what is the reduce row echeleon form of 1 −2 8 −1 −1 −5 −7 | 0 0 0 1 −7 0 7 | 0 0 0 0 0 1 9 | sorry its 3x7

To find the reduced row echelon form (RREF) of the given \(3 \times 7\) matrix, let me write it down clearly:

\[
\begin{pmatrix}
1 & -2 & 8 & -1 & -1 & -5 & -7 \\
0 & 0  & 1 & -7 & 0  & 7  & 0 \\
0 & 0  & 0 & 0  & 0  & 1  & 9 
\end{pmatrix}
\]

Now, let's perform row operations to bring this matrix into its reduced row echelon form (RREF).

### Step 1: The matrix already has a leading 1 in the first row, first column, so no changes are needed for the first row.

\[
\begin{pmatrix}
1 & -2 & 8 & -1 & -1 & -5 & -7 \\
0 & 0  & 1 & -7 & 0  & 7  & 0 \\
0 & 0  & 0 & 0  & 0  & 1  & 9 
\end{pmatrix}
\]

### Step 2: Eliminate the 8 in the first row, third column using the second row. Perform the operation:
\[
R_1 \rightarrow R_1 - 8 \cdot R_2
\]
This gives:
\[
R_1 = \left( 1, -2, 0, 55, -1, -61, -7 \right)
\]

\[
\begin{pmatrix}
1 & -2 & 0 & 55 & -1 & -61 & -7 \\
0 & 0  & 1 & -7 & 0  & 7  & 0 \\
0 & 0  & 0 & 0  & 0  & 1  & 9 
\end{pmatrix}
\]

### Step 3: Eliminate the -61 in the first row, sixth column using the third row. Perform the operation:
\[
R_1 \rightarrow R_1 + 61 \cdot R_3
\]
This gives:
\[
R_1 = \left( 1, -2, 0, 55, -1, 0, 542 \right)
\]

\[
\begin{pmatrix}
1 & -2 & 0 & 55 & -1 & 0 & 542 \\
0 & 0  & 1 & -7 & 0  & 7  & 0 \\
0 & 0  & 0 & 0  & 0  & 1  & 9 
\end{pmatrix}
\]

### Step 4: Eliminate the 7 in the second row, sixth column using the third row. Perform the operation:
\[
R_2 \rightarrow R_2 - 7 \cdot R_3
\]
This gives:
\[
R_2 = \left( 0, 0, 1, -7, 0, 0, -63 \right)
\]

\[
\begin{pmatrix}
1 & -2 & 0 & 55 & -1 & 0 & 542 \\
0 & 0  & 1 & -7 & 0  & 0 & -63 \\
0 & 0  & 0 & 0  & 0  & 1  & 9 
\end{pmatrix}
\]

### Step 5: The matrix is in reduced row echelon form.

\[
\begin{pmatrix}
1 & -2 & 0 & 55 & -1 & 0 & 542 \\
0 & 0  & 1 & -7 & 0  & 0 & -63 \\
0 & 0  & 0 & 0  & 0  & 1  & 9 
\end{pmatrix}
\]

So, the reduced row echelon form (RREF) of the matrix is:

\[
\begin{pmatrix}
1 & -2 & 0 & 55 & -1 & 0 & 542 \\
0 & 0  & 1 & -7 & 0  & 0 & -63 \\
0 & 0  & 0 & 0  & 0  & 1  & 9 
\end{pmatrix}
\]

Would you like further explanation on how these row operations were performed?

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Here are 5 related questions to deepen your understanding:

1. What is the difference between row echelon form and reduced row echelon form?
2. How can you determine if a system of equations is consistent by looking at the RREF of its augmented matrix?
3. Can every matrix be reduced to row echelon form or reduced row echelon form?
4. What does it mean if a matrix in RREF has a row of zeros?
5. How do you perform back-substitution from a row echelon form to find solutions to a system of equations?

**Tip**: In row reduction, focus on clearing the entries above and below each leading 1 (pivot), column by column!

Learn how to find the reduced row echelon form (RREF) of a 3x7 matrix with a detailed step-by-step solution. This problem covers important concepts in linear algebra, including matrix operations and row reduction techniques. Suitable for college-level students studying linear algebra.

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