We present the definition of a matrix in row echelon form and a matrix in reduced row echelon form. We then solve examples on how to write a given matrix in row echelon form and then in reduced row echelon form using the three row operations . More questions with detailed solutions are also included.
A matrix in row echelon form follows the following rules:
Example 1
For each matrix, use the rules above to explain whether it is row echelon form or not.
Solution to Example 1
a) The matrix in part a) is not in row echelon form because rule 2 is violated: the leading 1 in row 3 is to the left of the leading 1 in row 2; it should be to the right.
b) The matrix in part b) is not in row row echelon form because rule 3 is violated: row 2 has zeros only and is not at the bottom.
c) The matrix in part c) is in row echelon form.
d) The matrix in part d) is not in row echelon form because rules 2 and 3 are violated. The top row is all zeros and it should at the bottom, row 3 has a leading 1 that is to the left of the leading 1 in row 2. It should be to the right.
e) The matrix in part e) is in row echelon form.
A matrix is in reduced row echelon form if it is in row echelon form and with zeros above and below the leading 1's.
Example 2
Which of the following matrices is in row echelon form and which are in reduced row echelon form?
Solution to Example 2
Row echelon form: a) b) d) because they obey the rules of row echelon form
Reduced row echelon form: b) d) because they obey the rules of row echelon form and they have zeros below and above leading one's in each row.
Note that matrix in a) is in row echelon form but not reduced because above the leading 1 in row 2 there is a 1.
We now use the three row operations listed below to write a given matrix in row echelon form.
Example 3
Use any of the three row operations above to write the matrices in parts a) b) and d) of example 1 in row echelon form.
\( \)\( \)\( \)\( \)
Solution to Example 3
a) Interchange row 2 and 3 and rewrite the matrix as
\(\begin{bmatrix}
1 & 1 & - 1 & -3\\
0 & 1 & - 6 & - 12 \\
0 & 0 & 1 & 2
\end{bmatrix}
\)
b) Interchange row 2 and 3 and rewrite the matrix as
\(\begin{bmatrix}
1 & -1 & 2 & 0\\
0 & 0 & 1 & -9 \\
0 & 0 & 0 & 0
\end{bmatrix}
\)
d) Interchange row 1 and row 3 and rewrite the matrix as
\(\begin{bmatrix}
1 & 6 & - 1 & -1 \\
0 & 1 & -4 & 0\\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{bmatrix}
\)
Example 4
Use any of the three row operations above, or any combinations, to write the matrix \(\begin{bmatrix}
3 & 0 & -3 & 6 \\
2 & 0 & -4 & 2\\
0 & 1 & 1 & 3\\
1 & 1 & -2 & 1
\end{bmatrix}
\) in reduced row echelon form.
Solution to Example 4
Given \(\begin{bmatrix}
3 & 0 & -3 & 6 \\
2 & 0 & -4 & 2\\
0 & 1 & 1 & 3\\
1 & 1 & -2 & 1
\end{bmatrix}
\)
There are two parts in the process of rewriting a matrix in reduced row echelon form.
Part 1: We first rewrite the given matrix in row echelon form
We proceed per column starting from the leftmost one.Part 2: We now continue to rewrite it in reduced row echelon form
Part 1
a) Which of the following matrices are NOT in row echelon form? Explain why.
b) Of the matrices that are in row echelon, which are NOT in reduced row echelon form? Explain why.
Part 2
Rewrite in row reduced form the following matrices.
Part 1
a) Matrices 1. and 3.
In matrix 1., the 1 in column (2) of row(4) makes that matrix NOT a row echelon form.
In matrix 3., the 1 in column (1) of row (2) makes that matrix NOT a row echelon form
b) Matrices 2. and 5.
In matrix 2., the -1 and 2 in row (1) and the 2 in row (2) make that matrix NOT a reduced row echelon form.
In matrix 5., the 1 in column (3) of row (2) makes that matrix NOT a reduced row echelon form.
