Linear Algebra Final Exam
This exam is comprehensive over the entire course and includes 12
questions. You have 60 minutes to complete the exam.
The exam is worth 100 points. The 8 multiple choice questions are worth 5
points each (40 points total) and the 4 free response questions are worth
15 points each (60 points total).
Mark your multiple choice answers on this cover page. For the free
response questions, show your work and make sure to circle your final
answer.
1. (5 pts)
A
B
C
D
E
2. (5 pts)
A
B
C
D
E
3. (5 pts)
A
B
C
D
E
4. (5 pts)
A
B
C
D
E
5. (5 pts)
A
B
C
D
E
6. (5 pts)
A
B
C
D
E
7. (5 pts)
A
B
C
D
E
8. (5 pts)
A
B
C
D
E
1
1. (5 pts) Use Gauss-Jordan elimination to solve the system with a rref
matrix.
x − 4y + z = 20
−x + z = 10
4x + y − 2z = − 25
A
(x, y, z) = (1, − 3, − 9)
D
(x, y, z) = (−29, − 27, − 59)
B
(x, y, z) = (29,27,59)
E
(x, y, z) = (−85, − 75, − 195)
C
(x, y, z) = (−1, − 3,9)
2
2. (5 pts) Use the distributive property to find (A − C)B.
5 −4
A = −3 9
0
4
4 −1 7
B=
[3 7 8]
3 2
C = −1 7
[−3 3]
A
−10 −44 −34
−2 16
2
15
4
29
D
−9 −3
[−46 16 ]
B
2 −16 −2
−10 18 −12
−9 10 −12
E
31 −19
[16 4 ]
C
2 −16 −2
−2 16
2
15
4
29
3
3. (5 pts) Find the unit vector in the direction of u ⃗ = a ⃗ + 2 b ⃗ − 3 c ⃗ − d ,⃗ if
a ⃗ = (−3,5, − 1), b ⃗ = (4,2,7), c ⃗ = (0,2,1), and d ⃗ = (−1,3,5).
A
B
6
v⃗= 0
−5
v⃗=
−
D
4
v⃗= 6
15
6
6
11
61
0
E
5
11
v⃗=
0
5
61
6
C
61
v⃗=
−
0
5
61
4
4. (5 pts) Find the angle between u ⃗ and v .⃗
2
−3
u ⃗=
0
5
12
3
v⃗=
8
−3
A
θ = 0∘
D
θ = 180∘
B
θ = 90∘
E
θ = 60∘
C
θ = 45∘
5. (5 pts) Find the cross product of a ⃗ = (−2,3, − 1) and b ⃗ = (0, − 2,2).
A
a ⃗ × b ⃗ = (3, − 4,4)
D
a ⃗ × b ⃗ = (3, − 4, − 4)
B
a ⃗ × b ⃗ = (9,4,4)
E
a ⃗ × b ⃗ = (4,4,4)
C
a ⃗ × b ⃗ = (9, − 4,4)
5
6. (5 pts) Find the orthogonal complement of W, if W is a vector set in ℝ4.
y+z
W = −2x + y
x−z
x
A
W ⊥ = Span(
1
3
− 13
1
3
)
D
1
−1
1
W ⊥ = Span(
−1 )
1
− 13
B
W ⊥ = Span(
1
3
− 13
− 13
)
0
E
W ⊥ = Span(
− 13
1
3
)
0
− 13
C
W = Span(
⊥
1
3
− 13
)
1
6
7. (5 pts) For the matrix A, which choice is incorrect?
−1 −3
A= 2
4
[6
0]
A
−12
N(A ) = Span( −9 )
[ 1 ]
D
C(A T ) = Span(
B
Dim(C(A T )) = 2
E
C(A T ) = Span(
C
Dim(N(A T )) = 1
T
−1
2
,
[−3] [4])
−1
[−3])
7
8. (5 pts) Find the eigenvectors of the transformation matrix.
A=
−5 0
[ 3 1]
A
−5
0
and
[3]
[1]
D
0
−2
and
[1]
[1]
B
0
−1
and
[1]
[1]
E
1
−1
and
[1]
[0]
C
1
−2
and
[1]
[0]
8
9. (15 pts) Find the general solution to A x ⃗ = b .⃗
1 −2 −5 −3
1
A = 3 −1 −5 −4 with b ⃗ = 1
[2]
0 −5 −10 −5
10. (15 pts) If S : ℝ3 → ℝ3 and T : ℝ3 → ℝ3, then what are T(S( x ⃗ )) and
S(T( x ⃗ ))?
−x2 − 3x1
x2 − x3
S( x ⃗ ) =
x2
x1 − 2x2 + x3
x3 − x1
T( x ⃗ ) =
2x1 + x2 − x3
−2
x ⃗ = −1
[0]
9
11. (15 pts) If det(A) = 7, find det(B).
a b c
A= d e f
g h i
a 3g d
B = b 3h e
c 3i f
a b c
C = 3g 3h 3i
d e f
12. (15 pts) The subspace V is a plane in ℝ4. Use a Gram-Schmidt process
to change the basis of V into an orthonormal basis.
−2
1
−1
−1
0
1
V = Span(
,
,
0
1
−2 )
−1
1
−1
10
Linear Algebra Final Exam Answer Key
1. (5 pts)
A
B
C
D
E
2. (5 pts)
A
B
C
D
E
3. (5 pts)
A
B
C
D
E
4. (5 pts)
A
B
C
D
E
5. (5 pts)
A
B
C
D
E
6. (5 pts)
A
B
C
D
E
7. (5 pts)
A
B
C
D
E
8. (5 pts)
A
B
C
D
E
12
9. (15 pts)
10. (15 pts)
x ⃗ = c1
1
1
−2
−1
+ c2
+
1
0
0
1
1
5
− 25
0
0
8
T(S( x ⃗ )) = −8
[ 14 ]
−2
S(T( x ⃗ )) = 7
[2]
11. (15 pts)
12. (15 pts)
det(B) = − 21
V3 = Span(
1
2
−1
1
1
,
1
6
−1
−2
2
−1
,
0
14
1
1
−2
0
)
− 32
−1
13
Linear Algebra Final Exam Solutions
1.
