Midterm Study guide Answers

Disclaimer: this answers were created by humans. If friend find any kind of errors in them, or even suspect an error, you re welcome let her instructor recognize for your very own sake, and also for the advantage of her colleagues.

These room not complete solutions. Remember the you will need to present steps once carrying out row reductions and justify her answers.

1. (a) (i)

*

(ii)

*

(iii)

*

(iv) The equipment is

*

(b)
(i)

*

(ii)

*

(iii)

*

(iv) The equipment is x =

*

2.
(a)
*
~
*

Since there is not a pivot in every row as soon as the matrixis heat reduced,then the columns of the matrix will not expectations R3.

Note that there is no a pivot in every obelisk of thematrix.So, as soon as augmented to be a homogenous system, there will certainly be a freevariable(x3), and also the mechanism will have a nontrivial solution. So, thecolumnsof the matrix are linearly dependent.

(b)

*
~
*

Since there is a pivot in every row as soon as the procession isrow reduced,then the columns of the matrix will expectancy R3.

Note that there is not a pivot in every obelisk of thematrix.So, as soon as augmented to be a homogenous system, there will certainly be a freevariable(x4), and the system will have a nontrivial solution. Thus, thecolumnsof the matrix room linearly dependent. The is also feasible to seethat there will certainly be a totally free variable because there are more vectors thanentriesin every vector.

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(c)

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

(d)

*
~
*

Since over there is a pivot in every row once the procession isrow reduced,then the columns the the procession will expectancy R2.

Note that there is no a pivot in every pillar of thematrix.So, as soon as augmented to be a homogenous system, there will be a freevariable(x3), and the mechanism will have actually a nontrivial solution. So, thecolumnsof the matrix are linearly dependent. Again, that is also possibleto view that there will be a cost-free variable, due to the fact that there are more vectorsthan entries in each vector.

(e)

The vectors perform not span R2. You canconsider the 2x 2 matrix. Once 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 the column.)

(f)

Since over there are just two vectors, and also the vectors arenot multiplesof every other, then the vectors room linearly independent. Thus,therewill it is in a pivot in every column when the 2 x 2 procession is rowreduced.

Since the is well-known that there room 2 pivots because that this 2 x 2matrix (becausethere is one in each column), then we understand that there is a pivot ineveryrow (since there room two rows). Thus, the vectors expectations R2.

3.
(a) Echelon form:

*

(i) over there is no systems if h = 8. (ii) over there is a distinct solution if h is no 8. (iii) because that the device to have countless solutions, over there would need to bea cost-free variable. There is no means for the to take place in this system, sothere is no worth of h that renders this system have infinitelymanysolutions.

(b) Echelon form:

*

(i) there is no equipment if h – 1 = 0 and 5 – kis notzero, for this reason no equipment if h = 1 and also k is no 5. (ii) there is a unique solution if h – 1 is no zero and 5 – kis any kind of real number, for this reason a unique solution exists if h is not 1and kis any kind of real number. (iii) for the device to have plenty of solutions, there would need to bea totally free variable. The system has plenty of solutions if h = 1 and k= 5.

4.
(a) Echelon form:

*

The vector v3 is in the expectations v1, v2 as lengthy as thesystem is consistent,so we require h - 35 = 0 therefore h = 35. Also, for the vectors to be linearlydependent,the system would need a free variable. Therefore v1, v2, v3 is linearlydependentif h = 35.

(b) Echelon form:

*

The vector v3 is in the expectations v1, v2 as long asthe mechanism isconsistent, and this system is consistent for all h. Also, v1, v2, v3is linearly dependency for every h.

5.
(a) Echelon form:

*

This mechanism is regular when h = 12. Note: Thesolution wouldnot be unique.

(b) Echelon form:

*

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

(c) Echelon form:

*

Consistent as soon as h is not equal come -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 shaft vector that A is in R7, so thevectorsare not even in R4 . So, pivots have nothing come dowithit. The vectors room not in the space, much less able come spanit. (c) No, the columns that A will not span R7 .If there are 7 pivots (a pivot in every row), climate A will expectations R7 .However, due to the fact that A has only 4 columns, that is not possible to have actually morethan4 pivots.