Part 2
1)
Write the given matrix in row echelon form
\(
\begin{matrix}
\color{red}{\frac{1}{2} R_1} \\
\\
\end{matrix}
\begin{bmatrix}
1 & 2 & 1 \\
-1 & 2 & 0
\end{bmatrix}
\)
\(
\begin{matrix}
\\
\color{red}{R_2+ R_1} \\
\end{matrix}
\begin{bmatrix}
1 & 2 & 1 \\
0 & 4 & 1
\end{bmatrix}
\)
\(
\begin{matrix}
\\
\color{red}{\frac{1}{2} R_2} \\
\end{matrix}
\begin{bmatrix}
1 & 2 & 1 \\
0 & 1 & 1/4
\end{bmatrix}
\)
Write the above matrix in reduced row echelon form
\(
\begin{matrix}
\color{red}{R_1 - 2 R_2} \\
\\
\end{matrix}
\begin{bmatrix}
1 & 0 & 1/2 \\
0 & 1 & 1/4
\end{bmatrix}
\)
2)
Write the given matrix in row echelon form
\(
\begin{matrix}
\color{red}{ - R_1 } \\
\\
\\
\end{matrix}
\begin{bmatrix}
1 & -2 & -1 & 0\\
2 & 1 & 0 & -1\\
5 & 0 & -2 & 6
\end{bmatrix} \)
\(
\begin{matrix}
\\
\color{red}{ R_2 - 2 R_1 } \\
\color{red}{ R_3 - 5 R_1 } \\
\end{matrix}
\begin{bmatrix}
1 & -2 & -1 & 0\\
0 & 5 & 2 & -1\\
0 & 10 & 3 & 6
\end{bmatrix} \)
\(
\begin{matrix}
\\
\color{red}{\frac{1}{5} R_2} \\
\\
\end{matrix}
\begin{bmatrix}
1 & -2 & -1 & 0\\
0 & 1 & 2/5 & -1/5\\
0 & 10 & 3 & 6
\end{bmatrix} \)
\(
\begin{matrix}
\\
\\
\color{red}{ R_3 - 10 R_2 } \\
\end{matrix}
\begin{bmatrix}
1 & -2 & -1 & 0\\
0 & 1 & 2/5 & -1/5\\
0 & 0 & -1 & 8
\end{bmatrix} \)
\(
\begin{matrix}
\\
\\
\color{red}{- R_3}\\
\end{matrix}
\begin{bmatrix}
1 & -2 & -1 & 0\\
0 & 1 & 2/5 & -1/5\\
0 & 0 & 1 & -8
\end{bmatrix} \)
Write the above matrix in reduced row echelon form
\(
\begin{matrix}
\color{red}{R_1 + R_3} \\
\color{red}{ R_2 - (2/5) R_3 } \\
\\
\end{matrix}
\begin{bmatrix}
1 & -2 & 0 & -8\\
0 & 1 & 0 & 3\\
0 & 0 & 1 & -8
\end{bmatrix} \)
\(
\begin{matrix}
\color{red}{R_1 + 2 R_2} \\
\\
\\
\end{matrix}
\begin{bmatrix}
1 & 0 & 0 & - 2\\
0 & 1 & 0 & 3\\
0 & 0 & 1 & -8
\end{bmatrix} \)
3)
Write the given matrix in row echelon form
\(
\begin{matrix}
\color{red}{ - R_1 } \\
\\
\\
\\
\end{matrix}
\begin{bmatrix}
1 & -2 & 0 & -1 & -1\\
1 & -2 & -1 & 2 & 3\\
0 & 1 & 2 & -2 & 1 \\
-1 & 3 & 4 & 2 & 0
\end{bmatrix}
\)
\(
\begin{matrix}
\\
\color{red}{R_2 - R_1}\\
\\
\color{red}{R_4 + R_1}\\
\end{matrix}
\begin{bmatrix}
1 & -2 & 0 & -1 & -1\\
0 & 0 & -1 & 3 & 4\\
0 & 1 & 2 & -2 & 1 \\
0 & 1 & 4 & 1 & -1
\end{bmatrix}
\)
\(
\begin{matrix}
\\
\color{red}{\text{interchange} \; R_2 \; \text{and} \; R_3}\\
\color{red}{\text{interchange} \; R_3 \; \text{and} \; R_2}\\
\color{red}{R_4 - R_3}\\
\end{matrix}
\begin{bmatrix}
1 & -2 & 0 & -1 & -1\\
0 & 1 & 2 & -2 & 1\\
0 & 0 & -1 & 3 & 4 \\
0 & 0 & 2 & 3 & -2
\end{bmatrix}
\)
\(
\begin{matrix}
\\
\\
\color{red}{ - R_3}\\
\color{red}{R_4 +2 R_3}\\
\end{matrix}
\begin{bmatrix}
1 & -2 & 0 & -1 & -1\\
0 & 1 & 2 & -2 & 1\\
0 & 0 & 1 & -3 & -4 \\
0 & 0 & 0 & 9 & 6
\end{bmatrix}
\)
\(
\begin{matrix}
\\
\\
\\
\color{red}{ \frac{1}{9} R_4}\\
\end{matrix}
\begin{bmatrix}
1 & -2 & 0 & -1 & -1\\
0 & 1 & 2 & -2 & 1\\
0 & 0 & 1 & -3 & -4 \\
0 & 0 & 0 & 1 & 2/3
\end{bmatrix}
\)
Write the above matrix in reduced row echelon form
\(
\begin{matrix}
\color{red}{R_1 + R_4}\\
\color{red}{R_2 + 2 R_4}\\
\color{red}{R_3 + 3 R_4} \\
\\
\end{matrix}
\begin{bmatrix}
1 & -2 & 0 & 0 & -1/3\\
0 & 1 & 2 & 0 & 7/3\\
0 & 0 & 1 & 0 & - 2\\
0 & 0 & 0 & 1 & 2/3
\end{bmatrix}
\)
\(
\begin{matrix}
\\
\color{red}{R_2 - 2 R_3}\\
\\
\\
\end{matrix}
\begin{bmatrix}
1 & -2 & 0 & 0 & -1/3\\
0 & 1 & 0 & 0 & 19/3\\
0 & 0 & 1 & 0 & - 2\\
0 & 0 & 0 & 1 & 2/3
\end{bmatrix}
\)
\(
\begin{matrix}
\color{red}{R_1 + 2 R_2}\\
\\
\\
\\
\end{matrix}
\begin{bmatrix}
1 & 0 & 0 & 0 & 37/3\\
0 & 1 & 0 & 0 & 19/3\\
0 & 0 & 1 & 0 & - 2\\
0 & 0 & 0 & 1 & 2/3
\end{bmatrix}
\)