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The image shows a system of linear equations with a parameter \( m \in \mathbb{R} \), and the task is to determine the values of \( m \) such that the system has a unique solution. The system of equations is:

\[
\begin{cases}
x_1 + x_2 - 7x_3 = 1 \\
2x_1 - x_2 + mx_3 = 2 + m \\
-x_1 + 3x_2 - (m+1)x_3 = m^2 + 1
\end{cases}
\]

### Approach

To determine for which values of \( m \) the system has a unique solution, we need to check the determinant of the coefficient matrix. If the determinant is non-zero, the system has a unique solution.

The coefficient matrix \( A \) is:
\[
A =
\begin{pmatrix}
1 & 1 & -7 \\
2 & -1 & m \\
-1 & 3 & -(m+1)
\end{pmatrix}
\]

We can compute the determinant of matrix \( A \) and solve for the values of \( m \) that make this determinant non-zero.

### Determinant Calculation

\[
\text{det}(A) = 
\begin{vmatrix}
1 & 1 & -7 \\
2 & -1 & m \\
-1 & 3 & -(m+1)
\end{vmatrix}
\]

Expanding this determinant along the first row:

\[
\text{det}(A) = 1 \cdot 
\begin{vmatrix}
-1 & m \\
3 & -(m+1)
\end{vmatrix}
- 1 \cdot 
\begin{vmatrix}
2 & m \\
-1 & -(m+1)
\end{vmatrix}
+ (-7) \cdot 
\begin{vmatrix}
2 & -1 \\
-1 & 3
\end{vmatrix}
\]

Now, we can compute each of the 2x2 minors:

1. First minor:
\[
\begin{vmatrix}
-1 & m \\
3 & -(m+1)
\end{vmatrix}
= (-1)(-(m+1)) - (m)(3) = m+1 - 3m = -2m + 1
\]

2. Second minor:
\[
\begin{vmatrix}
2 & m \\
-1 & -(m+1)
\end{vmatrix}
= (2)(-(m+1)) - (m)(-1) = -2(m+1) + m = -2m - 2 + m = -m - 2
\]

3. Third minor:
\[
\begin{vmatrix}
2 & -1 \\
-1 & 3
\end{vmatrix}
= (2)(3) - (-1)(-1) = 6 - 1 = 5
\]

Substitute these back into the determinant expression:

\[
\text{det}(A) = 1 \cdot (-2m + 1) - 1 \cdot (-m - 2) + (-7) \cdot 5
\]

Simplifying:

\[
\text{det}(A) = (-2m + 1) + (m + 2) - 35 = -2m + m + 1 + 2 - 35 = -m - 32
\]

So, the determinant is:

\[
\text{det}(A) = -m - 32
\]

### Condition for a Unique Solution

For the system to have a unique solution, the determinant must be non-zero:

\[
-m - 32 \neq 0
\]

\[
m \neq -32
\]

Thus, the system has a unique solution for all \( m \in \mathbb{R} \) except \( m = -32 \).

Would you like further details or have any questions about this problem?

---

Here are five related questions you could explore:
1. What happens if \( m = -32 \)? Does the system have no solution or infinitely many solutions?
2. How do we determine whether a system has no solution or infinitely many solutions when the determinant is zero?
3. Can we apply Cramer's rule to solve this system for specific values of \( m \)?
4. How does the parameter \( m \) affect the behavior of the solution set geometrically?
5. What is the geometric interpretation of a system of three equations with three unknowns in 3D space?

**Tip:** When analyzing systems with parameters, checking the determinant is a quick and powerful method to determine the conditions for unique, infinite, or no solutions.

Cho m ∈ R và hệ phương trình (Given m ∈ R and the system of equations):

\[
\begin{cases}
x_1 + x_2 - 7x_3 = 1 \\
2x_1 - x_2 + mx_3 = 2 + m \\
-x_1 + 3x_2 - (m+1)x_3 = m^2 + 1
\end{cases}
\]

Find the values of m for which the system has a unique solution.

Find the values of m such that the given system of linear equations has a unique solution. This problem involves calculating the determinant of a 3x3 matrix and applying the condition for a unique solution. Suitable for students in Grades 10-12.

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