Solve the system of equations:

$
\begin{matrix} 
x^2 + x - 1 = y \\ 
y^2 + y - 1 = z \\ 
z^2 + z - 1 = x.
\end{matrix}
$
To solve the system of equations:
\[
\begin{cases}
x^2 + x - 1 = y \\
y^2 + y - 1 = z \\
z^2 + z - 1 = x
\end{cases}
\]

we will use substitution and algebraic manipulation.

1. **Substitution and Transformation:**
   Let \( a = x + 1 \), \( b = y + 1 \), and \( c = z + 1 \). Then we can rewrite the system as:
   \[
   \begin{cases}
   x = a - 1 \\
   y = b - 1 \\
   z = c - 1
   \end{cases}
   \]
   Substituting these into the original equations, we get:
   \[
   \begin{cases}
   (a-1)^2 + (a-1) - 1 = b - 1 \\
   (b-1)^2 + (b-1) - 1 = c - 1 \\
   (c-1)^2 + (c-1) - 1 = a - 1
   \end{cases}
   \]
   Simplifying each equation:
   \[
   \begin{cases}
   a^2 - 2a + 1 + a - 1 - 1 = b - 1 \\
   b^2 - 2b + 1 + b - 1 - 1 = c - 1 \\
   c^2 - 2c + 1 + c - 1 - 1 = a - 1
   \end{cases}
   \]
   \[
   \begin{cases}
   a^2 - a - 1 = b - 1 \\
   b^2 - b - 1 = c - 1 \\
   c^2 - c - 1 = a - 1
   \end{cases}
   \]
   Adding 1 to both sides of each equation:
   \[
   \begin{cases}
   a^2 - a = b \\
   b^2 - b = c \\
   c^2 - c = a
   \end{cases}
   \]

2. **Product of Variables:**
   Multiplying all three equations:
   \[
   (a^2 - a)(b^2 - b)(c^2 - c) = abc
   \]
   Since \( a^2 - a = b \), \( b^2 - b = c \), and \( c^2 - c = a \), we can substitute:
   \[
   b \cdot c \cdot a = abc
   \]
   This implies:
   \[
   abc = abc
   \]
   which is always true.

3. **Sum of Squares:**
   Adding the three equations:
   \[
   a^2 - a + b^2 - b + c^2 - c = a + b + c
   \]
   Rearranging:
   \[
   a^2 + b^2 + c^2 - a - b - c = a + b + c
   \]
   \[
   a^2 + b^2 + c^2 = 2(a + b + c)
   \]

4. **Solving for \(a, b, c\):**
   We know that \( a, b, c \) are roots of the equations \( a^2 - a = b \), \( b^2 - b = c \), and \( c^2 - c = a \). We can test possible values:
   - If \( a = 1 \), then \( b = 1^2 - 1 = 0 \), but this does not satisfy the system.
   - If \( a = -1 \), then \( b = (-1)^2 - (-1) = 2 \), but this does not satisfy the system.

   Testing \( a = 1 \), \( b = 1 \), \( c = 1 \):
   \[
   1^2 - 1 = 0 \quad \text{(not valid)}
   \]

   Testing \( a = -1 \), \( b = -1 \), \( c = -1 \):
   \[
   (-1)^2 - (-1) = 2 \quad \text{(not valid)}
   \]

   Therefore, the only valid solutions are \( x = y = z = 1 \) and \( x = y = z = -1 \).

The final answer is \( \boxed{ x = y = z = 1 } \) and \( x = y = z = -1 \).