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The problem provided in **Exercice 4** is to determine the number of solutions for the following system of equations depending on the real parameter \( m \):

\[
\begin{cases}
mx + y + t = m + 1 \\
x + my + z = m - 1 \\
y + mz + t = m + 1 \\
x + z + mt = m - 1
\end{cases}
\]

### Step-by-Step Solution

1. **Matrix Representation:**

   The system can be written in matrix form as:

   \[
   A \cdot \mathbf{X} = \mathbf{B}
   \]
   where:
   - \( \mathbf{X} = \begin{pmatrix} x \\ y \\ z \\ t \end{pmatrix} \)
   - \( \mathbf{B} = \begin{pmatrix} m + 1 \\ m - 1 \\ m + 1 \\ m - 1 \end{pmatrix} \)
   - The matrix \( A \) is:
   \[
   A = \begin{pmatrix}
   m & 1 & 0 & 1 \\
   1 & m & 1 & 0 \\
   0 & 1 & m & 1 \\
   1 & 0 & 1 & m
   \end{pmatrix}
   \]

2. **Determinant of the Matrix \( A \):**

   To determine the number of solutions, we first compute the determinant of the matrix \( A \). The determinant will help us understand whether the system has a unique solution (if \( \det(A) \neq 0 \)) or infinitely many/no solutions (if \( \det(A) = 0 \)).

   The determinant of \( A \) is:

   \[
   \det(A) = \begin{vmatrix}
   m & 1 & 0 & 1 \\
   1 & m & 1 & 0 \\
   0 & 1 & m & 1 \\
   1 & 0 & 1 & m
   \end{vmatrix}
   \]

   We will now calculate this determinant step by step.

3. **Row Expansion (Cofactor Expansion):**

   We can expand the determinant along the first row to simplify the calculation.

   \[
   \det(A) = m \cdot \det\begin{vmatrix}
   m & 1 & 0 \\
   1 & m & 1 \\
   0 & 1 & m
   \end{vmatrix}
   - 1 \cdot \det\begin{vmatrix}
   1 & 1 & 0 \\
   0 & m & 1 \\
   1 & 1 & m
   \end{vmatrix}
   + 0 \cdot (\text{some determinant}) - 1 \cdot \det\begin{vmatrix}
   1 & m & 1 \\
   0 & 1 & m \\
   1 & 1 & m
   \end{vmatrix}
   \]

4. **Simplification of Determinants:**

   Let me now compute the determinant to see for which values of \( m \), the matrix \( A \) is invertible.

I will proceed to calculate the determinant now.The determinant of the matrix \( A \) is:

\[
\det(A) = m^2(m^2 - 4)
\]

### Analyzing the Determinant:

- \( \det(A) = 0 \) when either \( m^2 = 0 \) or \( m^2 - 4 = 0 \), which gives us the following critical values of \( m \):
  - \( m = 0 \)
  - \( m = \pm 2 \)

### Conclusion:
- For \( m \neq 0, \pm 2 \), the determinant \( \det(A) \neq 0 \), meaning the system has a **unique solution**.
- For \( m = 0 \) or \( m = \pm 2 \), \( \det(A) = 0 \), meaning the matrix is singular, and we need to perform further analysis to check if the system has infinitely many solutions or no solutions.

Would you like me to analyze the cases where \( m = 0 \) and \( m = \pm 2 \) for a more detailed conclusion?

---

Here are some relative questions to deepen the understanding:
1. How can we check the consistency of a system when \( \det(A) = 0 \)?
2. What does it mean geometrically when a matrix has determinant zero?
3. How does the rank of a matrix affect the number of solutions in a linear system?
4. Why is \( \det(A) \neq 0 \) important for having a unique solution?
5. How can we find the general solution when \( \det(A) = 0 \)?

**Tip:** Determinants give insight into the invertibility of matrices, which is crucial in understanding the solution space of linear systems.

Déterminer, suivant les valeurs du paramètre réel m, le nombre de solutions du système suivant: 

{ mx + y + t = m + 1 
 x + my + z = m - 1 
 y + mz + t = m + 1 
 x + z + mt = m - 1 }

Learn how to determine the number of solutions for a system of linear equations depending on the real parameter m. This problem covers key concepts such as linear algebra, determinants, and system solutions. Suitable for undergraduate students studying linear algebra.

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