C. The matrix for the system is
1 −4 1
20
−1 0
1
10
4
1 −2 −25
Start by working on the first column.
1 −4 1
20
1 −4 1
20
0 −4 2
30 → 0 −4 2
30
4 1 −2 −25
0 17 −6 −105
Find the pivot entry in the second column, then zero out the rest of
the second column.
1 −4
0
1
− 12
1
20
1
− 15
2
→ 0
0 17 −6 −105
0
1
−1
− 12
−10
− 15
2
→
0 17 −6 −105
1 0 −1 −10
0 1 − 12 − 15
2
0 0
5
2
45
2
Find the pivot entry in the third column, then zero out the rest of the
third column.
14
1 0 −1 −10
0 1
0 0
− 12
1
− 15
2
1 0
→ 0 1
9
0 0
0
−1
− 12
− 15
2
1
9
1 0 0 −1
→ 0 1 0 −3
0 0 1 9
The third column is done, and we can see that the solution to the
linear system is (x, y, z) = (−1, − 3,9).
2.
A. To find A − C by subtracting matrices, we subtract corresponding
entries from each matrix.
5 −4
3 2
A − C = −3 9 − −1 7
[−3 3]
0
4
5−3
−4 − 2
A − C = −3 − (−1) 9 − 7
0 − (−3)
4−3
2 −6
A − C = −2 2
[3
1]
To find (A − C)B, multiply each row of A − C by the first column of B.
2 −6
4 −1 7
(A − C)B = −2 2
[3 7 8]
[3
1]
2(4) − 6(3) . . .
(A − C)B = −2(4) + 2(3) . . .
3(4) + 1(3) . . .
...
...
...
15
8 − 18 . . .
(A − C)B = −8 + 6 . . .
12 + 3 . . .
−10 . . .
(A − C)B = −2 . . .
15 . . .
...
...
...
...
...
...
Multiply each row of A − C by the second column of B.
−10 2(−1) − 6(7) . . .
(A − C)B = −2 −2(−1) + 2(7) . . .
15
3(−1) + 1(7) . . .
−10 −2 − 42 . . .
(A − C)B = −2 2 + 14 . . .
15 −3 + 7 . . .
−10 −44 . . .
(A − C)B = −2 16 . . .
15
4
...
Multiply each row of A − C by the third column of B.
−10 −44 2(7) − 6(8)
(A − C)B = −2 16 −2(7) + 2(8)
15
4
3(7) + 1(8)
−10 −44 14 − 48
(A − C)B = −2 16 −14 + 16
15
4
21 + 8
16
−10 −44 −34
(A − C)B = −2 16
2
15
4
29
3.
E. The vector sum is
u ⃗ = a ⃗+ 2 b ⃗ − 3 c ⃗− d ⃗
u ⃗ = (−3,5, − 1) + 2(4,2,7) − 3(0,2,1) − (−1,3,5)
Apply the scalars to each vector.
u ⃗ = (−3,5, − 1) + (8,4,14) + (0, − 6, − 3) + (1, − 3, − 5)
Add each of the vector components.
u ⃗ = (−3 + 8 + 0 + 1,5 + 4 − 6 − 3, − 1 + 14 − 3 − 5)
u ⃗ = (6,0,5)
Then, find the length of u .⃗
| | u ⃗| | =
u12 + u22 + u32
| | u ⃗| | =
62 + 02 + 52
| | u ⃗| | =
36 + 0 + 25
| | u ⃗| | =
61
Then the unit vector in the direction of u ⃗ = (6,0,5) is
17
v⃗=
v⃗=
1
u⃗
⃗
|| u ||
1
61
6
0
5
6
v⃗=
61
0
5
61
4.
B. The angle θ between two vectors is given by a relationship
between the dot product of the vectors and the lengths of the
vectors.
u ⃗ ⋅ v ⃗ = | | u ⃗ | | | | v ⃗ | | cos θ
u ⃗ ⋅ v ⃗ = (2, − 3,0,5) ⋅ (12,3,8, − 3)
u ⃗ ⋅ v ⃗ = (2)(12) + (−3)(3) + (0)(8) + (5)(−3)
u ⃗ ⋅ v ⃗ = 24 − 9 + 0 − 15
u ⃗⋅ v ⃗ = 0
Because the dot product is 0, u ⃗ and v ⃗ are orthogonal to one
another and the angle between them is θ = 90∘.
18
5.
E. The cross product would be
i
j
k
a ⃗ × b ⃗ = −2 3 −1
0 −2 2
a ⃗× b⃗ = i
−2 3
3 −1
−2 −1
−j
+k
0
2
0 −2
−2 2
a ⃗ × b ⃗ = i((3)(2) − (−1)(−2)) − j((−2)(2) − (−1)(0)) + k((−2)(−2) − (3)(0))
a ⃗ × b ⃗ = i(6 − 2) − j(−4 + 0) + k(4 − 0)
a ⃗ × b ⃗ = 4i + 4j + 4k
a ⃗ × b ⃗ = (4,4,4)
6.
C. We can rewrite W as
0
W = {x ⋅ −2 + y ⋅
1
1
1
1
1
0
+z⋅
0
−1
0
0
| x, y, z ∈ ℝ4}
The subspace W is a space in ℝ4, spanned by the three vectors
w 1⃗ = (0, − 2,1,1), w 2⃗ = (1,1,0,0) and w 3⃗ = (1,0, − 1,0). Therefore, its
orthogonal complement W ⊥ is the set of vectors which are
orthogonal to w 1⃗ = (0, − 2,1,1), w 2⃗ = (1,1,0,0) and w 3⃗ = (1,0, − 1,0).
19
0
W ⊥ = { x ⃗ ∈ ℝ4 | x ⃗ ⋅ −2 = 0 and x ⃗ ⋅
1
1
1
1
1
0
= 0 and x ⃗ ⋅
= 0}
0
−1
0
0
If we let x ⃗ = (x1, x2, x3, x4), we get three equations from these dot
products.
−2x2 + x3 + x4 = 0
x1 + x2 = 0
x1 − x3 = 0
Put these equations into an augmented matrix,
0 −2 1 1 | 0
1 1
0 0 | 0
1 0 −1 0 | 0
then put it into reduced row-echelon form.
