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Let's go through each problem step by step and solve them:

---

### 1. Solve the row echelon form matrix with back substitution:
The given matrix is:

\[
A = \begin{pmatrix}
3 & 1 & 9 \\
0 & -\frac{4}{3} & 0 \\
\end{pmatrix}
\]

**Back Substitution**:
We will solve this as a system of linear equations. Each row represents an equation in terms of \( x_1 \) and \( x_2 \).

**Step 1: Write equations** from the augmented matrix:

\[
3x_1 + x_2 = 9 \quad \text{(Row 1)}
\]
\[
-\frac{4}{3}x_2 = 0 \quad \text{(Row 2)}
\]

**Step 2: Solve for \( x_2 \) from Row 2:**
\[
-\frac{4}{3}x_2 = 0 \quad \Rightarrow \quad x_2 = 0
\]

**Step 3: Substitute \( x_2 = 0 \) into Row 1:**
\[
3x_1 + 0 = 9 \quad \Rightarrow \quad x_1 = \frac{9}{3} = 3
\]

So the solution is:
\[
x_1 = 3, \quad x_2 = 0
\]

---

### 2. Solve the row echelon form matrix with back substitution:
The given matrix is:

\[
A = \begin{pmatrix}
3 & 1 & 1 & 10 \\
0 & -2 & 0 & -10 \\
0 & 0 & 2 & -2 \\
\end{pmatrix}
\]

**Step 1: Write equations** from the augmented matrix:

\[
3x_1 + x_2 + x_3 = 10 \quad \text{(Row 1)}
\]
\[
-2x_2 = -10 \quad \text{(Row 2)}
\]
\[
2x_3 = -2 \quad \text{(Row 3)}
\]

**Step 2: Solve for \( x_3 \) from Row 3:**
\[
2x_3 = -2 \quad \Rightarrow \quad x_3 = -1
\]

**Step 3: Solve for \( x_2 \) from Row 2:**
\[
-2x_2 = -10 \quad \Rightarrow \quad x_2 = \frac{-10}{-2} = 5
\]

**Step 4: Substitute \( x_2 = 5 \) and \( x_3 = -1 \) into Row 1:**
\[
3x_1 + 5 + (-1) = 10 \quad \Rightarrow \quad 3x_1 + 4 = 10
\]
\[
3x_1 = 6 \quad \Rightarrow \quad x_1 = \frac{6}{3} = 2
\]

So the solution is:
\[
x_1 = 2, \quad x_2 = 5, \quad x_3 = -1
\]

---

### 3. Solve with row echelon form:
The given matrix is:

\[
A = \begin{pmatrix}
2 & 1 & 9 \\
3 & -1 & 6 \\
\end{pmatrix}
\]

We need to perform row operations to transform this matrix into row echelon form and then solve the system.

**Step 1: Eliminate the first element of Row 2.**
We can subtract \( \frac{3}{2} \) times Row 1 from Row 2 to make the element in the first column of Row 2 zero.

\[
R_2 \rightarrow R_2 - \frac{3}{2}R_1
\]

\[
\begin{pmatrix}
2 & 1 & 9 \\
0 & -\frac{5}{2} & -\frac{3}{2} \\
\end{pmatrix}
\]

**Step 2: Solve using back substitution.**
Now, the system of equations is:

\[
2x_1 + x_2 = 9 \quad \text{(Row 1)}
\]
\[
-\frac{5}{2}x_2 = -\frac{3}{2} \quad \text{(Row 2)}
\]

From Row 2:
\[
x_2 = \frac{-\frac{3}{2}}{-\frac{5}{2}} = \frac{3}{5}
\]

Substitute \( x_2 = \frac{3}{5} \) into Row 1:
\[
2x_1 + \frac{3}{5} = 9 \quad \Rightarrow \quad 2x_1 = 9 - \frac{3}{5} = \frac{45}{5} - \frac{3}{5} = \frac{42}{5}
\]
\[
x_1 = \frac{42}{10} = \frac{21}{5}
\]

So the solution is:
\[
x_1 = \frac{21}{5}, \quad x_2 = \frac{3}{5}
\]

---

Would you like to see details on any of these solutions? Here are some related questions you might explore:

1. How can we generalize back substitution for larger matrices?
2. What are the differences between row echelon form and reduced row echelon form?
3. How does Gaussian elimination differ from Gauss-Jordan elimination?
4. What is the significance of a system having no solution or infinitely many solutions in terms of the matrix form?
5. How can we handle inconsistent systems during back substitution?

**Tip**: Always double-check your row operations to ensure that each step brings the matrix closer to echelon form without introducing calculation errors.

Solve the row echelon form matrix with back substitution and row operations for three matrices.

Learn how to solve linear systems using the row echelon form and back substitution. This tutorial covers key linear algebra concepts with step-by-step solutions for solving systems of equations. Suitable for college-level linear algebra students.

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