Math 310, 1pm
Quiz 6 solutions: 3.4
Find a basis for R3:
~x1=
1
−4
−3
, ~x2=
2
−8
−6
, ~x3=
−1
1
2
, ~x4=
3
0
2
, ~x5=
−4
7
9
To form a basis for R3we need three linearly independent vectors. One could try row reduction
or trial and error using the determinant. I choose row reduction.
Form a matrix for which the columns are the vectors:
1 2 −1 3 −4
−4−8 1 0 7
−3−6 2 2 9
Row reduce until it is evident which columns contain lead ”ones”:
1 2 −1 3 −4
−4−8 1 0 7
−3−6 2 2 9
4R1+R2
−−−−−→
1 2 −1 3 −4
0 0 −3 12 −9
−3−6 2 2 9
3R1+R3
−−−−−→
1 2 −1 3 4
0 0 −3 12 −9
0 0 −1 11 −3
−
1
3R2
−−−−→
1 2 −134
0 0 1 −4 3
0 0 −1 11 −3
R2+R3
−−−−→
1 2 −1−3 4
0 0 1 −4 3
0 0 0 −7 0
Columns 1, 3, and 4 have lead ”ones”, therefore the vectors corresponding to these columns (~x1,
~x3, and ~x4) would be linearly independent.
To see, consider the matrix whose columns are the vectors ~x1,~x3, and ~x4:
1−1 3
−4 1 0
−3 2 2
Removing columns won’t change the result of row operations on the remaining columns,
therefore, from above, the reduced matrix is:
1−1 3
0 1 −4
0 0 −7
Solve the system for, say ~α:
1−1 3 0
0 1 −4 0
0 0 −7 0
Since each column of the coefficient matrix has a lead ”one”, there is a unique solution, which
must be the trivial solution, ~α =~
0.
1