1 1
0 0 | 0
1 1
0 0 | 0
0 −2 1 1 | 0 → 0 −2 1 1 | 0 →
1 0 −1 0 | 0
0 −1 −1 0 | 0
1
1
0
1
0
| 0
0
| 0 → 0 1
| 0
0 −1 −1
− 12 − 12
0 −1 −1
1
2
− 12
0
1
0
1
2
− 12
0
| 0
| 0 →
| 0
20
1
2
− 12
1
2
− 12
− 12
| 0
0 0
1
2
− 12
− 32
| 0
0 0
1 0
0
1
3
− 12
1
3
| 0
1 0 0
1 0
0 1
0 1 − 12
0 0
1
1 0
| 0 → 0 1
1
| 0 → 0 1 0
| 0
0 0 1
1
2
− 12
1
3
1
3
− 13
1
3
| 0
| 0 →
| 0
| 0
| 0
| 0
This rref form gives the system of equations
x1 +
1
x4 = 0
3
x2 −
1
x4 = 0
3
x3 +
1
x4 = 0
3
and we can solve the system for the pivot variables.
x1 = −
x2 =
1
x4
3
1
x
3 4
1
x3 = − x4
3
We can express this system as
21
1
−3
x1
1
x2
3
x3 = x4
1
−
x4
3
1
Which means the orthogonal complement is
− 13
W = Span(
⊥
1
3
− 13
)
1
7.
E. The transpose of A is
AT =
−1 2 6
[−3 4 0]
To find the null space, we’ll augment the matrix, and then put it into
reduced row-echelon form.
−1 2 6 | 0
1 −2 −6 | 0
→
→
[−3 4 0 | 0] [−3 4
0 | 0]
1 −2 −6 | 0
1 −2 −6 | 0
1 0 12 | 0
→
→
[0 −2 −18 | 0] [0 1
9 | 0] [0 1 9 | 0]
22
Because we have pivot entries in the first two columns, we’ll pull a
system of equations from the matrix,
x1 + 12x3 = 0
x2 + 9x3 = 0
and then solve the system’s equations for the pivot variables.
x1 = − 12x3
x2 = − 9x3
If we turn this into a vector equation, we get
x1
−12
x2 = x3 −9
[ 1 ]
x3
Therefore, the left null space is
−12
N(A T ) = Span( −9 )
[ 1 ]
The space of the null space of the transpose is always ℝm, where m
is the number of rows in the original matrix, A. The original matrix
has 3 rows, so the null space of the transpose N(A T ) is a subspace of
ℝ3.
The column space of the transpose A T, which is the same as the row
space of A, is simply given by the columns in A T that contain pivot
23
entries when A T is in reduced row-echelon form. So the column
space of A T is
C(A T ) = Span(
−1
2
,
[−3] [4])
The space of the column space of the transpose is always ℝn, where
n is the number of columns in the original matrix, A. The original
matrix has 2 columns, so the column space of the transpose C(A T ) is
a subspace of ℝ2.
Because there’s one vector that forms the basis of N(A T ), the
dimension of N(A T ) is Dim(N(A T )) = 1.
Because there are two vectors that form the basis of C(A T ), the
dimension of C(A T ) is Dim(C(A T )) = 2.
−12
N(A ) = Span( −9 ) in ℝ3
[ 1 ]
T
C(A T ) = Span(
8.
−1
2
,
in ℝ2
)
[−3] [4]
Dim(N(A T )) = 1
Dim(C(A T )) = 2
D. Find the determinant | λIn − A | .
λ
1 0
−5 0
−
[0 1] [ 3 1]
24
λ 0
−5 0
−
[0 λ] [ 3 1]
λ − (−5) 0 − 0
[ 0−3
λ − 1]
λ+5
0
[ −3 λ − 1]
The determinant is
(λ + 5)(λ − 1) − (−3)(0)
(λ + 5)(λ − 1)
λ = − 5 or λ = 1
With λ = − 5 and λ = 1, we’ll have two eigenspaces, given by
Eλ = N(λIn − A). With
Eλ = N
λ+5
0
([ −3 λ − 1])
we get
E−5 = N
−5 + 5
0
([ −3
−5 − 1])
E−5 = N
0
0
([−3 −6])
and
25
E1 = N
1+5
0
([ −3 1 − 1])
E1 = N
6 0
([−3 0])
Therefore, the eigenvectors in the eigenspace E−5 will satisfy
0
0
0
v⃗=
[−3 −6]
[0]
0
0 | 0
0 0 | 0
→
[−3 −6 | 0] [1 2 | 0]
0
0 0 v1
=
[1 2] [v2] [0]
v1 + 2v2 = 0
v1 = − 2v2
v1
−2
=
v
2
[v2]
[1]
So the eigenvector for E−5 will be
v1
−2
=
[v2] [ 1 ]
And the eigenvectors in the eigenspace E1 will satisfy
0
6 0
v⃗=
[−3 0]
[0]
26
6 0 | 0
1 0 | 0
1 0 | 0
→
→
[−3 0 | 0] [−3 0 | 0] [0 0 | 0]
1 0 v1
0
=
[0 0] [v2] [0]
v1 = 0
v1
0
=
t
[v2]
[1]
So the eigenvector for E1 will be
v1
0
=
[v2] [1]
Then the eigenvectors of the matrix are
0
−2
and
[1]
[1]
9.
Put the matrix A into reduced row-echelon form.
1 −2 −5 −3
1 −2 −5 −3
3 −1 −5 −4 → 0 5
10
5 →
0 −5 −10 −5
0 −5 −10 −5
1 −2 −5 −3
1 −2 −5 −3
0 1
2
1 → 0 1
2
1 →
0 0
0
0
0 −5 −10 −5
27
1 0 −1 −1
0 1 2
1
[0 0 0
0]
To find the complementary solution, augment rref(A) with the zero
vector to get a system of equations.
x1 − x3 − x4 = 0
x2 + 2x3 + x4 = 0
Solve for the pivot variables in terms of the free variables.
x1 = x3 + x4
x2 = − 2x3 − x4
The vectors that satisfy the null space are
x1
1
1
x2
−2
−1
=
x
+
x
3
4
x3
1
0
x4
0
1
We could therefore write the complementary solution as
x n⃗ = c1
1
1
−2
−1
+ c2
1
0
0
1
To find the particular solution, augment A with b ⃗ = (b1, b2, b3), then
put it in reduced row-echelon form.
