Reduced Row Echelon Form Of A 4x3 Matrix

To find the reduced row echelon form of this matrix:

(1)
\begin{align} \left(\begin{array}{cc} 5 & 4 & 5 & 2 \\ 6 & 5 & 3 & 2 \\ 7 & 8 & 5 & 6 \end{array}\right) \end{align}

1. We start by permuting the first two rows.

(2)
\begin{align} \left(\begin{array}{cc} 6 & 5 & 3 & 2\\ 5 & 4 & 5 & 2 \\ 7 & 8 & 5 & 6 \end{array}\right) \end{align}

2. Subtract the 2nd row from the 1st.

(3)
\begin{align} \left(\begin{array}{cc} 1 & 1 & -2 & 0\\ 5 & 4 & 5 & 2 \\ 7 & 8 & 5 & 6 \end{array}\right) \end{align}

3. Multiply the first row by -5 and add it to the 2nd.

(4)
\begin{align} \left(\begin{array}{cc} 1 & 1 & -2 & 0\\ 0 & -1 & 15 & 2 \\ 7 & 8 & 5 & 6 \end{array}\right) \end{align}

4. Multiply the first row by -7 and add it to the 3nd.

(5)
\begin{align} \left(\begin{array}{cc} 1 & 1 & -2 & 0\\ 0 & -1 & 15 & 2 \\ 0 & 1 & 19 & 6 \end{array}\right) \end{align}

Now we are done with the first column with a pivot in the first row and zeros under that pivot.

5. We now permute the 2nd and 3rd row.

(6)
\begin{align} \left(\begin{array}{cc} 1 & 1 & -2 & 0\\ 0 & 1 & 19 & 6 \\0 & -1 & 15 & 2 \end{array}\right) \end{align}

6. Add the 2nd row to the 3rd row.

(7)
\begin{align} \left(\begin{array}{cc} 1 & 1 & -2 & 0\\ 0 & 1 & 19 & 6 \\0 & 0 & 34 & 8 \end{array}\right) \end{align}

7. Divide the 3rd row by 34.

(8)
\begin{align} \left(\begin{array}{cc} 1 & 1 & -2 & 0\\ 0 & 1 & 19 & 6 \\0 & 0 & 1 & 4/17\end{array}\right) \end{align}

8. Multiply the 3rd row by -19 and add it to the 2nd row.

(9)
\begin{align} \left(\begin{array}{cc} 1 & 1 & -2 & 0\\ 0 & 1 & 0 & 26/17 \\0 & 0 & 1 & 4/17\end{array}\right) \end{align}

9. Multiply the 3rd row by 2 and add it to the first row.

(10)
\begin{align} \left(\begin{array}{cc} 1 & 1 & 0 & 8/17 \\ 0 & 1 & 0 & 26/17 \\0 & 0 & 1 & 4/17\end{array}\right) \end{align}

10. Multiply the 2nd row by -1 and add it to the first row.

(11)
\begin{align} \left(\begin{array}{cc} 1 & 0 & 0 & -18/17 \\ 0 & 1 & 0 & 26/17 \\0 & 0 & 1 & 4/17\end{array}\right) \end{align}

-Erick Rodriguez

Unless otherwise stated, the content of this page is licensed under Creative Commons Attribution-ShareAlike 3.0 License