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The given augmented matrix is:

\[
\begin{pmatrix}
1 & -1 & -7 & 2 & 1 & 3 & 12 \\
3 & 0 & 3 & -3 & -9 & 0 & -12 \\
-1 & 1 & 7 & 3 & 14 & 3 & 31 \\
1 & 1 & 9 & -2 & -1 & 1 & 2
\end{pmatrix}
\]

To solve this augmented matrix, we need to perform row reduction using Gaussian elimination or Gauss-Jordan elimination to reduce it to row echelon form (REF) or reduced row echelon form (RREF). This will allow us to find the solution to the system of linear equations.

### Step-by-Step Gaussian Elimination:
1. **Start with the first pivot element (Row 1, Column 1):**
   Use \( R1 = R1 \) as the leading row.

2. **Eliminate the first column entries below the pivot:**
   - \( R2 = R2 - 3R1 \)
   - \( R3 = R3 + R1 \)
   - \( R4 = R4 - R1 \)

3. **Continue with the second pivot (Row 2, Column 2):**
   Ensure that the element in Row 2, Column 2 is 1 (if necessary, scale the row). Then eliminate entries in the second column below it.

4. **Move to the third pivot (Row 3, Column 3):**
   Ensure the third pivot is 1 and eliminate entries below it.

5. **Back-substitute** to find the solutions.

I'll proceed with these steps to simplify the matrix.It appears the matrix isn't square, which is expected since it's an augmented matrix with six columns and four rows (corresponding to four equations with six variables). The system is likely underdetermined, meaning there are infinitely many solutions. I'll use Gaussian elimination to reduce the matrix manually, performing row operations.

Let’s proceed step-by-step:

### Step 1: Eliminate the entries below the pivot in column 1.

- \( R2 = R2 - 3R1 \)
- \( R3 = R3 + R1 \)
- \( R4 = R4 - R1 \)

After applying these operations, I’ll calculate the new matrix and proceed further.After performing the first set of row operations, the matrix becomes:

\[
\begin{pmatrix}
1 & -1 & -7 & 2 & 1 & 3 & 12 \\
0 & 3 & 24 & -9 & -12 & -9 & -48 \\
0 & 0 & 0 & 5 & 15 & 6 & 43 \\
0 & 2 & 16 & -4 & -2 & -2 & -10
\end{pmatrix}
\]

### Step 2: Eliminate entries below the pivot in column 2.

- \( R4 = R4 - \frac{2}{3} R2 \)

Next, I will perform this operation and continue the row reduction process.After performing the second set of row operations, the matrix now looks like:

\[
\begin{pmatrix}
1 & -1 & -7 & 2 & 1 & 3 & 12 \\
0 & 3 & 24 & -9 & -12 & -9 & -48 \\
0 & 0 & 0 & 5 & 15 & 6 & 43 \\
0 & 0 & 0 & 2 & 6 & 4 & 22
\end{pmatrix}
\]

### Step 3: Eliminate the entry in Row 4, Column 4 by scaling Row 4 to make the pivot element 1.

I will now normalize Row 4 to set the pivot in Column 4 equal to 1. Then, I will perform further steps for back substitution and solve the system.It seems like I can’t do more advanced data analysis right now. Please try again later.

However, you can continue solving the matrix by normalizing Row 4:

\[
R4 = \frac{1}{2} R4 \quad \text{(which makes the pivot element in Row 4, Column 4 equal to 1)}
\]

Once you do that, you can continue the Gaussian elimination process to solve the system.

Would you like more details on any of these steps?

Here are some related questions:
1. How does Gaussian elimination work for non-square matrices?
2. What does it mean for a system of linear equations to be underdetermined?
3. How can we interpret the solution of an augmented matrix geometrically?
4. What is the difference between row echelon form and reduced row echelon form?
5. How do we handle inconsistent systems in augmented matrices?

**Tip:** Always check for row consistency after performing row operations to avoid errors in interpreting solutions.

Solve the augmented matrix: [[1, -1, -7, 2, 1, 3, 12], [3, 0, 3, -3, -9, 0, -12], [-1, 1, 7, 3, 14, 3, 31], [1, 1, 9, -2, -1, 1, 2]]

Step-by-step solution of a system of linear equations represented as an augmented matrix. Learn how to use Gaussian elimination to reduce the matrix to row echelon form and back-substitute to find solutions.

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