1 −2 8 −1 −1 −5 =−7 0 0 0 1 −7 0 =7 0 0 0 0 0 1 =9 sorry its 3x6 matric

Ah, I see now! You're working with a \(3 \times 6\) matrix, which looks like this:

\[
A = \begin{bmatrix}
1 & -2 & 8 & -1 & -1 & -5 \\
0 & 0 & 0 & 1 & -7 & 0 \\
0 & 0 & 0 & 0 & 0 & 1
\end{bmatrix}
\]

We will reduce this matrix to **reduced row echelon form (RREF)** using Gaussian and Gauss-Jordan elimination. Here's the step-by-step process.

### Step 1: Start with the given matrix \( A \).

\[
A = \begin{bmatrix}
1 & -2 & 8 & -1 & -1 & -5 \\
0 & 0 & 0 & 1 & -7 & 0 \\
0 & 0 & 0 & 0 & 0 & 1
\end{bmatrix}
\]

### Step 2: Eliminate entries above the leading 1 in column 4.
In the second row, we have a leading 1 in column 4. We want to eliminate the entry in column 4 of the first row by adding row 2 multiplied by \(1\) to row 1:

\[
R_1 \to R_1 + R_2
\]

This gives:

\[
A = \begin{bmatrix}
1 & -2 & 8 & 0 & -8 & -5 \\
0 & 0 & 0 & 1 & -7 & 0 \\
0 & 0 & 0 & 0 & 0 & 1
\end{bmatrix}
\]

### Step 3: Eliminate the entries above the leading 1 in column 6.
In the third row, we have a leading 1 in column 6. To eliminate the entry in column 6 of row 1, add \(5\) times row 3 to row 1:

\[
R_1 \to R_1 + 5R_3
\]

This gives:

\[
A = \begin{bmatrix}
1 & -2 & 8 & 0 & -8 & 0 \\
0 & 0 & 0 & 1 & -7 & 0 \\
0 & 0 & 0 & 0 & 0 & 1
\end{bmatrix}
\]

No other row operations are necessary since the matrix already satisfies the properties of **reduced row echelon form (RREF)**:
- Each leading entry in a row is 1.
- The leading entries are the only nonzero entries in their columns.
- The leading 1s move from left to right as you move down the rows.

### Final RREF Matrix:
\[
A = \begin{bmatrix}
1 & -2 & 8 & 0 & -8 & 0 \\


Learn how to reduce a 3x6 matrix to its reduced row echelon form (RREF) using Gaussian and Gauss-Jordan elimination methods. Follow the detailed step-by-step process, including key matrix operations and elimination steps. Suitable for college-level students studying linear algebra.

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