Question

Let \(A=\begin{bmatrix} i & -i\\ -i & i\end{bmatrix},\ i=\sqrt{-1}\). Then, the system of linear equations \(A^8\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}8\\64\end{bmatrix}\) has :

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

Hint:

Write \(A=i\,B\) with \(B=\begin{bmatrix}1&-1\\-1&1\end{bmatrix}\); \(B^2=2B\), so \(A^8=128B\), which is singular and gives inconsistent equations.

Answer 1

The Correct Option is D

\(A=iB\), \(B=\begin{bmatrix}1&-1\\-1&1\end{bmatrix}\), and \(B^2=2B\Rightarrow B^n=2^{n-1}B\).
\(A^8=i^8B^8=2^7B=128B\).
The system becomes \(128\begin{bmatrix}x-y\\-(x-y)\end{bmatrix}=\begin{bmatrix}8\\64\end{bmatrix}\), i.e. \(x-y=\frac1{16}\) and \(x-y=-\frac12\).
These contradict each other, so there is no solution.
Answer: (4) No solution