28
1 −2 −5 −3 | b1
1 −2 −5 −3 |
b1
3 −1 −5 −4 | b2 → 0 5
10
5 | b2 − 3b1 →
0 −5 −10 −5 | b3
0 −5 −10 −5 |
b3
1 −2
0
−3 |
−5
1
2
|
1
0 −5 −10 −5 |
1
0
−1
0
1
2
1
(b
5 2
b1
− 3b1) →
b3
−1 | b1 + 25 (b2 − 3b1)
|
1
0 −5 −10 −5 |
1
(b
5 2
− 3b1)
→
b3
2
1 0 −1 −1 | b1 + 5 (b2 − 3b1)
0 1
2
1
|
0 0
0
0
|
1
(b
5 2
− 3b1)
b3 + b2 − 3b1
From the third row, the system is constrained.
−3b1 + b2 + b3 = 0
b2 = 3b1 − b3
We were asked to use b1 = 1, b2 = 1, and b3 = 2, which satisfies this
constraint equation.
1 = 3(1) − 2
1=1
29
Then the augmented matrix becomes
1 0 −1 −1 | 1 + 25 (1 − 3(1))
0 1
2
1
|
0 0
0
0
|
1 0 −1 −1 |
− 3(1))
→ 0 1
2
1
|
2 + 1 − 3(1)
0 0
0
0
|
1
(1
5
1
5
− 25
0
This gives a system of equations
x1 − x3 − x4 =
1
5
x2 + 2x3 + x4 = −
2
5
Because x3 and x4 are free variables, set x3 = 0 and x4 = 0.
x1 − 0 − 0 =
1
5
x2 + 2(0) + 0 = −
2
5
The system becomes
x1 =
1
5
x2 = −
2
5
So the particular solution is
30
x p⃗ =
1
5
− 25
0
0
The general solution is the sum of the complementary and particular
solutions.
x ⃗ = x p⃗ + x n⃗
x ⃗ = c1
10.
1
1
−2
−1
+ c2
+
1
0
0
1
1
5
− 25
0
0
Apply the transformation S to each column of the I3 identity matrix.
−0 − 3(1)
−3
1
S( 0 ) =
= 0
0−0
[0]
[0]
0
−1 − 3(0)
0
−1
S( 1 ) =
= 1
1−0
[0]
[1]
1
−0 − 3(0)
0
0
S( 0 ) =
= −1
0−1
[1]
[0]
0
31
So the transformation S can be written as
−3 −1 0
S( x ⃗ ) = 0
1 −1 x ⃗
[0
1
0]
Apply the transformation T to each column of the I3 identity matrix.
1 − 2(0) + 0
1
1
T( 0 ) =
= −1
0−1
[0]
[2]
2(1) + 0 − 0
0 − 2(1) + 0
0
−2
T( 1 ) =
= 0
0−0
[0]
[1]
2(0) + 1 − 0
0 − 2(0) + 1
0
1
T( 0 ) =
= 1
1−0
[1]
[−1]
2(0) + 0 − 1
So the transformation T can be written as
1 −2 1
T( x ⃗ ) = −1 0
1 x⃗
[2
1 −1]
Then the composition T ∘ S can be written as
−3 −1 0
1 −2 1
T(S( x ⃗ )) = −1 0
1
0
1 −1 x ⃗
[2
1 −1] [ 0
1
0]
32
−3 + 0 + 0 −1 − 2 + 1 0 + 2 + 0
T(S( x ⃗ )) = 3 + 0 + 0
1+0+1 0+0+0
−6 + 0 + 0 −2 + 1 − 1 0 − 1 + 0
−3 −2 2
T(S( x ⃗ )) = 3
2
0
−6 −2 −1
x⃗
x⃗
Transform x ⃗ = (−2, − 1,0).
−3 −2 2
−2
−2
T S( −1 ) = 3
2
0
−1
( [0] )
[
0]
−6 −2 −1
−3(−2) − 2(−1) + 2(0)
−2
3(−2) + 2(−1) + 0(0)
T S( −1 ) =
( [0] )
−6(−2) + (−2)(−1) + (−1)(0)
6+2+0
−2
T S( −1 ) = −6 − 2 + 0
( [0] )
12 + 2 + 0
8
−2
T S( −1 ) = −8
( [ 0 ] ) [ 14 ]
Then the composition S ∘ T can be written as
−3 −1 0
1 −2 1
S(T( x ⃗ )) = 0
1 −1 −1 0
1 x⃗
[0
]
[
1
0
2
1 −1]
33
−3 + 1 + 0 6 − 0 + 0 −3 − 1 + 0
S(T( x ⃗ )) = 0 − 1 − 2 0 + 0 − 1 0 + 1 + 1
0−1+0 0+0+0 0+1+0
−2 6 −4
S(T( x ⃗ )) = −3 −1 2
−1 0
1
x⃗
x⃗
Transform x ⃗ = (−2, − 1,0).
−2 6 −4
−2
−2
S T( −1 ) = −3 −1 2
−1
( [0] )
[
0]
−1 0
1
(−2)(−2) + 6(−1) − 4(0)
−2
S T( −1 ) = (−3)(−2) + (−1)(−1) + 2(0)
( [0] )
(−1)(−2) + 0(−1) + 1(0)
4−6−0
−2
S T( −1 ) = 6 + 1 + 0
( [0] )
2+0+0
−2
−2
S T( −1 ) = 7
( [0] ) [2]
11.
The matrices A and C are identical, other than two changes. Matrix A
has rows 2 and 3 that are swapped, compared to matrix C. When
matrices are identical other than a swapped row, the determinant of
one is equal to the negative determinant of the other.
34
The second change is that the second row of C has been multiplied
by 3, compared to matrix A. If we have a row multiplied by a
constant k, then the determinant of the new matrix is multiplied by k.
Putting these two changes together, we get
det(C) = − 3det(A)
det(C) = − 3(7) = − 21
We also see that B = C T. The determinant of a transpose of a square
matrix will always be equal to the determinant of the original matrix,
which means det(B) = det(C) = − 21.
