Question

If \(A = [\begin{array}{cc}2i & i^3 \\ i^2 & 1\end{array}]\), then \(A^{-1}\) is equal to

• A. \([\begin{array}{cc}-i & 1 \\ -i & 2\end{array}]\)
• B. \([\begin{array}{cc}i & 1 \\ i & -2\end{array}]\)
• C. \([\begin{array}{cc}-i & -1 \\ i & -2\end{array}]\)
• D. \([\begin{array}{cc}i & -1 \\ -i & 2\end{array}]\)

Hint:

Simplify the powers of $i$ first, then use $A^{-1}=\frac{1}{\det A}\text{adj}A$.

Answer 1

The Correct Option is A

Step 1: Check by multiplication
Try $B=\begin{bmatrix}-i&1\\-i&2\end{bmatrix}$. Then $AB$ has first row $(2i)(-i)+(-i)(-i)=2+(-1)=1$ and $(2i)(1)+(-i)(2)=0$.
Second row: $(-1)(-i)+(1)(-i)=0$ and $(-1)(1)+(1)(2)=1$. So $AB=I$ and $B=A^{-1}$.

Final Answer:
Option (A). \[ \boxed{\text{(A)}} \]