91Ó°ÊÓ

The identity matrix \(I\) in the set of 2x2 matrices is \( \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \). Thus, we need to find matrices \(A\) for which \(A^2 = I\). Four such matrices can be: \( \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \), \( \begin{pmatrix} -1 & 0 \ 0 & -1 \end{pmatrix} \),\( \begin{pmatrix} 0 & -1 \ 1 & 0 \end{pmatrix} \), \( \begin{pmatrix} 0 & 1 \ -1 & 0 \end{pmatrix} \)

The general form of a 2x2 matrix is: \(A= \begin{pmatrix} a & b \ c & d \end{pmatrix} \)If \(A^2=I\), then \(A^2 = \begin{pmatrix} a & b \ c & d \end{pmatrix} \begin{pmatrix} a & b \ c & d \end{pmatrix} = \begin{pmatrix} a^2 + b*c & a*b + b*d \ a*c + c*d & b*c + d^2 \end{pmatrix} = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \).This gives us a system of equations:1. \(a^2 + b*c = 1\)2. \(a*b + b*d = 0\), which gives \(b*(a+d) = 0\)3. \(a*c + c*d = 0\), which gives \(c*(a+d) = 0\)4. \(b*c + d^2 = 1\)From equations 2 and 3, either \(b = 0\), or \(c = 0\), or \(a + d = 0\).Case 1: If \(b = 0\) and \(c = 0\), from equations 1 and 4 we get \(a^2 = d^2 = 1\), hence \(a,d\) can be either 1 or -1. Case 2: If \(a+d = 0\), we substitute \(a = -d\) into equation 1 to get \(-d^2 + d^2 = 1\), which is impossible. Hence \(a+d \neq 0\).Therefore the only solutions to \(A^2 = I\) are matrices where \(a = d = ±1\) and \(b = c = 0\), or \(a = -d\) and \(b = -c = ±1\).

With the above derivations, it can be determined that all 2x2 matrices \(A\) with \(A^{2}=I\) are: 1. \( \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \)2. \( \begin{pmatrix} -1 & 0 \ 0 & -1 \end{pmatrix} \)3. \( \begin{pmatrix} 1 & a \ a & 1 \end{pmatrix} \), \(a \in \mathbb{R}\setminus\{0\}\)4. \( \begin{pmatrix} -1 & a \ a & -1 \end{pmatrix} \), \(a \in \mathbb{R}\setminus\{0\}\)