Question

Let $A=\left[\begin{array}{cc}i & -i \\ -i & i\end{array}\right], i=\sqrt{-1}$. Then, the system of linear equations $A^{8}\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{c}8 \\ 64\end{array}\right]$ has :

• A. A unique solution
• B. Infinitely many solutions
• C. No solution
• D. Exactly two solutions

Answer 1

Answer: (C)

$A=\left[\begin{array}{cc}i & -i \\ -i & i\end{array}\right]$
$A ^{2}=\left[\begin{array}{cc}-2 & 2 \\ 2 & -2\end{array}\right]=2\left[\begin{array}{cc}-1 & 1 \\ 1 & -1\end{array}\right]$
$A ^{4}=2^{2}\left[\begin{array}{cc}2 & -2 \\ -2 & 2\end{array}\right]=8\left[\begin{array}{cc}1 & -1 \\ -1 & 1\end{array}\right]$
$A ^{8}=64\left[\begin{array}{cc}2 & -2 \\ -2 & 2\end{array}\right]=128\left[\begin{array}{cc}1 & -1 \\ -1 & 1\end{array}\right]$
$A ^{8}\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{c}8 \\ 64\end{array}\right]$
$\Rightarrow 128\left[\begin{array}{cc}1 & -1 \\ -1 & 1\end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{c}8 \\ 64\end{array}\right]$
$\Rightarrow 128\left[\begin{array}{c}x-y \\ -x+y\end{array}\right]=\left[\begin{array}{c}8 \\ 64\end{array}\right]$
$\Rightarrow x-y=\frac{1}{16} \ldots . .(1)$
$\&-x+y=\frac{1}{2}\ldots . .(2)$
$\Rightarrow $ From (1) & (2): No solution.