Midterm Study Guide Answers
Disclaimer: These answers were created by humans. If you find
any errors in them, or even suspect an error, please let your
instructor know for your own sake, and for the benefit of your
colleagues.
These are not complete solutions. Remember that you will need
to show steps when carrying out row reductions and justify your
answers.
1. (a)
(i)

(ii)

(iii)

(iv) The solution is

(b)
(i)

(ii)

(iii)

(iv) The solution is x =

2.
(a)
~
Since there is not a pivot in every row when the matrix
is row reduced,
then the columns of the matrix will not span R^{3}.
Note that there is not a pivot in every column of the
matrix.
So, when augmented to be a homogenous system, there will be a free
variable
(x3), and the system will have a nontrivial solution. So, the
columns
of the matrix are linearly dependent.

(b)
~
Since there is a pivot in every row when the matrix is
row reduced,
then the columns of the matrix will span R^{3}.
Note that there is not a pivot in every column of the
matrix.
So, when augmented to be a homogenous system, there will be a free
variable
(x4), and the system will have a nontrivial solution. Thus, the
columns
of the matrix are linearly dependent. It is also possible to see
that there will be a free variable since there are more vectors than
entries
in each vector.

(c)
Since there are only two vectors, it is not possible to
span R^{4}.
Consider the 4 x 2 matrix. It would not be possible to have a
pivot
in every row when the matrix is row reduced.
Also, since there are only two vectors in the set, it can be noted
upon visual inspection that the vectors are linearly independent, since
neither is a multiple of the other. Thus, the homogeneous system will
not
have any free variables, and the system will have only the trivial
solution.

(d)
~
Since there is a pivot in every row when the matrix is
row reduced,
then the columns of the matrix will span R^{2}.
Note that there is not a pivot in every column of the
matrix.
So, when augmented to be a homogenous system, there will be a free
variable
(x3), and the system will have a nontrivial solution. So, the
columns
of the matrix are linearly dependent. Again, it is also possible
to see that there will be a free variable, since there are more vectors
than entries in each vector.

(e)
The vectors do not span R^{2}. You can
consider the 2
x 2 matrix. When row reduced, there will not be a pivot in every
row.
Since the zero vector is in the set, the vectors are not
linearly independent.
(There is no pivot in that column.)

(f)
Since there are only two vectors, and the vectors are
not multiples
of each other, then the vectors are linearly independent. Thus,
there
will be a pivot in every column when the 2 x 2 matrix is row
reduced.
Since it is known that there are 2 pivots for this 2 x 2
matrix (because
there is one in each column), then we know that there is a pivot in
every
row (since there are two rows). Thus, the vectors span R^{2}.

3.
(a) Echelon form:
(i) There is no solution if h = 8.
(ii) There is a unique solution if h is not 8.
(iii) For the system to have many solutions, there would have to be
a free variable. There is no way for that to happen in this system, so
there is no value of h that makes this system have infinitely
many
solutions.

(b) Echelon form:
(i) There is no solution if h – 1 = 0 and 5 – k
is not
zero, so no solution if h = 1 and k is not 5.
(ii) There is a unique solution if h – 1 is not zero and 5 – k
is any real number, so a unique solution exists if h is not 1
and k
is any real number.
(iii) For the system to have many solutions, there would have to be
a free variable. The system has many solutions if h = 1 and k
= 5.

4.
(a) Echelon form:
The vector v3 is in the Span {v1, v2} as long as the
system is consistent,
so we need h  35 = 0 so h = 35. Also, for the vectors to be linearly
dependent,
the system would need a free variable. So {v1, v2, v3} is linearly
dependent
if h = 35.

(b) Echelon form:
The vector v3 is in the Span {v1, v2} as long as
the system is
consistent, and this system is consistent for all h. Also, {v1, v2, v3}
is linearly dependent for all h.

5.
(a) Echelon form:
This system is consistent when h = 12. Note: The
solution would
not be unique.

(b) Echelon form:
This system is consistent for all h. Note: The
solution is unique.

(c) Echelon form:
Consistent when h is not equal to 14. Note: The
solution would
be unique.

6.
(a) A must have 4 pivots in order for its columns to be linearly
independent (a pivot in every column).
(b) No, each column vector of A is in R^{7}, so the
vectors
are not even in R^{4 }. So, pivots have nothing to do
with
it. The vectors are not in the space, much less able to span
it.
(c) No, the columns of A will not span R^{7 }.
If there are 7 pivots (a pivot in every row), then A will span R^{7 }.
However, since A has only 4 columns, it is not possible to have more
than
4 pivots.
7.
(a) The columns of B are linearly dependent regardless of the
number of pivots.
B must have 8 pivots in order for its columns to be linearly
independent
(a pivot in every column). However, it is not possible for this
to
happen, since there are only 5 rows. Since it is not possible to
have a pivot in every column, it is not possible for the columns of B
to
form a linearly independent set.
(b) Yes, the columns of B will span R^{5} if there are 5 pivots
(a pivot in every row).
(c) No, each column vector of B is in R^{5}, so the
vectors
are not even in R^{8}. So, pivots have nothing to do with
it. The vectors are not in the space, much less able to span
it.
8. For x_{1}v_{1}+ x_{2}v_{2}
+ x_{3}v_{3} + x_{4}v_{4}
+ x_{5}v_{5} = 0, it is possible to have a
nontrivial
solution.
For instance, x_{1}= x_{2} = x_{4}
= x_{5} = 0 and x_{3} = 1 is a
nontrivial
solution to this equation.
Since x_{1}v_{1}+ x_{2}v_{2}
+ x_{3}v_{3} + x_{4}v_{4}
+ x_{5}v_{5} = 0 does not have only the trivial
solution, then the vectors v_{1}, v_{2},
v_{3}, v_{4}, v_{5}
form a linearly dependent set.
Or, if you augment v_{1}, v_{2},
v_{3}, v_{4}, v_{5}and
the zero vector to form a matrix, that matrix can not have a pivot in
the
third column since the third column is all zeros. Thus the system will
have a free variable so the columns of the matrix are linearly
dependent.
9. (a) Reduced echelon form:
The solution is
x = .
(Trivial Solution)

