Matrices of a certain form whose square is the identity matrix

Question

Find all $2\times 2$ matrices of the form $$A = \begin{bmatrix}0&b\\a&0\end{bmatrix},$$ where $a$ and $b$ are complex numbers, such that $$A^2=\begin{bmatrix}1&0\\0&1\end{bmatrix}$$ (note that this implies that $A^{-1} = A$). If you take $A^{-1}$ and multiply it with $A$ you will get $I$. Is that the answer? Is it enough to just multiply $A$ and $A^{-1}$ to show that it becomes the identity matrix?

Answer 1

If you compute $A^2$, you should find the following matrix: $$\begin{pmatrix}ab & 0\\ 0 & ab \end{pmatrix}.$$ This matrix is equal to the identity matrix if and only if $ab = 1$. Hence $a \neq 0$ and $b = 1/a$. Therefore, the matrices that satisfy the criterion $A^2 = I$ are exactly those matrices $$\begin{pmatrix}0 & 1/a\\ a & 0 \end{pmatrix}$$ with $a \in \mathbb{C}\setminus\{0\}$.

Answer 2

$$A =\begin{bmatrix}0&b\\a&0\end{bmatrix} \implies A^2 =\begin{bmatrix} ab&0\\0&ab\end{bmatrix}$$

Thus we need to have ab=1 for complex numbers a and b.

Let $ a=x_1 + i y_1$ and $ b=x_2 + i y_2$, $$ab=1 \implies a=1/b$$ $$ x_1 + i y_1 = \frac {1}{x_2 + i y_2}= \frac {x_2 - i y_2 }{x_2^2 + y_2^2}$$

Therefore we have $$ x_1 = \frac {x_2 }{x_2^2 + y_2^2}$$ h and $$ y_1 = \frac {-y_2 }{x_2^2 + y_2^2}$$

In complex form we have $$A=\begin{bmatrix}0&z\\1/z&0\end{bmatrix}$$