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3 and 4 . determinant and metrices — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths3 and 4 . determinant and metrices1 Mark
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 $A^4=A^2, A^2=(-I)(-I)=I=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$

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

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

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

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

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

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

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