7. (a) The columns of B are linearly dependent regardless of thenumber that pivots. B must have 8 pivots in order for its columns to be linearlyindependent(a pivot in every column). However, that is not feasible for thistohappen, since there are only 5 rows. Because it is not feasible tohave a pivot in every column, it is not feasible for the columns that Btoform a linearly live independence set. (b) Yes, the columns the B will expectations R5 if there space 5 pivots(a pivot in every row). (c) No, each shaft vector of B is in R5, so thevectorsare not also in R8. So, pivots have actually nothing to carry out withit. The vectors space not in the space, much much less able come spanit.

8. Because that x1v1+ x2v2+ x3v3 + x4v4+ x5v5 = 0, it is feasible to have actually anontrivialsolution.

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

Or, if girlfriend augment v1, v2, v3, v4, v5andthe zero vector to form a matrix, the matrix have the right to not have a pivot inthethird column since the 3rd column is every zeros. Hence the system willhave a cost-free variable so the columns the the matrix are linearlydependent.
9. (a) decreased echelon form:
*

The systems is

x =

*
.

(Trivial Solution)

(b) lessened echelon form:

*

The equipment is

x =

*

(Trivial Solution)

(c) diminished echelon form:

*

The systems is

*

(Nontrivial Solution)

10.
(a) Augment and reduce to lessened echelon form:

*
~
*

This represents a continual system.

The systems is x =

*
.

Since there are totally free variables, the systems is notunique.

(b) Augment and also reduce to reduced echelon form:

*
~
*
This represents a consistent system with distinctive solution: x=
*
.

(c)

*
~
*
The last heat leads to a contradiction, 0 = 1. No solution is possible,so the device is inconsistent.

(d)

*
~
*

Homogeneous systems are always consistent. This systemhas a complimentary variable,so there are nontrivial solutions: x =

*
. The systems is not unique.

11. One echelon kind is:

*

The last row leads to a contradiction, 0 = 1. No equipment ispossible,so the mechanism is inconsistent.

12. (a) Echelon (b) Echelon (c)Reduced echelon(d) no

13.

*

14. This matrix is already in diminished echelon form. Thegeneral solution:

x1 = 2x2 - 2 x2 is free x3 = 6 x4 = 1
15. (a) Yes. One echelon kind of A is
*

so the system has actually a solution. We can uncover weights thatallow us to create bas a linear combination of A’s columns.

(b) No. An echelon type of A is

*
.

there is a contradiction in the last row, 0 = 1. For this reason thesystem has actually nosolution. Us cannot discover weights that enable us come write b together alinearcombination the A’s columns.

16. Augment < v1 v3 v2>and reduce to reduced echelon form:

*

This device is regular so v2 is in spanv1, v3withc1 = -3 and c3 = 2. This permits us to create v2as a linear combination of v1 and also 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 indistinguishable to saying that v2 is inSpanv1, v2, v3, v4, v5
18. (a) abdominal =
*

(b)

Ab is undefined, due to the fact that A is 2X3 and also b is2X1.

19. Lots of options for this. Below 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 also reduce to reduced echelon form:
*
~
*
So, x =
*
,and x is unique.

21. (a) due to the fact that T is a mapping native R2 into R7 by therule T(x) = Ax, climate T plot upon anarbitraryvector x in R2 and transforms it into a vector in R7.Thus, x is 2 x 1 and also Ax is 7 x 1. In order because that thematrix multiplication to be defined, A must have actually 2columns.Since the result vector is 7 x 1, then A must have 7rows.Thus, A should be a 7 x 2 matrix.

(b) due to the fact that T is a mapping indigenous R4 into R3by therule T(x) = Ax, then T action 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 result vector is 3 x 1, climate A must have 3rows.Thus, A must be a 3 x 4 matrix.

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

*
~
*
sox =
*

23. (a) x1v + x2w + x3z= <v w z>

*
= Ax for this reason 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 conventional matrix is

A =

*
.

27. The conventional matrix is

A =

*
.

28.

(a)When A is heat reduced, over there is no a pivot in everyrow. So, thecolumns of A carry out not expectations R4. Thus, T does no map R3onto R4.

(b)When A is heat reduced, over there is a pivot in everycolumn, for this reason the columnsof A are linearly independent.Thus, T is one-to-one.

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29.

(a)Since A has actually a pivot in every row, the columns that Aspan R3;thus, T maps R3 ~ above R3.

(b) due to the fact that A has actually a pivot in every column, the columns ofA space linearlyindependent; thus, T is a one-to-one mapping.