Midterm Study guide Answers
Disclaimer: these answers were developed by humans. If girlfriend find any kind of errors in them, or even suspect one error, please let your instructor know for your very own sake, and for the advantage of her colleagues.
These are not complete solutions. Remember the you will require to present steps when carrying out row reductions and also justify her answers.
1. (a) (i)
|(i) ||(ii) |
|(iii) ||(iv) The solution is x = |
Since over there is not a pivot in every row once the matrixis heat reduced,then the columns the the matrix will not span R3.
Note that there is not a pivot in every obelisk of thematrix.So, as soon as augmented to it is in a homogenous system, there will certainly be a freevariable(x3), and also the device will have actually a nontrivial solution. So, thecolumnsof the matrix space linearly dependent.
Since over there is a pivot in every row when the matrix isrow reduced,then the columns the the procession will span R3.
Note that there is not a pivot in every column of thematrix.So, once augmented to it is in a homogenous system, there will be a freevariable(x4), and also the system will have a nontrivial solution. Thus, thecolumnsof the matrix room linearly dependent. The is also possible to seethat there will be a complimentary variable due to the fact that there are more vectors thanentriesin each vector.
Since over there are just two vectors, it is not feasible tospan R4.Consider the 4 x 2 matrix. It would not be possible to have actually apivotin every row as soon as the procession is row reduced. Also, because there are only two vectors in the set, it deserve to be notedupon visual inspection that the vectors room linearly independent, sinceneither is a multiple of the other. Thus, the homogeneous device willnothave any free variables, and the system will have actually only the trivialsolution.
Since there is a pivot in every row as soon as the matrix isrow reduced,then the columns that the procession will expectations R2.
Note the there is no a pivot in every column of thematrix.So, as soon as augmented to it is in a homogenous system, there will be a freevariable(x3), and the mechanism will have actually a nontrivial solution. So, thecolumnsof the matrix room linearly dependent. Again, that is likewise possibleto watch that there will be a totally free variable, since there are more vectorsthan entries in every vector.
The vectors perform not span R2. You canconsider the 2x 2 matrix. As soon as row reduced, there will not it is in a pivot in everyrow.
Since the zero vector is in the set, the vectors are notlinearly independent.(There is no pivot in that column.)
Since over there are just two vectors, and also the vectors arenot multiplesof every other, climate the vectors are linearly independent. Thus,therewill it is in a pivot in every pillar when the 2 x 2 procession is rowreduced.
Since the is well-known that there are 2 pivots for this 2 x 2matrix (becausethere is one in each column), then we understand that over there is a pivot ineveryrow (since there room two rows). Thus, the vectors expectancy R2.
|(a) Echelon form: |
(i) there is no equipment if h = 8. (ii) there is a distinct solution if h is no 8. (iii) for the device to have countless solutions, over there would have to bea totally free variable. Over there is no way for the to take place in this system, sothere is no worth of h that provides this system have actually infinitelymanysolutions.
|(b) Echelon form: |
(i) there is no equipment if h – 1 = 0 and also 5 – kis notzero, so no systems if h = 1 and k is not 5. (ii) there is a distinct solution if h – 1 is not zero and 5 – kis any real number, so a distinct solution exist if h is no 1and kis any type of real number. (iii) for the device to have numerous solutions, over there would need to bea cost-free variable. The system has numerous solutions if h = 1 and also k= 5.
|(a) Echelon form: |
The vector v3 is in the expectations v1, v2 as lengthy as thesystem is consistent,so we need h - 35 = 0 therefore h = 35. Also, for the vectors to it is in linearlydependent,the system would need a totally 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 system isconsistent, and also this device is continual for every h. Also, v1, v2, v3is linearly dependency for every h.
|(a) Echelon form: |
This system is regular when h = 12. Note: Thesolution wouldnot it is in unique.
|(b) Echelon form: |
This system is continual for all h. Note: Thesolution is unique.
|(c) Echelon form: |
Consistent when h is no equal to -14. Note: Thesolution wouldbe unique.
6. (a) A must have actually 4 pivots in order for its columns to be linearlyindependent (a pivot in every column). (b) No, each obelisk vector of A is in R7, therefore thevectorsare not even in R4 . So, pivots have nothing come dowithit. The vectors room not in the space, much much less able to spanit. (c) No, the columns that A will certainly not span R7 .If there are 7 pivots (a pivot in every row), then A will span R7 .However, due to the fact that A has actually only 4 columns, that is not feasible to have morethan4 pivots.
