math111_logo Row Echelon Form


Exercise Compute the (reduced) row echelon form and list the pivot columns.

1)

[ 0 ]

Answer The matrix is a reduced row echelon form, with no pivot columns.

2)

[ 0 1 ]

Answer The matrix is a reduced row echelon form, with [col 2] pivot.

3)

[ 1 2 ]

Answer The matrix is a reduced row echelon form, with [col 1] pivot.

4)

[ 2 1 ]

Answer The matrix is a row echelon form, but not reduced. By (1/2)[row 1], we get the reduced row echelon form.

[ 1 1/2 ]

[col 1] is pivot.

5)

[ 0 ]
1

Answer The matrix is not a row echelon form. By exchanging the two rows, we get a row echelon form

[ 1 ]
0

which is in fact reduced. [col 1] is pivot.

6)

[ 1 ]
2

Answer The matrix is not a row echelon form. By (-2)[row 1] + [row 2], we get a row echelon form

[ 1 ]
0

which is in fact reduced. [col 1] is pivot.

7)

[ 2 ]
1

Answer The matrix is not a row echelon form. By (-1/2)[row 1] + [row 2], we get a row echelon form

[ 2 ]
0

which is not yet reduced. By further doing (1/2)[row 1], we get the reduced row echelon form.

[ 1 ]
0

[col 1] is pivot.

8)

[ 0 -1 ]
0 0
0 0

Answer The matrix is a row echelon form, but not reduced. By (-1)[row 1], we get the reduced row echelon form.

[ 0 1 ]
0 0
0 0

[col 2] is pivot.

9)

[ 0 0 0 1 ]
0 0 1 0
0 1 0 0
1 0 0 0

Answer The matrix is not a row echelon form. By [row 1] ↔ [row 4] and [row 2] ↔ [row 3], we get the reduced row echelon form.

[ 1 0 0 0 ]
0 1 0 0
0 0 1 0
0 0 0 1

All columns are pivot.

10)

[ 1 2 ]
0 0
0 0
0 0

Answer The matrix is a reduced row echelon form, with [col 1] pivot.

11)

[ 0 1 0 2 0 ]
0 0 1 0 2
0 0 0 1 0
0 0 0 0 0

Answer The matrix is a reduced row echelon form, with [col 2], [col 3], [col 4] pivot.

12)

[ 1 0 0 0 ]
-1 1 0 0
0 -1 1 0
0 0 -1 1

Answer The matrix is not a row echelon form. By [row 1] + [row 2], [row 2] + [row 3], [row 3] + [row 4], we get the reduced row echelon form.

[ 1 0 0 0 ]
0 1 0 0
0 0 1 0
0 0 0 1

All columns are pivot.

13)

[ 1 2 1 1 ]
2 1 1 0
0 3 1 0
3 0 1 0

Answer The matrix is not a row echelon form. By (-2)[row 1] + [row 2] and (-3)[row 1] + [row 4], we have

[ 1 2 1 1 ]
0 -3 -1 -2
0 3 1 0
0 -6 -2 -3

Then by [row 3] + [row 2], 2[row 3] + [row 4], and [row 2] ↔ [row 3], we get

[ 1 2 1 1 ]
0 3 1 0
0 0 0 -2
0 0 0 1

Finally, by 2[row 4] + [row 3] and [row 3] ↔ [row 4], we get a row echelon form.

[ 1 2 1 1 ]
0 3 1 0
0 0 0 1
0 0 0 0

[col 1], [col 2], [col 4] are pivot.

To compute the reduced row echelon form, we use (-1)[row 3] + [row 1] to make [col 4] into the right form.

[ 1 2 1 0 ]
0 3 1 0
0 0 0 1
0 0 0 0

By (-2/3)[row 2] + [row 1] and (1/3)[row 2], we make [col 2] into the right form.

[ 1 0 1/3 0 ]
0 1 1/3 0
0 0 0 1
0 0 0 0

This is the reduced row echelon form.

14)

[ 1 0 1 2 ]
1 4 -3 -2
1 -1 2 3

Answer The matrix is not a row echelon form. By (-1)[row 1] + [row 2] and (-1)[row 1] + [row 3], we have

[ 1 0 1 2 ]
0 4 -4 -4
0 -1 1 1

By 4[row 3] + [row 2], [row 2] ↔ [row 3], we have

[ 1 0 1 2 ]
0 -1 1 1
0 0 0 0

This is a row echelon form. [col 1] and [col 2] are pivot.

By (-1)[row 2], we get the reduced row echelon form.

[ 1 0 1 2 ]
0 1 -1 -1
0 0 0 0

15)

[ 2 -3 6 2 5 ]
-2 3 -3 -3 -4
4 -6 9 5 9
-2 3 3 -4 1

Answer The matrix is not a row echelon form. By [row 1] + [row 2], (-2)[row 1] + [row 3], [row 1] + [row 4], we have

[ 2 -3 6 2 5 ]
0 0 3 -1 1
0 0 -3 1 -1
0 0 9 -2 6

By [row 2] + [row 3], (-3)[row 2] + [row 4], [row 3] ↔ [row 4], we have

[ 2 -3 6 2 5 ]
0 0 3 -1 1
0 0 0 1 3
0 0 0 0 0

This is a row echelon form. [col 1], [col 3], [col 4] are pivot.

