8.5. LINEAR INDEPENDENCE AND BASES 163
Form the matrix mentioned above.1 2 0 32 1 1 23 0 1 20 1 2 −1
Then the row reduced echelon form of this matrix is
1 0 0 10 1 0 10 0 1 −10 0 0 0
.
Thus not all the columns are pivot columns and so the vectors are not linear independent.Note the fourth column is of the form
1
1000
+1
0100
+(−1)
0010
From Lemma 8.2.5, the same linear relationship exists between the columns of the originalmatrix. Thus
1
1230
+1
2101
+(−1)
0112
=
322−1
.
Note the usefulness of the row reduced echelon form in discovering hidden linear rela-tionships in collections of vectors.
Example 8.5.7 Determine whether the vectors
1230
2101
0112
3220
are
linearly independent. If they are linearly dependent, exhibit one of the vectors as a linearcombination of the others.
The matrix used to find this is 1 2 0 32 1 1 23 0 1 20 1 2 0