https://file.aiquickdraw.com/mathbot/file/1727375754610ok8ac7gy.jpg

The matrix shown in the image is:

\[
A = 
\begin{pmatrix}
0 & 0 & 2 & 0 & 7 & 12 \\
2 & 4 & -10 & 6 & 12 & 28 \\
4 & 4 & -5 & 6 & -5 & -1
\end{pmatrix}
\]

The goal is to bring this matrix to Reduced Row Echelon Form (RREF).

Let's follow the steps:

### Step 1: Swap Rows (if necessary)
Since the first pivot should be in the top-left position, we see that the first column in the first row has a zero. So, we need to swap rows to ensure we have a non-zero pivot.

Swap row 1 and row 2:
\[
A = 
\begin{pmatrix}
2 & 4 & -10 & 6 & 12 & 28 \\
0 & 0 & 2 & 0 & 7 & 12 \\
4 & 4 & -5 & 6 & -5 & -1
\end{pmatrix}
\]

### Step 2: Create leading 1 in Row 1
To create a leading 1 in the first row, divide the entire first row by 2:
\[
A = 
\begin{pmatrix}
1 & 2 & -5 & 3 & 6 & 14 \\
0 & 0 & 2 & 0 & 7 & 12 \\
4 & 4 & -5 & 6 & -5 & -1
\end{pmatrix}
\]

### Step 3: Eliminate entries below the leading 1 in Row 1
Subtract 4 times Row 1 from Row 3 to eliminate the leading entry in the third row:
\[
R_3 = R_3 - 4R_1
\]

\[
A = 
\begin{pmatrix}
1 & 2 & -5 & 3 & 6 & 14 \\
0 & 0 & 2 & 0 & 7 & 12 \\
0 & -4 & 15 & -6 & -29 & -57
\end{pmatrix}
\]

### Step 4: Create leading 1 in Row 2
In the second row, divide by 2 to make the leading 1 for the third column:
\[
A = 
\begin{pmatrix}
1 & 2 & -5 & 3 & 6 & 14 \\
0 & 0 & 1 & 0 & 3.5 & 6 \\
0 & -4 & 15 & -6 & -29 & -57
\end{pmatrix}
\]

### Step 5: Eliminate entries above and below the leading 1 in Row 2
Subtract (-5) times Row 2 from Row 1 to eliminate the third column entry in the first row:
\[
R_1 = R_1 + 5R_2
\]

\[
A = 
\begin{pmatrix}
1 & 2 & 0 & 3 & 23.5 & 44 \\
0 & 0 & 1 & 0 & 3.5 & 6 \\
0 & -4 & 15 & -6 & -29 & -57
\end{pmatrix}
\]

Now subtract 15 times Row 2 from Row 3 to eliminate the third column entry in the third row:
\[
R_3 = R_3 - 15R_2
\]

\[
A = 
\begin{pmatrix}
1 & 2 & 0 & 3 & 23.5 & 44 \\
0 & 0 & 1 & 0 & 3.5 & 6 \\
0 & -4 & 0 & -6 & -81.5 & -147
\end{pmatrix}
\]

### Step 6: Create leading 1 in Row 3
To eliminate the -4 entry in Row 3, first divide the third row by -4:
\[
A = 
\begin{pmatrix}
1 & 2 & 0 & 3 & 23.5 & 44 \\
0 & 0 & 1 & 0 & 3.5 & 6 \\
0 & 1 & 0 & 1.5 & 20.375 & 36.75
\end{pmatrix}
\]

Subtract 2 times Row 3 from Row 1:
\[
R_1 = R_1 - 2R_3
\]

Finally, subtract 2 times Row 3 from Row 2.

This will leave the matrix fully reduced.

Would you like to continue solving this step-by-step or ask for clarifications?

Here are some related questions:
1. What is the significance of a matrix being in RREF?
2. Can RREF of a matrix be unique?
3. How does RREF help in solving systems of linear equations?
4. What conditions make a matrix invertible?
5. Can every matrix be reduced to RREF?

**Tip:** Always try to work with smaller entries in a matrix to simplify your row operations!

Learn how to bring a matrix to its Reduced Row Echelon Form (RREF) with a detailed step-by-step solution. This problem involves core linear algebra concepts, matrix operations, and row transformations. Suitable for high school or college students.

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