12.
Define v 1⃗ = (−1,1,1, − 1), v 2⃗ = (−2, − 1,0,1), and v 3⃗ = (1,0, − 2, − 1).
V = Span( v 1⃗ , v 2⃗ , v 3⃗ )
The length of v 1⃗ is
| | v 1⃗ | | =
(−1)2 + (1)2 + (1)2 + (−1)2 =
1+1+1+1 =
4=2
Then if u 1⃗ is the normalized version of v 1⃗ , we can say
u 1⃗ =
1
2
−1
1
1
−1
So we can say that V is spanned by u 1⃗ , v 2⃗ , and v 3⃗ .
35
V1 = Span( u 1⃗ , v 2⃗ , v 3⃗ )
We’ll name w 2⃗ as the vector that connects ProjV1 v 2⃗ to v 2⃗ .
w 2⃗ = v 2⃗ − ProjV1 v 2⃗
w 2⃗ = v 2⃗ − ( v 2⃗ ⋅ u 1⃗ ) u1 ⃗
Plug in the values we already have.
−2
−2
1
−1
−1
w 2⃗ =
−(
⋅
0
0
2
1
1
−1
1
1
1 )2
−1
−1
1
1
−1
−2
−2
−1
−1
1 −1
−1
1
1
w 2⃗ =
− (
⋅
0
0
1 ) 1
4
−1
−1
1
1
−2
−1
1
−1
1
w 2⃗ =
− ((−2)(−1) + (−1)(1) + (0)(1) + (1)(−1))
0
1
4
−1
1
−2
−1
1
−1
1
w 2⃗ =
− (0)
0
1
4
−1
1
36
−2
−1
w 2⃗ =
0
1
So w 2⃗ is orthogonal to u 1⃗ , but it hasn’t been normalized, so let’s
normalize it. The length of w 2⃗ is
| | w 2⃗ | | =
(−2)2 + (−1)2 + (−0)2 + (1)2
| | w 2⃗ | | =
4+1+0+1
| | w 2⃗ | | =
6
Then the normalized version of w 2⃗ is u 2⃗ .
u 2⃗ =
1
6
−2
−1
0
1
So we can say that V is spanned by u 1⃗ , u 2⃗ , and v 3⃗ . Then the vector
w 3⃗ is given by
w 3⃗ = v 3⃗ − ProjV1 v 3⃗ − ProjV2 v 3⃗
w 3⃗ = v 3⃗ − ( v 3⃗ ⋅ u 1⃗ ) u1 ⃗ − ( v 3⃗ ⋅ u 2⃗ )u2⃗
Plug in the values we already have.
37
1
1
1
0
0
w 3⃗ =
−(
⋅
−2
−2
2
−1
−1
−1
1
1
1 )2
−1
1
−1
1
0
1
−(
⋅
1
−2
6
−1
−1
−2
1
−1
0 ) 6
1
−2
−1
0
1
1
1
1
−2
−2
−1
−1
1
1
0
0
0
−1
−1
1
1
w 3⃗ =
− (
⋅
−
⋅
0 ) 0
1 ) 1
−2
4 −2
6 ( −2
−1
−1
−1
−1
−1
1
1
1
−1
1
0
1
w 3⃗ =
− ((1)(−1) + (0)(1) + (−2)(1) + (−1)(−1))
1
−2
4
−1
−1
−2
1
−1
− ((1)(−2) + (0)(−1) + (−2)(0) + (−1)(1))
0
6
1
1
−2
−1
1
1
0
−1
1
w 3⃗ =
− (−2)
− (−3)
0
1
−2
4
6
−1
−1
1
1
1
0
⃗
w3 =
+
−2
2
−1
−1
1
1
+
1
2
−1
−2
−1
0
1
38
1
1
0
w 3⃗ =
+
−2
−1
w 3⃗ =
−2
1
2
1
2
− 12
+
−1
− 12
0
1
2
− 12
0
− 32
−1
The length of w 3⃗ is
| | w 3⃗ | | =
1
3
−
+ 02 + −
+ (−1)2
( 2)
( 2)
| | w 3⃗ | | =
1
9
+0+ +1
4
4
| | w 3⃗ | | =
14
2
2
2
Then the normalized version of w 3⃗ is u 3⃗ :
u 3⃗ =
2
14
− 12
0
− 32
−1
39
Therefore, we can say that u 1⃗ , u 2⃗ , and u 3⃗ form an orthonormal
basis for V.
V3 = Span(
1
2
−1
1
1
,
1
6
−1
−2
2
−1
,
0
14
1
− 12
0
)
− 32
−1
40
41
Linear Algebra Final Exam
This exam is comprehensive over the entire course and includes 12
questions. You have 60 minutes to complete the exam.
The exam is worth 100 points. The 8 multiple choice questions are worth 5
points each (40 points total) and the 4 free response questions are worth
15 points each (60 points total).
Mark your multiple choice answers on this cover page. For the free
response questions, show your work and make sure to circle your final
answer.
1. (5 pts)
A
B
C
D
E
2. (5 pts)
A
B
C
D
E
3. (5 pts)
A
B
C
D
E
4. (5 pts)
A
B
C
D
E
5. (5 pts)
A
B
C
D
E
6. (5 pts)
A
B
C
D
E
7. (5 pts)
A
B
C
D
E
8. (5 pts)
A
B
C
D
E
42
1. (5 pts) Determine whether the system has one solution, no solutions, or
infinitely many solutions.
−x + y + 2z + 7w = 4
3x − y − 4z + 5w = 8
2x + 4y + 3z − w = 4
−y − z − 13w = 5
A
No solutions
B
(x, y, z) = (−1,5, − 3,0)
C
Infinitely many solutions
D
(x, y, z) = (−4,0,2,2)
E
(x, y, z) = (0,4, − 2,0)
43
2. (5 pts) Find the product B(AC).
