# Prove the following

Question:

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

1. A unique solution

2. Infinitely many solutions

3. No solution

4. Exactly two solutions

Correct Option: , 3

Solution:

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

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

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

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

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

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

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

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

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

$\Rightarrow$ From (1) & (2) : No solution.