(b) Reduced echelon form:
The solution is
x =
(Trivial Solution)

(c) Reduced echelon form:

The solution is
(Nontrivial Solution)

10.
(a) Augment and reduce to reduced echelon form:
~
This represents a consistent system.
The solution is x =
.
Since there are free variables, the solution is not
unique.


(b) Augment and reduce to reduced echelon form:
~
This represents a consistent system with unique solution: x
= .


(c)
~
The last row leads to a contradiction, 0 = 1. No solution is possible,
so the system is inconsistent.


(d)
~
Homogeneous systems are always consistent. This system
has a free variable,
so there are nontrivial solutions: x =
.
The solution is not unique.


11. An echelon form is:
The last row leads to a contradiction, 0 = 1. No solution is
possible,
so the system is inconsistent.
12. (a) Echelon (b) Echelon (c)
Reduced echelon
(d) Neither
13.
14. This matrix is already in reduced echelon form. The
general solution:
x_{1} = 2x_{2}  2
x_{2} is free
x_{3} = 6
x_{4} = 1
15. (a) Yes. An echelon form of A is
so the system has a solution. We can find weights that
allow us to write b
as a linear combination of A’s columns.

(b) No. An echelon form of A is
.
there is a contradiction in the last row, 0 = 1. So the
system has no
solution. We cannot find weights that allow us to write b as a
linear
combination of A’s columns.

16. Augment [ v_{1 }v_{3
}v_{2}]
and reduce to reduced echelon form:
This system is consistent so v_{2} is in span{v_{1},
v_{3}}with
c_{1} = 3 and c_{3} = 2. This allows us to write v_{2}
as a linear combination of v_{1} and v_{3}:
v_{2}
= c_{1}v_{1} + c_{3}v_{3}
= 3v_{1} + 2v_{3}, or,
.
17. 0v_{1}+ v_{2} + 0v_{3}
+ 0v_{4} + 0v_{5} = v_{2},
so v_{2} is a linear combination of the vectors v_{1},
v_{2}, v_{3}, v_{4},
v_{5}
. This is equivalent to saying that v_{2} is in
Span
{v_{1}, v_{2}, v_{3}, v_{4},
v_{5}}
18. (a)
Ab =

(b)
Ab is undefined, since A is 2X3 and b is
2X1.

19. Lots of options for this. Here are five of them:
<>0v_{1} + 0v_{2} = 0 = (0,
0, 0,
0)
1v_{1} + 0v_{2} = v_{1}
= (5, 0, 1, 3)
0v_{1} + 3v_{2} = 3v_{2}
= (0, 12, 6, 3)
1v_{1} + 1v_{2} = (5, 4, 1, 4)
3v_{1}  2v_{2} = (15, 0, 3, 9)  (0,
8, 4, 2) = (15, 8, 7, 7)
20. Augment A with b and reduce to reduced echelon form:
~
So, x = ,
and x is unique.
21.
(a) Since T is a mapping from R^{2} into R^{7} by the
rule T(x) = Ax, then T acts upon an
arbitrary
vector x in R^{2} and transforms it into a vector in R^{7}.
Thus, x is 2 x 1 and Ax is 7 x 1. In order for the
matrix multiplication to be defined, A must have 2
columns.
Since the resulting vector is 7 x 1, then A must have 7
rows.
Thus, A must be a 7 x 2 matrix.
(b) Since T is a mapping from R^{4} into R^{3}
by the
rule T(x) = Ax, then T acts upon an
arbitrary
vector x in R^{4} and transforms it into a vector in R^{3}.
Thus, x is 4 x 1 and Ax is 3 x 1. In order for the
matrix multiplication to be defined, A must have 4
columns.
Since the resulting vector is 3 x 1, then A must have 3
rows.
Thus, A must be a 3 x 4 matrix.
22. Augment A with the zero vector and reduce to reduced
echelon
form:
~ so
x =
23.
(a) x_{1}v + x_{2}w + x_{3}z
= [v w z]
= Ax so A =
(b)
T(u) = Au =
24.
T(x) = Ax = [ T(e1) T(e2) T(e3) ] =
=
T(2, 8, 1) = A
= [ T(e1) T(e2) T(e3) ]
=
=
25. The standard matrix is
A = [ T(e1) T(e2) ]
=
.

26. The standard matrix is
A =
.

27. The standard matrix is
A =
.

28.
(a)When A is row reduced, there is not a pivot in every
row. So, the
columns of A do not span R^{4}. Thus, T does not map R^{3}
onto R^{4}.
(b)When A is row reduced, there is a pivot in every
column, so the columns
of A are linearly independent.Thus, T is onetoone.

29.
(a)Since A has a pivot in every row, the columns of A
span R^{3};
thus, T maps R^{3 }onto R^{3}.
(b) Since A has a pivot in every column, the columns of
A are linearly
independent; thus, T is a onetoone mapping.