7
3 −5
A=
[−5 −8 −4]
−2 3
B=
[ 4 −1]
6
1
C = −7 −2
[2
3]
A
B(AC) =
−78 47
[−40 55]
D
B(AC) =
78
47
[−40 −55]
B
B(AC) =
−32 25
[ 26 −55]
E
B(AC) =
32 25
[26 −55]
C
The product isn’t defined
44
3. (5 pts) Simplify w ⃗ ⋅ (−2 u ⃗ + 4 v ⃗ ).
u ⃗ = (−3,2,1,0)
v ⃗ = (1, − 5, − 4,1)
w ⃗ = (0, − 1,1,2)
A
B
C
14
2
D
E
2
14
(0,24, − 18,8)
4. (5 pts) Find the equation of the plane passing through A and
perpendicular to AB, given A(2, − 1,4) and B(0, − 3,2).
A
2x − y + 4z = 5
D
x+y+z =7
B
x+y+z =5
E
x+y+z =−5
C
2x − y + 4z = − 5
45
5. (5 pts) Find the general solution to A x ⃗ = b .⃗
2 −4 6
A = 3 −5 −2
−5 7 18
b ⃗ = (1,1, − 1)
A
x⃗=
37
2
21
2
D
B
x ⃗ = −1
2
0
C
x ⃗ = −1
2
0
1
− 12
− 12
−19
+ c1 −11
[ 0 ]
E
19
+ c1 11
[1]
− 12
x ⃗ = −1
2
0
19
x ⃗ = c1 11
[1]
46
6. (5 pts) Transform x ⃗ = (−2,5) with S ∘ T, if S : ℝ2 → ℝ2 and T : ℝ2 → ℝ2.
S( x ⃗ ) =
x1 + x2
[2x2 − x1]
T( x ⃗ ) =
x1 + 3x2
[ 2x2 ]
A
(S ∘ T )( x ⃗ ) = (7,23)
D
(S ∘ T )( x ⃗ ) = (24,39)
B
(S ∘ T )( x ⃗ ) = (39,24)
E
(S ∘ T )( x ⃗ ) = (−23,7)
C
(S ∘ T )( x ⃗ ) = (23,7)
47
7. (5 pts) Find the orthogonal complement of V.
−1
−1
−2
0
V = Span(
,
0 )
−2
3
−5
A
−2
3
V ⊥ = Span(
1 )
−4
B
−3
2
−1
4
V ⊥ = Span(
,
1
0 )
0
1
C
−2
1
V ⊥ = Span(
1 )
0
D
1
0
0
V ⊥ = Span(
, 1 )
2
−1
−3
4
E
3
−2
1
−4
V ⊥ = Span(
,
1
0 )
0
1
48
8. (5 pts) Use the transformation T : ℝ2 → ℝ2 to transform [ x ⃗ ]B = (−1,4) in
the basis B in the domain to a vector in the basis B in the codomain.
T( x ⃗ ) =
−3 10
x⃗
[ 0 4]
−2 −4
B = Span(
,
[ 3 ] [ 0 ])
A
−60
[T( x ⃗ )]B =
[−48]
B
[T( x ⃗ )]B =
−4
[−1]
C
[T( x ⃗ )]B =
3
[6]
D
E
23
[T( x ⃗ )]B =
49
[− 2 ]
[T( x ⃗ )]B =
4
3
5
− 12
49
9. (15 pts) Find the four fundamental subspaces of M, including their
spaces and dimensions.
3
0
6
−3 1
0
M=
6 −2 0
0 −1 −6
10. (15 pts) Find the least squares solution to the system.
x + 3y = − 6
y−x =4
y=1
50
11. (15 pts) The subspace V is a space in ℝ3. Use a Gram-Schmidt process to
change the basis of V into an orthonormal basis.
2
0
−2
V = Span( 0 , −2 , −2 )
[−2] [ 4 ] [ 3 ]
12. (15 pts) Find the Eigenvalues and Eigenvectors of the matrix, then
describe what’s happening in the Eigenbases.
−1 4 −2
A= 0 2 0
[ 0 0 −1]
51
Linear Algebra Final Exam Answer Key
1. (5 pts)
A
B
C
D
E
2. (5 pts)
A
B
C
D
E
3. (5 pts)
A
B
C
D
E
4. (5 pts)
A
B
C
D
E
5. (5 pts)
A
B
C
D
E
6. (5 pts)
A
B
C
D
E
7. (5 pts)
A
B
C
D
E
8. (5 pts)
A
B
C
D
E
53
9. (15 pts)
10. (15 pts)
3
0
−3
C(M ) = Span(
, 1 )
6
−2
−1
0
ℝ4
Dim = 2
−2
N(M ) = Span( −6 )
[1]
ℝ3
Dim = 1
3
−3
C(M T ) = Span( 0 , 1 )
6 [0]
ℝ3
Dim = 2
0
2
N(M T ) = Span(
,
1
0
ℝ4
Dim = 2
x *⃗ =
1
1
0 )
1
14
1
,−
( 3
3)
−
1
1
11. (15 pts)
V3 = Span(
2
−
0
1
2
3
, −
1
3
1
3
−
, −
−
1
6
2
3
)
1
6
54
4
12. (15 pts)
1
3
E−1 = Span( 0 ) and E2 = Span( 1 )
[0]
0
• since λ = 2 in the eigenspace E2, any vector v ⃗ in
E2, under the transformation T, will be scaled by 2,
meaning that T( v ⃗ ) = λ v ⃗ = 2 v ,⃗ and
• since λ = − 1 in the eigenspace E−1, any vector v ⃗
in E−1, under the transformation T, will be scaled
by −1, meaning that T( v ⃗ ) = λ v ⃗ = − v .⃗
55
Linear Algebra Final Exam Solutions
1.
A. Rewrite the system as an augmented matrix.
−1 1
2
7
3 −1 −4 5
2
4
3 −1
0 −1 −1 −13
|
|
|
|
4
8
4
5
To put A into reduced row-echelon form, start by working on the
first column.
1 −1 −2 −7
3 −1 −4 5
2 4
3 −1
0 −1 −1 −13
| −4
1 −1 −2 −7
| 8
0 2
2
26
→
| 4
2 4
3 −1
| 5
0 −1 −1 −13
1 −1 −2 −7
0 2
2
26
0 6
7
13
0 −1 −1 −13
| −4
| 20
| 12
| 5
| −4
| 20
→
| 4
| 5
Find the pivot entry in the second column, then zero out the rest of
the second column.
