Math 1553 Worksheet (Reading Day) Problem Set, Fall 2025/26 Solutions
1. T/F: If {u, v, w} is a set of linearly dependent vectors, then w must be a linear com-
bination of u and v.
Solution.
False. For example, in R2 take u = v = e1 and w = e2.
2. Find the value of k that makes the following vectors linearly dependent:
! ! !
−3 3 3
0 , −3 , −1 .
3 k −1
Solution.
Take the matrix A whose columns are the vectors and row-reduce a bit:
! !
−3 3 3 R3=R3+R1
−3 3 3
A= 0 −3 −1 −−−−−→ 0 −3 −1 .
3 k −1 0 k+3 2
This matrix will have three pivots unless the second and third rows are multiples
of each other. This means the third row is −2 times the second, so −3(−2) = k + 3,
hence k = 3. Alternatively, the student could have computed det(A) = 9 − 3k, so
k = 3.
3. T/F: If {u, v} is a basis for a subspace W , then {u − v, u + v} is also a basis for W .
Solution.
True. This was almost directly taken from a Webwork.
4. Which of the follo w ing are subspaces of R ?
4
xy
(1) The set W = in R4 : 2x − y − z = 0 .
z
w • ‹ • ‹
1 0 1 0 1
(2) The set of solutions to x= .
0 3 0 −1 0
Solution.
(1) is a subspace, since W = Nul 2 −1 −1 0 . But (2) is not, since we see
immediately that the solution set does not contain the zero vector.
1
, 2 SOLUTIONS
!
a
5. T/F: Let W be the set of vectors b in R3 with abc = 0. Then W is closed under
c
addition.
Solution.
False. Taken directly from Webwork.
• ‹
0 −1
6. Counterclockise rotation by 90◦:
1 0
• ‹
0 1 .
Reflection about y = x : 1 0
• ‹
0 1
Clockwise rotation by 90◦: −1 0 .
• ‹
Reflection across the x -axis: 10 −10 .
• ‹
−1 0
Reflection across the y-axis: .
0 1
7. Find k so that the matrix transformation corresponding to the following matrix is
not onto: •1 3 9 ‹
.
2 6 k
Solution.
The columns will span R2 unless they are all collinear, in which case k = 18. Alter-
natively we could find k = 18 by row-reducing and determining when the matrix
will fail to have a pivot in the second row.
8. Find the nonzero value of k so that the matrix is not invertible:
!
1 −1 0
k k2 0 .
−1 1 5
Solution.
We compute the determinant to be 5k(k + 1), which is zero when k = −1. (Note
this problem asked for a nonzero value of k, so k = 0 is incorrect)
9. Taken almost verbatim from Webwork.
“For each y in Rn, there is at most one x in Rm so that T (x ) = y.” This means T is
1 − 1.
“For each y in Rn, there is at least one x in Rm so that T (x ) = y.” This means
1. T/F: If {u, v, w} is a set of linearly dependent vectors, then w must be a linear com-
bination of u and v.
Solution.
False. For example, in R2 take u = v = e1 and w = e2.
2. Find the value of k that makes the following vectors linearly dependent:
! ! !
−3 3 3
0 , −3 , −1 .
3 k −1
Solution.
Take the matrix A whose columns are the vectors and row-reduce a bit:
! !
−3 3 3 R3=R3+R1
−3 3 3
A= 0 −3 −1 −−−−−→ 0 −3 −1 .
3 k −1 0 k+3 2
This matrix will have three pivots unless the second and third rows are multiples
of each other. This means the third row is −2 times the second, so −3(−2) = k + 3,
hence k = 3. Alternatively, the student could have computed det(A) = 9 − 3k, so
k = 3.
3. T/F: If {u, v} is a basis for a subspace W , then {u − v, u + v} is also a basis for W .
Solution.
True. This was almost directly taken from a Webwork.
4. Which of the follo w ing are subspaces of R ?
4
xy
(1) The set W = in R4 : 2x − y − z = 0 .
z
w • ‹ • ‹
1 0 1 0 1
(2) The set of solutions to x= .
0 3 0 −1 0
Solution.
(1) is a subspace, since W = Nul 2 −1 −1 0 . But (2) is not, since we see
immediately that the solution set does not contain the zero vector.
1
, 2 SOLUTIONS
!
a
5. T/F: Let W be the set of vectors b in R3 with abc = 0. Then W is closed under
c
addition.
Solution.
False. Taken directly from Webwork.
• ‹
0 −1
6. Counterclockise rotation by 90◦:
1 0
• ‹
0 1 .
Reflection about y = x : 1 0
• ‹
0 1
Clockwise rotation by 90◦: −1 0 .
• ‹
Reflection across the x -axis: 10 −10 .
• ‹
−1 0
Reflection across the y-axis: .
0 1
7. Find k so that the matrix transformation corresponding to the following matrix is
not onto: •1 3 9 ‹
.
2 6 k
Solution.
The columns will span R2 unless they are all collinear, in which case k = 18. Alter-
natively we could find k = 18 by row-reducing and determining when the matrix
will fail to have a pivot in the second row.
8. Find the nonzero value of k so that the matrix is not invertible:
!
1 −1 0
k k2 0 .
−1 1 5
Solution.
We compute the determinant to be 5k(k + 1), which is zero when k = −1. (Note
this problem asked for a nonzero value of k, so k = 0 is incorrect)
9. Taken almost verbatim from Webwork.
“For each y in Rn, there is at most one x in Rm so that T (x ) = y.” This means T is
1 − 1.
“For each y in Rn, there is at least one x in Rm so that T (x ) = y.” This means
