math 3c practice midterm solutions prepared by vince zaccone for campus learning assistance services...

Math 3CPractice Midterm

Solutions

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB



1) Solve the following system of linear equations:

2z4yx3

5z2yx

1zyx

Augmented matrix form

2

5

1

413

211

111

5

4

1

000

120

111

2R3R*3R

1

4

1

120

120

1111R2R*2R

1R33R*3R

Row reduction:

The 3rd row represents the equation 0x+0y+0z = -5. This equation has no solution, so the original system of equations is inconsistent (not solvable).



2) Consider the following matrix:

4000

01300

4010

30226

A

a)Find the determinant of this matrix.b)Is this matrix invertible? If so, find the inverse.c)What is the determinant of the inverse of this matrix?d)Using this information, solve the following system of linear equations:

12w4

26z13

3yw4

2w3y22x6

a) Since A is a triangular matrix, the determinant is the product of the diagonal elements – det(A) = -312

b) Since det(A) is not zero, A is invertible, and we can use row reduction to find the inverse.

1A

41

131

2485

311

61

Identity

131*61*

41

485

*

41

43

*

*

41

41*

*

000

000

1010

0

1000

0100

0010

0001

3R3R

1R1R

000

0100

1010

0221

1000

01300

0010

0006

2R221R1R

000

0100

1010

001

1000

01300

0010

00226

4R42R2R

4R31R1R

000

0100

0010

0001

1000

01300

4010

30226

4R4R

2R2R

1000

0100

0010

0001

4000

01300

4010

30226



2) Consider the following matrix:

4000

01300

4010

30226

A

a)Find the determinant of this matrix.b)Is this matrix invertible? If so, find the inverse.c)What is the determinant of the inverse of this matrix?d)Using this information, solve the following system of linear equations:

12w4

26z13

3yw4

2w3y22x6

c) Since A-1is a triangular matrix, the determinant is the product of the diagonal elements: det(A -1) = -1/312

d) Notice that we have the inverse for the coefficient matrix, so we just need to multiply:

3

2

9

w

z

y

x

12

26

3

2

000

000

1010

0

w

z

y

x

6187

41

131

2485

311

61

3) Consider the following set of vectors in R3

3

2

1

a,

2

1

1

a,

1

1

0

a,

1

0

1

a 4321

a)How many linearly independent vectors are in this set?b)Is the span of this set of vectors R3? c)Is this a basis for R3?d)Is each of the following vectors within the span of this set? If so, express them as a linear combination of the vectors in the set.

4

2

2

v,

3

0

2

v 21

a) Put the column vectors in a matrix and perform elementary row operations:

0000

2110

1101

2R3R3R

2110

2110

1101

1R3R3R

3211

2110

1101**

This has 2 pivot columns, so there are 2 linearly independent vectors in the set.

In other words, the span of this set is 2-dimensional.

b) No, the span is 2-dimensional, and R3 is 3-dimensional.

c) No, this is not a basis because it does not span R3, and there are 4 vectors, not 3.

d) We have several options for this part. One way is to make a good guess. If you notice that v2 is just 2 times a3 then we have our answer for that one:

2

1

1

2

4

2

2

For v1 there is no obvious answer, so we need to try to solve for it. Since we know that our span is only 2-dimensional, we only need two basis vectors. Choose any 2 independent vectors from our set, say a1 and a2, and try to write v1 as a linear combination of them.

ntInconsiste

cc3

c0

c2

c

c

0

c

0

c

3

0

2

acacv

21

2

1

2

2

1

1

22111

We get an inconsistent set of equations, so v1 is not in the span. Prepared by Vince Zaccone


4) Consider the following set of vectors in P2:

x2a,xx1a,xx1a 32

22

1

a)How many linearly independent vectors are in this set?b)Does this set span P2?

c)Is this set a basis for P2?

d)Express each of the standard basis vectors {1, x, x2} as a linear combination of the vectors in the set.

a) These are 3 independent vectors. We can see this if we try to form a linear combination of them, and find that there is no way to cancel them out:

tindependen0ccc

0c2cc

0ccc

0cc

?0)c2cc(x)ccc(x)cc(

?0)x2(c)xx1(c)xx1(c

?0acacac

321

321

321

21

3213212

21

32

22

1

332211

b) P2 is a 3-dimensional space, and we have 3 independent vectors in P2 – yes they span P2.

c) P2 is a 3-dimensional space, and we have 3 independent vectors in P2 – yes they form a basis for P2.

d) Form a linear combination of a1, a2 and a3:

3221

121

321

221

1

321

321

21

2321321

221332211

aaa1

1c;c;c

1c2cc

0ccc

0cc

1x0x0)c2cc(x)ccc(x)cc(1acacac

Similar analysis yields the other 2 combinations.

3221

1212

321

a0aax

aaax



math 3c practice midterm solutions prepared by vince zaccone for campus learning assistance services...

Documents