Vector Worksheets With Answers - Math 125, Exam 3 Page 4
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1
9
3. Let A =
and b = (2, 1, 2).
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0
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(a) (6 pts) Determine if b is in the column space of A. Clearly explain your answer. If it
is, write it as a linear combination of the column vectors.
Solution: If b is in the column space of A, then b can be written as a linear combination of
the vectors. That is b = c
a
+ c
a
+ c
a
where a is the i-th column vector. To solve for
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1
2
2
3
3
the coefficients, we solve the augmented matrix [A b]. [A b] in reduced row echelon form
is
1 0 0
2.5
.
0 1 0
0
0 0 1
2
Therefore, c
= 2.5, c
= 0, and c
= 2. Hence b is in the column space of A and can be
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2
3
written as b = 2.5 a
+ 2 a
.
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3
(b) (4 pts) Determine if b is in the null space of A.
Solution: Recall that if a vector v is in the null space of a matrix A then Av = 0. Here,
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1
9
2
1
A b =
=
.
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4
8
1
0
4
0
4
2
0
Therefore, b is not in the null space of A.
Another approach would have been to find the null space of A. The null space of A
is only the trivial solution (0, 0, 0). Clearly, b = (2, 1, 2) can not be written as a linear
combination of the zero vector.
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