1 −1 −2 −7
0 1
1
13
0 6
7
13
0 −1 −1 −13
| −4
1 0 −1 6
| 10
0 1
1
13
→
| 12
0 6
7
13
| 5
0 −1 −1 −13
| 6
| 10
→
| 12
| 5
56
|
|
1 0 −1 6
6
1 0 −1 6
6
0 1
1
13 | 10
0 1 1
13 | 10
→
0 0
1 −65 | −48
0 0 1 −65 | −48
| 15
0 −1 −1 −13 |
5
0 0 0
0
Find the pivot entry in the third column, then zero out the rest of the
third column.
1
0
0
0
0
1
0
0
0 −59 | −42
1 0
1 13 | 10
0 1
→
1 −65 | −48
0 0
| 15
0 0
0 0
0 −59 | −42
0 78 | 58
1 −65 | −48
| 15
0 0
The fourth row shows us that 0 = 15, which can’t be true. Therefore,
the system has no solutions.
2.
E. First multiply the matrix A by the matrix C, multiplying each row
of A by each column of C.
6
1
7
3 −5
AC =
[−5 −8 −4] [−7 −2]
2
3
AC =
7(6) + 3(−7) − 5(2)
7(1) + 3(−2) − 5(3)
[−5(6) − 8(−7) − 4(2) −5(1) − 8(−2) − 4(3)]
AC =
42 − 21 − 10
7 − 6 − 15
[−30 + 56 − 8 −5 + 16 − 12]
57
AC =
11 −14
[18 −1 ]
Now multiply the matrix B by the matrix AC, multiplying each row of
B by each column of AC.
3.
B(AC) =
−2 3
11 −14
[ 4 −1] [18 −1 ]
B(AC) =
−2(11) + 3(18) −2(−14) + 3(−1)
[ 4(11) − 1(18)
4(−14) − 1(−1) ]
B(AC) =
−22 + 54 28 − 3
[ 44 − 18 −56 + 1]
B(AC) =
32 25
[26 −55]
A. First, we need to apply the scalars to the vectors to find −2 u ⃗ and
4 v .⃗
−2 u ⃗ = − 2(−3,2,1,0) = (6, − 4, − 2,0)
4 v ⃗ = 4(1, − 5, − 4,1) = (4, − 20, − 16,4)
Then the sum −2 u ⃗ + 4 v ⃗ is
4
6
−20
−2 u ⃗ + 4 v ⃗ = −4 +
−2
−16
0
4
58
6+4
−4 − 20
−2 u ⃗ + 4 v ⃗ =
−2 − 16
0+4
10
−24
−2 u ⃗ + 4 v ⃗ =
−18
4
Calculate the dot product of w ⃗ and −2 u ⃗ + 4 v .⃗
10
−24
w ⃗ ⋅ (−2 u ⃗ + 4 v ⃗ ) = [0 −1 1 2]
−18
4
w ⃗ ⋅ (−2 u ⃗ + 4 v ⃗ ) = 0(10) − 1(−24) + 1(−18) + 2(4)
w ⃗ ⋅ (−2 u ⃗ + 4 v ⃗ ) = 0 + 24 − 18 + 8
w ⃗ ⋅ (−2 u ⃗ + 4 v ⃗ ) = 14
4.
B. First find the normal vector to the plane.
AB⃗ = (0 − 2, − 3 − (−1),2 − 4)
AB⃗ = (−2, − 2, − 2)
Plugging the normal vector and the point on the plane into the
plane equation gives
59
a(x − x0) + b(y − y0) + c(z − z0) = 0
−2(x − 2) − 2(y − (−1)) − 2(z − 4) = 0
Now we’ll simplify to get the equation of the plane into standard
form.
−2(x − 2) − 2(y + 1) − 2(z − 4) = 0
−2x + 4 − 2y − 2 − 2z + 8 = 0
−2x − 2y − 2z + 10 = 0
x+y+z =5
5.
D. Put the matrix A, augmented with the zero vector, into reduced
row-echelon form.
2 −4 6 | 0
1 −2 3 | 0
3 −5 −2 | 0 → 3 −5 −2 | 0 →
−5 7 18 | 0
−5 7 18 | 0
| 0
| 0
1 −2 3
1 −2 3
0
1 −11 | 0 → 0 1 −11 | 0 →
−5 7
18 | 0
0 −3 33 | 0
1 0 −19 | 0
1 0 −19 | 0
0 1 −11 | 0 → 0 1 −11 | 0
| 0
0 −3 33 | 0
0 0 0
60
To find the complementary solution, pull out a system of equations
from rref(A).
x1 − 19x3 = 0
x2 − 11x3 = 0
Solve for the pivot variables in terms of the free variable.
x1 = 19x3
x2 = 11x3
The vector that satisfies the null space is
x1
19
x2 = x3 11
[1]
x3
We could therefore write the complementary solution as
19
x n⃗ = c1 11
[1]
To find the particular solution, augment A with b ⃗ = (b1, b2, b3), then
put it in reduced row-echelon form.
2 −4 6 | b1
1 −2 3 |
3 −5 −2 | b2 → 3 −5 −2 |
−5 7 18 | b3
−5 7 18 |
1
b
2 1
b2
b3
→
61
1
−2
0
1
−5
7
1
b
2 1
|
3
−11 |
|
18
− 32 b1
+ b2
0
−19 | − 52 b1 + 2b2
0
1
−11 |
|
33
→ 0
b3
1
0 −3
1 −2
− 32 b1 + b2
5
b
2 1
1
b
2 1
|
3
−11 | − 32 b1 + b2 →
1
0 −3
5
b
2 1
|
33
5
1 0 −19 |
− 2 b1 + 2b2
→ 0 1 −11 |
+ b3
0 0
+ b3
− 32 b1 + b2
| −2b1 + 3b2 + b3
0
Substitute the values from b ⃗ = (1,1, − 1).
