Vector Worksheets With Answers - Math 125, Exam 3 Page 13
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193. (2 pts each) Let A denote an m
n matrix. Mark each statement True or False. Justify
each answer.
(a) The null space of A is the solution set of the equation Ax = 0.
Solution: TRUE. This is exactly how we defined the null space on page 68 of the notes.
(b) The column space is the set of all solutions to Ax = b.
Solution: FALSE. The columns space is not the solution vector x. The column space is the
set of vectors b that can written b = Ax. That is b = x
a
+ x
a
+ . . . + x a .
1
1
2
2
(c) If the null space of a 7
6 matrix is 5 dimensional, then the dimension of the column
space is 2.
Solution: FALSE. This is an application of the rank + nullity = the number of columns
theorem. Recall that the column rank equals the row rank and, by definition, the nullity is
the rank of the null space. Therefore, rank + 5 = 6. Hence, the column rank (or dimension
of the column space) is 1. Not 2.
(d) Ax = b is consistent if and only if b is in the column space of A.
Solution: TRUE. Using the augmented matrix [A B] is exactly method used to determine
if b is in the column space of A. The augmented matrix [A B] is equivalent to the matrix
equation Ax = b.
(e) If S = span u
, u
, . . . , u , then u
, u
, . . . , u
is a basis for S.
1
2
1
2
Solution: FALSE. u
, u
, . . . , u
is a basis for S if and only if all the vectors in the set
1
2
are linearly independent.
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