![0 2 4. [6 pts) (a) (4pts) Find a basis for the span of vectors ui -2 | u,-|-1 | , and u3 | 5 ,u2 = 0 (b) (2 pts) Find the rank and nullity for the matrix A-u u us].](http://img.homeworklib.com/questions/42e07040-0473-11ec-a4d3-795deb8e1d61.png?x-oss-process=image/resize,w_560)
Homework Answers
4. Let A = [u1,u2,u3] =
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3 |
2 |
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1 |
1 |
1 |
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-2 |
-1 |
5 |
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1 |
0 |
1 |
To answer the questions, we will reduce A to its RREF as under:
Interchange the 1st row and the 2nd row
Add 2 times the 1st row to the 3rd row
Add -1 times the 1st row to the 4th row
Multiply the 2nd row by 1/3
Add -1 times the 2nd row to the 3rd row
Add 1 times the 2nd row to the 4th row
Multiply the 3rd row by 3/19
Add -2/3 times the 3rd row to the 4th row
Add -2/3 times the 3rd row to the 2nd row
Add -1 times the 3rd row to the 1st row
Add -1 times the 2nd row to the 1st row
Then the RREF of A is
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1 |
0 |
0 |
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0 |
1 |
9 |
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0 |
1 |
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0 |
0 |
0 |
(a). It is implied by the RREF of A that the vectors u1,u2,u3 are linearly independent so that { u1,u2,u3} is a basis for Span(u1,u2,u3).
(b). The rank of A isequal to the number of non-zero rows in its RREF, i.e. 3.
As per the dimension theorem( rank-nullity theorem), the nullity of A = No. of columns in A - rank(A) = 3-3 = 0.
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