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We are given a system of linear equations with a parameter \(a\) and asked to determine the values of \(a\) for which:

1. The system has infinitely many solutions.
2. The system has no solutions.
3. The system has a unique solution.

The system of equations is:

\[
\begin{aligned}
1. & \quad x + ay + 2z = 5, \\
2. & \quad ax + y + z = 1, \\
3. & \quad 2x - 2y + (a + 2)z = 8.
\end{aligned}
\]

We'll use **Gaussian elimination** (row operations) to analyze the system and classify its solutions based on the value of \(a\). Let’s set up the augmented matrix first:

\[
\left[ \begin{array}{ccc|c}
1 & a & 2 & 5 \\
a & 1 & 1 & 1 \\
2 & -2 & a+2 & 8
\end{array} \right]
\]

### Step 1: Row Reduction
We will perform row operations to bring this matrix to row echelon form.

1. **Row 2 (R2):** Subtract \(a\) times Row 1 (R1) from Row 2.
2. **Row 3 (R3):** Subtract 2 times Row 1 (R1) from Row 3.

After performing these operations, we will identify the different cases for \(a\).

Let me compute the row reduction to find the conditions for each case.The row-reduced form of the augmented matrix is:

\[
\left[
\begin{array}{ccc|c}
1 & 0 & 0 & \frac{a - 11}{a^2 + 2a + 1} \\
0 & 1 & 0 & \frac{5a - 7}{a^2 + 2a + 1} \\
0 & 0 & 1 & \frac{8}{a + 1}
\end{array}
\right]
\]

From this row-reduced matrix, we can analyze the different cases for \(a\):

### Case (a): Infinitely many solutions
Infinitely many solutions occur when a row of zeros appears in the coefficient matrix while the corresponding constant in the augmented matrix is also zero. For this to happen, the denominator in all entries must be undefined, meaning the system has dependent equations.

From the reduced form, the denominators are \(a^2 + 2a + 1 = (a + 1)^2\), so the system has infinitely many solutions if \(a = -1\). 

### Case (b): No solutions
No solutions occur if there is a row of zeros in the coefficient matrix but a non-zero value in the augmented matrix. This could occur if \(a = -1\) makes the denominator in any term zero but the corresponding numerator non-zero.

For \(a = -1\), the entry in the last row becomes:
\[
\frac{8}{a+1} = \frac{8}{0},
\]
which is undefined. Thus, for \(a = -1\), there is no solution.

### Case (c): Unique solution
For all other values of \(a\) except \(a = -1\), the system has a unique solution because the matrix remains full rank (with 3 pivot columns) and no division by zero occurs.

### Conclusion:
- \(a = -1\) leads to no solutions.
- For any \(a \neq -1\), the system has a unique solution.
- There are no values of \(a\) that lead to infinitely many solutions.

#### Special case for \(x + y + z = 0\) when \(a = -1\):
In case (a), there are no values of \(a\) leading to infinitely many solutions, so there is no need to find a particular solution for this.

Would you like more details or have any questions? 

Here are five related questions to explore further:
1. How does Gaussian elimination ensure that we identify the types of solutions?
2. What does the rank of a matrix reveal about the solutions of a system?
3. How would you interpret a system with a row of all zeros in the augmented matrix?
4. Can a system with a unique solution have any parameters left undetermined?
5. How do changes in the coefficient matrix affect the consistency of the system?

**Tip**: In general, the number of solutions to a linear system is determined by comparing the rank of the coefficient matrix and the augmented matrix (rank theorem).

Consider the following system of linear equations: x + ay + 2z = 5, ax + y + z = 1, 2x - 2y + (a + 2)z = 8. Using row operations, find the values of a for which the system has infinitely many solutions, no solutions, or a unique solution.

This problem explores the conditions for infinitely many solutions, no solutions, and a unique solution for a system of linear equations involving a parameter a. The solution involves Gaussian elimination and row operations, suitable for college-level linear algebra.

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