Find all solutions of the matrix equation `X^2=1,` where 1 is the 2*2 unit matrix, and X is a real matrix,i.e. a matrix all of whose elements are real.

Answer

To find all solutions of the matrix equation \( X^2 = I \), where \( I \) is the \( 2 \times 2 \) identity matrix and \( X \) is a real matrix, we can follow these steps:

### Step 1: Define the Matrix
Let \( X \) be a \( 2 \times 2 \) matrix of the form:
\[ X = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \]
where \( a, b, c, d \) are real numbers.

### Step 2: Compute \( X^2 \)
Now, we compute \( X^2 \):
\[ X^2 = X \cdot X = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} a^2 + bc & ab + bd \\ ac + dc & bc + d^2 \end{pmatrix} \]

### Step 3: Set \( X^2 \) Equal to \( I \)
We know that \( X^2 = I \), where \( I \) is the identity matrix:
\[ I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]
Thus, we have:
\[ \begin{pmatrix} a^2 + bc & ab + bd \\ ac + dc & bc + d^2 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

### Step 4: Set Up the Equations
From the equality of the matrices, we can set up the following equations:
1. \( a^2 + bc = 1 \) (Equation 1)
2. \( ab + bd = 0 \) (Equation 2)
3. \( ac + dc = 0 \) (Equation 3)
4. \( bc + d^2 = 1 \) (Equation 4)

### Step 5: Solve the Equations
From Equation 2, we can factor out \( b \):
\[ b(a + d) = 0 \]
This gives us two cases:
- Case 1: \( b = 0 \)
- Case 2: \( a + d = 0 \) (i.e., \( d = -a \))

#### Case 1: \( b = 0 \)
Substituting \( b = 0 \) into the equations:
- From Equation 1: \( a^2 = 1 \) → \( a = \pm 1 \)
- From Equation 4: \( d^2 = 1 \) → \( d = \pm 1 \)
- Equations 3 becomes \( 0 = 0 \), which is always true.

Thus, we have the following solutions for \( X \):
1. \( X = \begin{pmatrix} 1 & 0 \\ c & -1 \end{pmatrix} \) for any real \( c \)
2. \( X = \begin{pmatrix} -1 & 0 \\ c & 1 \end{pmatrix} \) for any real \( c \)

#### Case 2: \( d = -a \)
Substituting \( d = -a \) into the equations:
- From Equation 1: \( a^2 + bc = 1 \)
- From Equation 4: \( bc + a^2 = 1 \)

Both equations are identical, so we can just use one:
\[ a^2 + bc = 1 \quad \text{(Equation 5)} \]
Equation 3 becomes:
\[ ac - ac = 0 \quad \text{(always true)} \]

Thus, we can express \( c \) in terms of \( a \) and \( b \):
\[ bc = 1 - a^2 \quad \Rightarrow \quad c = \frac{1 - a^2}{b} \quad \text{(for } b \neq 0\text{)} \]

### Final Solutions
Combining both cases, the solutions for \( X \) can be expressed as:
1. \( X = \begin{pmatrix} 1 & 0 \\ c & -1 \end{pmatrix} \) for any real \( c \)
2. \( X = \begin{pmatrix} -1 & 0 \\ c & 1 \end{pmatrix} \) for any real \( c \)
3. \( X = \begin{pmatrix} a & b \\ \frac{1 - a^2}{b} & -a \end{pmatrix} \) for any real \( a \) and \( b \neq 0 \)