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Question:

Let $\mathrm{A}=\left[\begin{array}{cc}\mathrm{i} & -\mathrm{i} \\ -\mathrm{i} & \mathrm{i}\end{array}\right], \mathrm{i}=\sqrt{-1}$.Then, the system of linear equations

$\mathrm{A}^{8}\left[\begin{array}{l}\mathrm{x} \\ \mathrm{y}\end{array}\right]=\left[\begin{array}{c}8 \\ 64\end{array}\right]$ has :

  1. (1) A unique solution

  2. (2) Infinitely many solutions

  3. (3) No solution

  4. (4) Exactly two solutions


Correct Option: , 3

Solution:

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

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

$A^{4}=2^{2}\left[\begin{array}{cc}2 & -2 \\ -2 & 2\end{array}\right]=8\left[\begin{array}{cc}1 & -1 \\ -1 & 1\end{array}\right]$

$A^{8}=64\left[\begin{array}{cc}2 & -2 \\ -2 & 2\end{array}\right]=128\left[\begin{array}{cc}1 & -1 \\ -1 & 1\end{array}\right.$

$A^{8}\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{l}8 \\ 64\end{array}\right]$

$\Rightarrow \quad 128\left[\begin{array}{cc}1 & -1 \\ -1 & 1\end{array}\right]\left[\begin{array}{l}\mathrm{x} \\ \mathrm{y}\end{array}\right]_{0}=\left[\begin{array}{c}8 \\ 64\end{array}\right]$

$\Rightarrow \quad 128\left[\begin{array}{c}x-y \\ -x+y\end{array}\right]=\left[\begin{array}{c}8 \\ 64\end{array}\right]$

$\Rightarrow \quad x-y=\frac{1}{16}$

$\& \quad-x+y=\frac{1}{2}$

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