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We are given a row-echelon form of an augmented matrix for a system of 3 equations with 3 unknowns \(x_1\), \(x_2\), and \(x_3\):

\[
\begin{pmatrix}
1 & 2 & 3 & 5 \\
0 & 1 & 2 & 0 \\
0 & 0 & k^2 - 1 & k + 1
\end{pmatrix}
\]

We need to find the values of \(k\) for which there are:

(a) No solutions  
(b) An infinite number of solutions  
(c) Exactly one solution  

### Step-by-step analysis:

#### 1. **Case (c): Exactly one solution**
For there to be exactly one solution, the matrix must have a unique solution, meaning none of the equations should become inconsistent, and the system must be fully determined.

To avoid inconsistency and ensure a unique solution, the third row must not be zero. Hence, \( k^2 - 1 \neq 0 \) (the pivot element in the third row should be non-zero). 

This gives:
\[
k^2 - 1 \neq 0
\]
\[
k \neq \pm 1
\]

Thus, for exactly one solution, \( k \neq \pm 1 \).

#### 2. **Case (b): Infinite number of solutions**
For an infinite number of solutions, we require the third row to be entirely zero, implying that the system is underdetermined.

For the third row to be zero, both conditions must hold:
\[
k^2 - 1 = 0 \quad \text{and} \quad k + 1 = 0
\]

Solving these:
\[
k^2 - 1 = 0 \quad \Rightarrow \quad k = \pm 1
\]
Now, checking each case:

- If \( k = 1 \):
  \[
  k + 1 = 1 + 1 = 2 \quad (\text{not zero, so no infinite solutions for } k = 1)
  \]
  
- If \( k = -1 \):
  \[
  k + 1 = -1 + 1 = 0 \quad (\text{zero, so there are infinite solutions for } k = -1)
  \]

Therefore, there is an infinite number of solutions when \( k = -1 \).

#### 3. **Case (a): No solutions**
For there to be no solutions, the system must become inconsistent. This occurs when the third row leads to a contradiction, i.e., when the coefficient of \(x_3\) is zero but the constant on the right-hand side is non-zero. 

This happens when:
\[
k^2 - 1 = 0 \quad \text{and} \quad k + 1 \neq 0
\]

From \( k^2 - 1 = 0 \), we know \( k = \pm 1 \).

- For \( k = 1 \):
  \[
  k + 1 = 1 + 1 = 2 \quad (\text{non-zero, so this gives no solutions})
  \]

- For \( k = -1 \):
  \[
  k + 1 = 0 \quad (\text{so this does not give a contradiction})
  \]

Thus, there are no solutions when \( k = 1 \).

### Conclusion:

- (a) **No solutions**: \( k = 1 \)
- (b) **Infinite number of solutions**: \( k = -1 \)
- (c) **Exactly one solution**: \( k \neq \pm 1 \)

---

Do you need any further explanation or details?

Here are some related questions for practice:

1. How do you determine if a matrix is in row-echelon form?
2. What is the significance of the determinant in systems of linear equations?
3. How would the solution change if the system had more variables than equations?
4. What methods are commonly used to solve systems of linear equations?
5. How does Gaussian elimination differ from Gauss-Jordan elimination?

**Tip:** When analyzing systems of equations, always check for the possibility of inconsistent rows that can cause no solutions.

Find all values of k for a system of equations represented by the following augmented matrix:

\[
\begin{pmatrix}
1 & 2 & 3 & 5 \\
0 & 1 & 2 & 0 \\
0 & 0 & k^2 - 1 & k + 1
\end{pmatrix}
\]

The system has the following cases:
(a) No solutions
(b) Infinite number of solutions
(c) Exactly one solution.

Learn how to find the values of k for a system of linear equations represented by an augmented matrix in row-echelon form. The problem covers key linear algebra concepts, including cases for no solutions, infinite solutions, and exactly one solution. Suitable for advanced high school students or undergraduates.

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