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 R3. 

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 R3. 

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 R4.  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 R2. 

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 R2.  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 R2. 
 
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 R7, so the vectors are not even in R4 .  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 R7 .  If there are 7 pivots (a pivot in every row), then A will span R7 .  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 R5 if there are 5 pivots (a pivot in every row).
(c)  No, each column vector of B is in R5, so the vectors are not even in R8.  So, pivots have nothing to do with it.  The vectors are not in the space, much less able to span it. 

8. For x1v1+ x2v2 + x3v3 + x4v4 + x5v5 = 0, it is possible to have a nontrivial solution. 

For instance, x1= x2 = x4 = x5 = 0 and x3 = 1 is a nontrivial solution to this equation. 
Since x1v1+ x2v2 + x3v3 + x4v4 + x5v5 = 0 does not have only the trivial solution, then the vectors v1, v2, v3, v4, v5 form a linearly dependent set.

Or, if you augment v1, v2, v3, v4, v5and 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:

x1 = 2x2 - 2
x2 is free
x3 = 6
x4 = 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 [ v1  v3  v2] and reduce to reduced echelon form:

This system is consistent so v2 is in span{v1, v3}with c1 = -3 and c3 = 2. This allows us to write v2 as a linear combination of v1 and v3: v2 = c1v1 +  c3v3 = -3v1 + 2v3, or,

.

17. 0v1+ v2 + 0v3 + 0v4 + 0v5 =  v2, so v2 is a linear combination of the vectors v1, v2, v3, v4, v5 . This is equivalent to saying that  v2 is in Span {v1, v2, v3, v4, v5}
 
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:

<>0v1 + 0v2 = 0 = (0, 0, 0, 0) 
1v1 + 0v2 = v1 = (5, 0, -1, 3) 
0v1 + 3v2 = 3v2 = (0, 12, 6, 3)
1v1 + 1v2 = (5, 4, 1, 4)
3v1 - 2v2 = (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 R2 into R7 by the rule T(x) = Ax, then T acts upon an arbitrary vector x in R2 and transforms it into a vector in R7.  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 R4 into R3 by the rule  T(x) = Ax, then T acts upon an arbitrary vector x in R4 and transforms it into a vector in R3.  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) x1v + x2w + x3z  =  [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 R4. Thus, T does not map R3 onto R4. 

(b)When A is row reduced, there is a pivot in every column, so the columns of A are linearly independent.Thus, T is one-to-one. 

 29.

(a)Since A has a pivot in every row, the columns of A span  R3; thus, T maps  R3 onto  R3. 

(b) Since A has a pivot in every column, the columns of A are linearly independent; thus, T is a one-to-one mapping.