Midterm Study Guide Answers

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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 matrixis row reduced,then the columns of the matrix will not span R3.

Note that there is not a pivot in every column of thematrix.So, when augmented to be a homogenous system, there will be a freevariable(x3), and the system will have a nontrivial solution. So, thecolumnsof the matrix are linearly dependent.

(b)

*
~
*

Since there is a pivot in every row when the matrix isrow reduced,then the columns of the matrix will span R3.

Note that there is not a pivot in every column of thematrix.So, when augmented to be a homogenous system, there will be a freevariable(x4), and the system will have a nontrivial solution. Thus, thecolumnsof the matrix are linearly dependent. It is also possible to seethat there will be a free variable since there are more vectors thanentriesin each vector.

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(c)

Since there are only two vectors, it is not possible tospan R4.Consider the 4 x 2 matrix. It would not be possible to have apivotin every row when the matrix is row reduced. Also, since there are only two vectors in the set, it can be notedupon visual inspection that the vectors are linearly independent, sinceneither is a multiple of the other. Thus, the homogeneous system willnothave any free variables, and the system will have only the trivialsolution.

(d)

*
~
*

Since there is a pivot in every row when the matrix isrow reduced,then the columns of the matrix will span R2.

Note that there is not a pivot in every column of thematrix.So, when augmented to be a homogenous system, there will be a freevariable(x3), and the system will have a nontrivial solution. So, thecolumnsof the matrix are linearly dependent. Again, it is also possibleto see that there will be a free variable, since there are more vectorsthan entries in each vector.

(e)

The vectors do not span R2. You canconsider the 2x 2 matrix. When row reduced, there will not be a pivot in everyrow.

Since the zero vector is in the set, the vectors are notlinearly independent.(There is no pivot in that column.)

(f)

Since there are only two vectors, and the vectors arenot multiplesof each other, then the vectors are linearly independent. Thus,therewill be a pivot in every column when the 2 x 2 matrix is rowreduced.

Since it is known that there are 2 pivots for this 2 x 2matrix (becausethere is one in each column), then we know that there is a pivot ineveryrow (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 bea free variable. There is no way for that to happen in this system, sothere is no value of h that makes this system have infinitelymanysolutions.

(b) Echelon form:

*

(i) There is no solution if h – 1 = 0 and 5 – kis notzero, 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 – kis any real number, so a unique solution exists if h is not 1and kis any real number. (iii) For the system to have many solutions, there would have to bea 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 thesystem is consistent,so we need h - 35 = 0 so h = 35. Also, for the vectors to be linearlydependent,the system would need a free variable. So {v1, v2, v3} is linearlydependentif h = 35.

(b) Echelon form:

*

The vector v3 is in the Span {v1, v2} as long asthe system isconsistent, 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: Thesolution wouldnot be unique.

(b) Echelon form:

*

This system is consistent for all h. Note: Thesolution is unique.

(c) Echelon form:

*

Consistent when h is not equal to -14. Note: Thesolution wouldbe unique.

6. (a) A must have 4 pivots in order for its columns to be linearlyindependent (a pivot in every column). (b) No, each column vector of A is in R7, so thevectorsare not even in R4 . So, pivots have nothing to dowithit. The vectors are not in the space, much less able to spanit. (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 morethan4 pivots.

7. (a) The columns of B are linearly dependent regardless of thenumber of pivots. B must have 8 pivots in order for its columns to be linearlyindependent(a pivot in every column). However, it is not possible for thistohappen, since there are only 5 rows. Since it is not possible tohave a pivot in every column, it is not possible for the columns of Btoform 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 thevectorsare not even in R8. So, pivots have nothing to do withit. The vectors are not in the space, much less able to spanit.

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

For instance, x1= x2 = x4= x5 = 0 and x3 = 1 is anontrivialsolution to this equation. Since x1v1+ x2v2+ x3v3 + x4v4+ x5v5 = 0 does not have only the trivialsolution, then the vectors v1, v2, v3, v4, v5form a linearly dependent set.

Or, if you augment v1, v2, v3, v4, v5andthe zero vector to form a matrix, that matrix can not have a pivot inthethird column since the third column is all zeros. Thus the system willhave a free variable so the columns of the matrix are linearlydependent.
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 notunique.

(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 systemhas 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 ispossible,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. Thegeneral 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 thatallow us to write bas 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 thesystem has nosolution. We cannot find weights that allow us to write b as alinearcombination 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}withc1 = -3 and c3 = 2. This allows us to write v2as 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 inSpan{v1, v2, v3, v4, v5}
18. (a) Ab =
*

(b)

Ab is undefined, since A is 2X3 and b is2X1.

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 therule T(x) = Ax, then T acts upon anarbitraryvector 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 thematrix multiplication to be defined, A must have 2columns.Since the resulting vector is 7 x 1, then A must have 7rows.Thus, A must be a 7 x 2 matrix.

(b) Since T is a mapping from R4 into R3by therule T(x) = Ax, then T acts upon anarbitraryvector 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 thematrix multiplication to be defined, A must have 4columns.Since the resulting vector is 3 x 1, then A must have 3rows.Thus, A must be a 3 x 4 matrix.

22. Augment A with the zero vector and reduce to reducedechelonform:

*
~
*
sox =
*

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 everyrow. So, thecolumns of A do not span R4. Thus, T does not map R3onto R4.

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

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29.

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

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