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We are given an augmented matrix representing a linear system, and the questions are related to the values of \(a\) and \(b\) that determine whether the system has:

1. A unique solution.
2. Infinitely many solutions.
3. No solution (inconsistent system).

The augmented matrix is:

\[
\begin{bmatrix}
1 & 1 & 0 & 2 \\
1 & 2 & 0 & 1 \\
3 & 5 & a & b
\end{bmatrix}
\]

Let's analyze the system for each question.

---

### (a) **Unique Solution**

To find when the system has a unique solution, we need to ensure that the matrix has full rank, i.e., no row is a linear combination of the others. This typically involves performing Gaussian elimination and checking for any contradictions or dependencies between the rows.

Let’s start by subtracting Row 1 from Row 2 to eliminate the first column from Row 2:
\[
R_2 \rightarrow R_2 - R_1: \quad
\begin{bmatrix}
1 & 1 & 0 & 2 \\
0 & 1 & 0 & -1 \\
3 & 5 & a & b
\end{bmatrix}
\]

Next, we subtract \(3R_1\) from Row 3 to eliminate the first column from Row 3:
\[
R_3 \rightarrow R_3 - 3R_1: \quad
\begin{bmatrix}
1 & 1 & 0 & 2 \\
0 & 1 & 0 & -1 \\
0 & 2 & a & (b - 6)
\end{bmatrix}
\]

Now, subtract \(2R_2\) from Row 3 to simplify the second column:
\[
R_3 \rightarrow R_3 - 2R_2: \quad
\begin{bmatrix}
1 & 1 & 0 & 2 \\
0 & 1 & 0 & -1 \\
0 & 0 & a & (b - 4)
\end{bmatrix}
\]

For the system to have a unique solution, the matrix must be of full rank, meaning \(a \neq 0\). When \(a \neq 0\), we can perform back substitution to find the unique solution.

Thus, for a unique solution, **\(a \neq 0\)**. The solution depends on the values of \(b\), and it can be determined through back substitution if required.

---

### (b) **Infinitely Many Solutions**

The system will have infinitely many solutions when one of the rows becomes dependent, leading to a free variable. This occurs if \(a = 0\).

Substituting \(a = 0\) into the matrix:
\[
\begin{bmatrix}
1 & 1 & 0 & 2 \\
0 & 1 & 0 & -1 \\
0 & 0 & 0 & (b - 4)
\end{bmatrix}
\]

For infinitely many solutions, the last row must correspond to the equation \(0 = 0\), meaning \(b = 4\).

Therefore, the system has infinitely many solutions when **\(a = 0\) and \(b = 4\)**.

---

### (c) **Inconsistent System**

The system is inconsistent if it leads to a contradictory equation like \(0 = c\), where \(c\) is a non-zero constant. This occurs if the last row is of the form:
\[
0 \ 0 \ 0 \ | \ (b - 4)
\]
which gives the equation \(0 = b - 4\). For inconsistency, \(b \neq 4\) when \(a = 0\).

Thus, the system is inconsistent when **\(a = 0\) and \(b \neq 4\)**.

---

### Summary:
- (a) Unique solution: \(a \neq 0\).
- (b) Infinitely many solutions: \(a = 0\) and \(b = 4\).
- (c) Inconsistent: \(a = 0\) and \(b \neq 4\).

Would you like a detailed step-by-step explanation or further clarification?

Here are 5 related questions to extend your understanding:
1. What is the significance of matrix rank in determining the type of solutions?
2. How does Gaussian elimination help identify the solution types?
3. What happens geometrically when a system has infinitely many solutions?
4. How do dependent equations affect the structure of the matrix?
5. How does back substitution work in solving linear systems?

**Tip:** Always check for both zero rows and contradictions (like \(0 = 1\)) when determining inconsistency in a system.

Consider a linear system whose augmented matrix is of the form [[1, 1, 0, 2], [1, 2, 0, 1], [3, 5, a, b]]. For what values of a will the system have a unique solution? For what values of a and b will the system have infinitely many solutions? For what values of a and b will the system be inconsistent?

Explore how to find the values of a and b in a linear system using an augmented matrix. Learn when the system has a unique solution, infinitely many solutions, or is inconsistent through Gaussian elimination and matrix rank analysis.

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