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: Simplify
\(i^2=-1\) and \(i^3=-i\), so \(A=\begin{bmatrix}2i&-i\\-1&1\end{bmatrix}\).

Step 2: Determinant
\(\det A=2i\cdot1-(-i)(-1)=2i-i=i\).

Step 3: Adjoint
Swap the diagonal entries and change the signs of the others: \(\text{adj}A=\begin{bmatrix}1&i\\1&2i\end{bmatrix}\).

Step 4: Inverse
Divide by \(i\), using \(\frac1i=-i\): \(A^{-1}=-i\begin{bmatrix}1&i\\1&2i\end{bmatrix}=\begin{bmatrix}-i&1\\-i&2\end{bmatrix}\). This is option (A).

Final Answer:
\(A^{-1}=\begin{bmatrix}-i&1\\-i&2\end{bmatrix}\), option (A). \[ \boxed{\text{(A)}} \]