To get the reduced row echelon form, we do [row 3] + [row 2], (-2)[row 3] + [row 1] to get

[ 2 -3 6 0 -1 ]
0 0 3 0 4
0 0 0 1 3
0 0 0 0 0

Then by (-2)[row 2] + [row 1], (1/3)[row 2], we have

[ 2 -3 0 0 -9 ]
0 0 1 0 4/3
0 0 0 1 3
0 0 0 0 0

Finally, (1/2)[row 1] gives us the reduced row echelon form.

[ 1 -3/2 0 0 -9/2 ]
0 0 1 0 4/3
0 0 0 1 3
0 0 0 0 0

16)

[ 1 -2 1 1 2 ]
-1 3 0 2 -2
0 1 1 3 4
1 2 5 13 5

Answer The matrix is not a row echelon form. By [row 1] + [row 2], (-1)[row 1] + [row 4], we have

[ 1 -2 1 1 2 ]
0 1 1 3 0
0 1 1 3 4
0 4 4 12 3

By (-1)[row 2] + [row 3], (-4)[row 2] + [row 4], (-3/4)[row 3] + [row 4], we get a row echelon form.

[ 1 -2 1 1 2 ]
0 1 1 3 0
0 0 0 0 4
0 0 0 0 0

[col 1], [col 2], [col 5] are pivot.

To get the reduced row echelon form, we do (-1/2)[row 3] + [row 1], (1/4)[row 3] to get

[ 1 -2 1 1 0 ]
0 1 1 3 0
0 0 0 0 1
0 0 0 0 0

Then 2[row 2] + [row 1] give us the reduced row echelon form.

[ 1 0 3 7 0 ]
0 1 1 3 0
0 0 0 0 1
0 0 0 0 0

17)

[ 1 -1 0 1 ]
-2 3 1 2
1 0 1 5
1 2 3 13
2 -2 4 5

Answer The matrix is not a row echelon form. By 2[row 1] + [row 2], (-1)[row 1] + [row 3], (-1)[row 1] + [row 4],(-2)[row 1] + [row 5], we have

[ 1 -1 0 1 ]
0 1 1 4
0 1 1 4
0 3 3 12
0 0 4 3

By (-1)[row 2] + [row 3], (-3)[row 2] + [row 4], [row 5], [row 3] ↔ [row 5], we get a row echelon form.

[ 1 -1 0 1 ]
0 1 1 4
0 0 4 3
0 0 0 0
0 0 0 0

[col 1], [col 2], [col 3] are pivot.

We may further use (-1/4)[row 3] + [row 2], [row 2] + [row 1], 1/4[row 3] to get the reduced row echelon form.

[ 1 0 0 17/4 ]
0 1 0 13/4
0 0 1 3/4
0 0 0 0
0 0 0 0

18)

[ 1 2 3 0 1 1 ]
0 3 2 1 -1 0
-1 1 2 -2 1 0
1 -2 -2 1 0 0
4 1 3 2 2 1

Answer The matrix is not a row echelon form. By [row 3] + [row 4], [row 1] + [row 4], (-4)[row 1] + [row 5], we have

[ 1 2 3 0 1 1 ]
0 3 2 1 -1 0
0 3 5 -2 2 1
0 -1 0 -1 1 0
0 -7 -9 2 -2 -3

By 3[row 4] + [row 2], 3[row 4] + [row 3], (-7)[row 4] + [row 5], [row 2] ↔ [row 4], we get

[ 1 2 3 0 1 1 ]
0 -1 0 -1 1 0
0 0 5 -5 5 1
0 0 2 -2 2 0
0 0 -9 9 -9 -3

By (-5/2)[row 4] + [row 3], (9/2)[row 4] + [row 5], (1/2)[row 4], [row 3] ↔ [row 4], we get

[ 1 2 3 0 1 1 ]
0 -1 0 -1 1 0
0 0 1 -1 1 0
0 0 0 0 0 1
0 0 0 0 0 -3

By 3[row 4] + [row 5], we get a row echelon form.

[ 1 2 3 0 1 1 ]
0 -1 0 -1 1 0
0 0 1 -1 1 0
0 0 0 0 0 1
0 0 0 0 0 0

[col 1], [col 2], [col 3] [col 6] are pivot.

We may use (-1)[row 4] + [row 1], (-3)[row 3] + [row 1], 2[row 2] + [row 1], (-1)[row 2] to get the reduced row echelon form.

[ 1 0 0 1 0 0 ]
0 1 0 1 -1 0
0 0 1 -1 1 0
0 0 0 0 0 1
0 0 0 0 0 0