7. (a) The columns the B room linearly dependent nevertheless of thenumber that pivots. B must have 8 pivots in order for its columns to it is in linearlyindependent(a pivot in every column). However, it is not feasible for thistohappen, since there are only 5 rows. Since it is not feasible tohave a pivot in every column, it is not feasible for the columns of Btoform a linearly live independence set. (b) Yes, the columns the B will expectancy R5 if there room 5 pivots(a pivot in every row). (c) No, each pillar vector that B is in R5, for this reason thevectorsare not also in R8. So, pivots have actually 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 feasible 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 dependence set.
Or, if friend augment v1, v2, v3, v4, v5andthe zero vector to form a matrix, that matrix have the right to not have a pivot inthethird column since the 3rd column is all zeros. Thus the system willhave a free variable so the columns that the matrix room linearlydependent. The solution is x = (Trivial Solution) The systems is x = (Trivial Solution) (Nontrivial Solution) This represents a continual system. The solution is x = Since over there are cost-free variables, the systems is notunique. Homogeneous systems are always consistent. This systemhas a free variable,so there space nontrivial solutions: x =
9. (a) decreased echelon form: (b) reduced echelon form: (c) reduced echelon form: The solution is (a) Augment and also reduce to lessened echelon form: (b) Augment and also reduce to decreased echelon form: (c) (d)
The solution is
The systems is
This represents a continual system.
The solution is x =
Since over there are cost-free variables, the systems is notunique.
Homogeneous systems are always consistent. This systemhas a free variable,so there space nontrivial solutions: x =
The last heat leads to a contradiction, 0 = 1. No systems ispossible,so the system is inconsistent.
12. (a) Echelon (b) Echelon (c)Reduced echelon(d) neither
14. This procession is already in diminished echelon form. Thegeneral solution:
x1 = 2x2 - 2 x2 is free x3 = 6 x4 = 1
|15. (a) Yes. One echelon type of A is |
so the system has actually a solution. We can discover weights thatallow us to compose bas a linear combination of A’s columns.
|(b) No. One echelon form of A is |
there is a contradiction in the critical row, 0 = 1. So thesystem has nosolution. Us cannot find weights that permit us to write b as alinearcombination that A’s columns.
16. Augment < v1 v3 v2>and minimize to reduced echelon form:
This mechanism is continuous so v2 is in spanv1, v3withc1 = -3 and c3 = 2. This allows us to write v2as a linear mix 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 identical to saying the v2 is inSpanv1, v2, v3, v4, v5 Ab is undefined, because A is 2X3 and also b is2X1.
18. (a) ab = (b)
Ab is undefined, because A is 2X3 and also b is2X1.
19. Too many of choices for this. Right here are 5 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 diminished echelon form:
21. (a) due to the fact that T is a mapping indigenous R2 into R7 by therule T(x) = Ax, then T plot upon anarbitraryvector x in R2 and also transforms it right into a vector in R7.Thus, x is 2 x 1 and also Ax is 7 x 1. In order for thematrix multiplication to it is in defined, A must have 2columns.Since the resulting vector is 7 x 1, then A must have 7rows.Thus, A should be a 7 x 2 matrix.
(b) because T is a mapping native R4 right into R3by therule T(x) = Ax, climate T acts upon anarbitraryvector x in R4 and transforms it right 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 need to be a 3 x 4 matrix.
22. Augment A v the zero vector and reduce to reducedechelonform:
23. (a) x1v + x2w + x3z= <v w z>
24. T(x) = Ax = < T(e1) T(e2) T(e3) > ==
|25. The typical matrix is |
A = < T(e1) T(e2) >
|26. The typical matrix is |
|27. The typical matrix is |
(a)When A is row reduced, over there is not a pivot in everyrow. So, thecolumns that A carry out not expectations R4. Thus, T does no map R3onto R4.
(b)When A is row reduced, there is a pivot in everycolumn, therefore the columnsof A are linearly independent.Thus, T is one-to-one.
(a)Since A has a pivot in every row, the columns that Aspan R3;thus, T maps R3 ~ above R3.
(b) due to the fact that A has a pivot in every column, the columns ofA are linearlyindependent; thus, T is a one-to-one mapping.