} {\left\{ {r\left( {r + 2} \right) + 1} \right\}}  \cdot r\ ! = k\ !$ then the number of divisors of $k$ is
If $x = \cos 10^\circ \cos 20^\circ \cos 40^\circ ,$ then the value of $x$ is
In a third order determinant, each element of the first column consists of sum of two terms, each element of the second column consists of sum of three terms and each element of the third column consists of sum of four terms. Then it can be decomposed into $n $determinants, where $ n$  has the value
Among the statements
(S1) : The set $\left\{\mathrm{z} \in \mathbb{C}-\{-\mathrm{i}\}:|\mathrm{z}|=1\right.$ and $\frac{\mathrm{z}-\mathrm{i}}{\mathrm{z}+\mathrm{i}}$ is purely real} contains exactly two elements, and
(S2) : The set $\left\{\mathrm{z} \in \mathbb{C}-\{-1\}:|z|=1\right.$ and $\frac{\mathrm{z}-1}{\mathrm{z}+1}$ is purely imaginary} contains infinitely many elements.
$\mathop {\lim }\limits_{x \to \infty } \frac{{\sin x}}{x} = $
How many six-digit numbers are there in which no digit is repeated, even digits appear at even places, odd digits appear in odd places and the number is divisible by $4$ ?
If $\alpha + \beta - \gamma = \pi ,$ then ${\sin ^2}\alpha + {\sin ^2}\beta - {\sin ^2}\gamma = $