The Final Exam will be accumulative, about 35 % for the material covered in Chapter 1 to Chapter 5, about 65 % for the material covered after the second midterm. We didn't really talk about Chapter 7, so those sections in Chapter 7 is not required, e
Homework 11 | 4/28/2016
7.3.6
I find the eigenvalues in the typical fashion:
det [ A I n ]=0 det
[ ] [ ] ]
[
]
2 3
0
2
3
=0 det
= 0
4 5
0
4
5
( 2 )( 5 ) 12=0 27 2=0 1 =
7 57
7+ 57
; 2=
2
2
I now find the eigenvectors associated with the values above:
Homework 9 | 4/7/2016
6.1.8
The determinant of the given matrix can be calculated as:
1 2 3
1 1
1 1
1 1
| 1 1 1| = 1 |
|2|
|+3|
| = 1 (1) 2 (2) + 3 (1) = 0
2 1
3 1
3 2
3 2 1
6.1.18
A matrix is invertible if and only if its determinant is not zero; I find
Homework 10 | 4/7/2016
7.1.2
Note that by definition of being an eigenvector, it satisfies = . Multiplying both sides by 1
yields the solution proof:
1
1
1
1 = 1 () = 1 () =
1
1
Clearly, is an eigenvalue of and its eigenvalue is clearly
7.1.6
Denote the
Homework 2 | 2/2/2016
2.1.4
The linear transformation is given by the matrix equation:
[ ] [ ][ ]
y1
9 3 3 x
1
y
y = A x 2 = 2 9 1 x 2
4 9 2
y3
5 1
5 x3
y4
The matrix of the linear transformation is therefore:
[ ]
9 3 3
2 9 1
4 9 2
5 1
5
2.1.6
From the de
Homework 8 | 4/7/2016
5.3.4
A matrix is orthogonal if its product with its transpose yields the identity; that is to say that
A A T =I n
Must be true for orthogonality; using the matrix A that was provided:
[
]
2 6 3
6 3 2 1
2
6 3
6
1 3 2
1 49 12 0
T
AA =
Homework 4 | 2/16/2016
3.2.2
To determine if the set is a subspace in 3 , I need to check three distinct properties:
1. A zero vector must exist
2. The set must be closed under addition
3. The set must be closed under scalar multiplication
Only if all thr
Homework 6 | 3/10/2016
4.1.20
Algebra dictates that + = 0 implies = which in turn means that the matrix A can be written as
=[
1
] = [
0
0
0 1
0 0
]+[
]+[
]
1
0 0
1 0
1 0
0 1 0 0
],[
],[
]; as there are three linearly
0 1 0 0 1 0
independent vectors that
Homework 2 | 2/2/2016
2.1.4
The linear transformation is given by the matrix equation:
1
9
2
2
= [ ] = [
4
3
4
5
9
2
The matrix of the linear transformation is therefore: [
4
5
2.1.6
3 3
1
9 1
] [ 2]
9 2
3
1
5
3 3
9 1
]
9 2
1
5
From the definition in
Homework 6 | 3/10/2016
4.1.20
Algebra dictates that
a+ d=0 implies d=a which in turn means that the matrix A can be
written as
[
] [
] [ ] [ ]
A= a b = a 1 0 +b 0 1 +c 0 0
c a
0 1
0 0
1 0
At this stage, it is clear that a basis for the space is
[
][ ][ ]
Homework 8 | 4/7/2016
5.3.4
A matrix is orthogonal if its product with its transpose yields the identity; that is to say that
=
Must be true for orthogonality; using the matrix A that was provided:
1 2
= [6
7
3
6 3 1 2
6 3
1 49 12 0
[12 49 24]
3 2 ] [
Homework 4 | 2/16/2016
3.2.2
To determine if the set is a subspace in
R3 , I need to check three distinct properties:
1. A zero vector must exist
2. The set must be closed under addition
3. The set must be closed under scalar multiplication
Only if all th
Homework 3 | 2/2/2016
2.4.10
To determine if the matrix is invertible, first I find the determinant of the given:
0
det [0
1
0 1
1
1 0] = 0 det [
0
0 0
0
0 0
0
] 0 det [
] + 1 det [
0
1 0
1
1
] = 0 0 + 1 (0 0 1 1) = 1
0
Because the determinant is not zero
Homework 7 | 3/24/2016
5.1.4
From a theorem presented in the chapter, the angle between the vectors
and is given by
= cos1 (
)
By definition, = and
=
which I can calculate as:
= = [ 7 ] [ 7 ] = 170
11 11
1 1
=
= [ ] [ ] = 2
1 1
Which me
Homework 3 | 2/2/2016
2.4.10
To determine if the matrix is invertible, first I find the determinant of the given:
[ ]
0 0 1
1 0
0 0
0 1
det 0 1 0 =0 det
0 det
+1 det
= 00+ 1 ( 0 01 1 )=1
0 0
1 0
1 0
1 0 0
[ ]
[ ]
[ ]
Because the determinant is not zero, t
Homework 9 | 4/7/2016
6.1.8
The determinant of the given matrix can be calculated as:
| |
1 2 3
1 1 2 1 1 +3 1 1 =1 (1 )2 (2 ) +3 (1 )=0
1 1 1 =1
2 1
3 1
3 2
3 2 1
| | | | | |
6.1.18
A matrix is invertible if and only if its determinant is not zero; I fin
Practice Exam , November
,
Practice Exam
Linear Algebra, Dave Bayer, November
,
Name:
Uni:
[1]
[2]
[3]
[4]
[5]
Total
If you need more than one page for a problem, clearly indicate on each page where to look next for
your work.
