Question

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

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

Answer 1

The correct option is A No solution

$A=\left[\begin{array}{cc}i& -i\\ -i& i\end{array}\right]A^2=\left[\begin{array}{cc}i& -i\\ -i& i\end{array}\right]\left[\begin{array}{cc}i& -i\\ -i& i\end{array}\right]=\left[\begin{array}{cc}-2& 2\\ 2& -2\end{array}\right]=2\left[\begin{array}{cc}-1& 1\\ 1& -1\end{array}\right]A^4=4\left[\begin{array}{cc}-1& 1\\ 1& -1\end{array}\right]\left[\begin{array}{cc}-1& 1\\ 1& -1\end{array}\right]=4\left[\begin{array}{cc}2& -2\\ -2& 2\end{array}\right]=8\left[\begin{array}{cc}1& -1\\ -1& 1\end{array}\right]A^8=64\left[\begin{array}{cc}1& -1\\ -1& 1\end{array}\right]\left[\begin{array}{cc}1& -1\\ -1& 1\end{array}\right]=64\left[\begin{array}{cc}2& -2\\ -2& 2\end{array}\right]=128\left[\begin{array}{cc}1& -1\\ -1& 1\end{array}\right]128\left[\begin{array}{cc}1& -1\\ -1& 1\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}8\\ 64\end{array}\right]128\left[\begin{array}{c}x-y\\ -x+y\end{array}\right]=\left[\begin{array}{c}8\\ 64\end{array}\right]\Rightarrow 128(x-y)=8\Rightarrow x-y=\frac{1}{16}\cdots \left(1\right)$
and
$128(-x+y)=64\Rightarrow x-y=-\frac{1}{2}\cdots \left(2\right)$
From $\left(1\right)$ and $\left(2\right)$ no solution exists