[1] Find the determinant of
Exam , November
Exam
Linear Algebra, Dave Bayer, November
,
[1] Find the determinant of the matrix
2
1
2
1
3
4
3
4
1
1
2
3
1
1
3
2
[2] Find the determinant of the matrix
2
1
0
0
0
0
0
3
2
1
0
0
0
0
0
3
2
1
0
0
0
b
1
0
0
c
1
1
0
d
w
a
x
b
1
=
c
1 y
Exam 3, November 19, 2013
Exam 3
Linear Algebra, Dave Bayer, November 19, 2013
Name:
Uni:
[1]
[2]
[3]
[4]
[5]
Total
If you need more than one page for a problem, clearly indicate on each page where to look next for
your work.
[1] Find the determinant of t
Exam , October
,
Exam
Linear Algebra, Dave Bayer, October
,
Name:
Uni:
[1]
[2]
[3]
[4]
[5]
Total
If you need more that one page for a problem, clearly indicate on each page where to look next for
your work.
[1] Find a basis for the subspace V of R4 spanne
Exam , March ,
Exam
Linear Algebra, Dave Bayer, March ,
[1] Find the row space and the column space of the matrix
01234
0 2 4 6 8
0 3 6 9 2
04826
[2] By least squares, nd the equation of the form y = ax + b that best ts the data
x1 y1
01
x2 y2
= 1 2
x
Exam 2, March 6, 2014
Exam 2
Linear Algebra, Dave Bayer, March 6, 2014
Name:
Uni:
[1]
[2]
[3]
[4]
[5]
Total
If you need more that one page for a problem, clearly indicate on each page where to look next for
your work.
[1] Find the row space and the column
Exam 1, February 11, 2014
Exam 1
Linear Algebra, Dave Bayer, February 11, 2014
Name:
Uni:
[1]
[2]
[3]
[4]
[5]
Total
If you need more that one page for a problem, clearly indicate on each page where to look next for
your work.
[1] Using matrix multiplicati
Exam , September
Exam
Linear Algebra, Dave Bayer, September
,
[1] Using matrix multiplication, count the number of paths of length ten from w to itself.
w
x
y
z
[2] Solve the following system of equations.
w
1011
2
2 0 0 3 x = 2
y
1121
4
z
[3] Express
Exam 1, September 24, 2013
Exam 1
Linear Algebra, Dave Bayer, September 24, 2013
Name:
Uni:
[1]
[2]
[3]
[4]
[5]
Total
If you need more that one page for a problem, clearly indicate on each page where to look next for
your work.
[1] Using matrix multiplica
Homework 10 | 4/7/2016
7.1.2
Note that by definition of
v
being an eigenvector, it satisfies
A v = v . Multiplying both sides by
A1 yields the solution proof:
1
1
1
A1 v = A1 ( v )= A1 ( A v ) = v
Clearly,
v is an eigenvalue of
A
1
and its eigenvalue is c
Homework 7 | 3/24/2016
5.1.4
From a theorem presented in the chapter, the angle between the vectors
=cos1
u and v is given by
(uu vv)
By definition,
v= v v and u= u u which I can calculate as:
v= v v =
[ ][ ]
7 7 = 170
11 11
[ ][ ]
u= u u = 1 1 = 2
1 1
Wh