− 52 (1) + 2(1)
1 0 −19 |
− 32 (1) + 1
0 1 −11 |
0 0
0
| −2(1) + 3(1) − 1
1 0 −19 | − 12
→ 0 1 −11 | − 1
2
0 0
0
|
0
Rewrite the matrix as a system of equations.
x1 − 19x3 = −
1
2
1
x2 − 11x3 = −
2
Now, because x3 is a free variable, set x3 = 0.
x1 − 19(0) = −
1
2
1
x2 − 11(0) = −
2
62
The system becomes
x1 = −
1
2
x2 = −
1
2
So the particular solution is
− 12
xp ⃗ = − 1
2
0
The general solution is the sum of the complementary and particular
solutions.
x ⃗ = x p⃗ + x n⃗
− 12
x ⃗ = −1
2
0
6.
19
+ c1 11
[1]
C. Given x ⃗ = (−2,5) and
S( x ⃗ ) =
x1 + x2
[2x2 − x1]
63
T( x ⃗ ) =
x1 + 3x2
[ 2x2 ]
start by using S to transform the standard basis vectors.
S
1+0
1
1
=
=
([0]) [2(0) − 1] [−1]
S
0+1
0
1
=
=
([1]) [2(1) − 0] [2]
So the transformation S can be written as the matrix-vector product
S( x ⃗ ) =
1 1
x⃗
[−1 2]
Use T to transform the standard basis vectors.
T
1 + 3(0)
1
1
=
=
([0]) [ 2(0) ] [0]
T
0 + 3(1)
0
3
=
=
([1]) [ 2(1) ] [2]
So the transformation T can be written as the matrix-vector product
T( x ⃗ ) =
1 3
x⃗
[0 2]
If we call the matrix from S
A=
1 1
[−1 2]
64
and we call the matrix from T
B=
1 3
[0 2]
then the composition of the transformations is
S(T( x ⃗ )) = AB x ⃗
S(T( x ⃗ )) =
1 1 1 3
x⃗
[−1 2] [0 2]
S(T( x ⃗ )) =
1(1) + 1(0)
1(3) + 1(2)
x⃗
[−1(1) + 2(0) −1(3) + 2(2)]
S(T( x ⃗ )) =
1 5
x⃗
[−1 1]
To transform x ⃗ = (−2,5), multiply this transformation matrix by
x ⃗ = (−2,5).
S
−2
1 5 −2
=
( ([ 5 ])) [−1 1] [ 5 ]
S
1(−2) + 5(5)
−2
=
( ([ 5 ])) [−1(−2) + 1(5)]
S
−2
23
=
( ([ 5 ])) [ 7 ]
T
T
T
65
7.
E. The subspace V is a plane in ℝ4, spanned by the two vectors
v 1⃗ = (−1,0, − 2,3) and v 2⃗ = (−1, − 2,0, − 5). Its orthogonal complement
V ⊥ is the set of vectors which are orthogonal to both
v 1⃗ = (−1,0, − 2,3) and v 2⃗ = (−1, − 2,0, − 5).
−1
−1
−2
0
V ⊥ = { x ⃗ ∈ ℝ4 | x ⃗ ⋅
= 0 , x ⃗⋅
= 0}
0
−2
3
−5
Let x ⃗ = (x1, x2, x3, x4) to get two equations from these dot products.
−x1 − 2x3 + 3x4 = 0
−x1 − 2x2 − 5x4 = 0
Put these equations into an augmented matrix,
−1 0 −2 3 | 0
[−1 −2 0 −5 | 0]
then put it into reduced row-echelon form.
1
0 2 −3 | 0
1 0 2 −3 | 0
→
→
[−1 −2 0 −5 | 0] [0 −2 2 −8 | 0]
1 0 2 −3 | 0
[0 1 −1 4 | 0]
The rref form gives the system of equations
x1 + 2x3 − 3x4 = 0
66
x2 − x3 + 4x4 = 0
Solve the system for the pivot variables, x1 and x2.
x1 = − 2x3 + 3x4
x2 = x3 − 4x4
So we could also express the system as
x1
3
−2
x2
1
−4
=
x
+
x
3
4
x3
1
0
x4
0
1
The orthogonal complement is
3
−2
1
−4
V ⊥ = Span(
,
1
0 )
0
1
8.
B. In order to transform a vector in the alternate basis in the domain
into a vector in the alternate basis in the codomain, we need to find
the transformation matrix M.
[T( x ⃗ )]B = M[ x ⃗ ]B
We know that M = C −1AC, and A was given to us in the problem as
part of T( x ⃗ ), so we just need to find C and C −1.
67
The change of basis matrix C for the basis B is made of the column
vectors that span B, v 1⃗ = (−2,3) and v 2⃗ = (−4,0), so
C=
−2 −4
[3
0]
Now we’ll find C −1.
[C | I ] =
[C | I ] =
−2 −4 | 1 0
[3
0 | 0 1]
1 2 | − 12 0
[3 0 |
1
[C | I ] =
2
| − 12 0
3
2
0 −6 |
1 2 | − 12
[C | I ] =
1]
0
1
0
0 1 | − 14 − 16
1 0 |
[C | I ] =
0
0 1 | − 14
1
3
− 16
So,
C −1 =
0
− 14
1
3
− 16
68
With A, C, and C −1, we can find M = C −1AC.
0
M=
− 14
0
M=
− 14
0
M=
− 14
0
M=
− 14
1
3
− 16
−3 10 −2 −4
[ 0 4 ][ 3
0]
1
3
− 16
−3(−2) + 10(3) −3(−4) + 10(0)
[ 0(−2) + 4(3)
0(−4) + 4(0) ]
1
3
− 16
6 + 30 12 + 0
[0 + 12 0 + 0 ]
1
3
− 16
36 12
[12 0 ]
0(36) + 13 (12)
M=
0(12) + 13 (0)
− 14 (36) − 16 (12) − 14 (12) − 16 (0)
M=
0+4
0+0
[−9 − 2 −3 − 0]
M=
4
0
[−11 −3]
We’ve been asked to transform [ x ⃗ ]B = (−1,4), so we’ll multiply M by
this vector.
[T( x ⃗ )]B = M[ x ⃗ ]B
69
9.
[T( x ⃗ )]B =
4
0
−1
[−11 −3] [ 4 ]
[T( x ⃗ )]B =
4(−1) + 0(4)
[−11(−1) − 3(4)]
[T( x ⃗ )]B =
−4 + 0
[11 − 12]
[T( x ⃗ )]B =
−4
[−1]
First put M into reduced row-echelon form.
1 0
2
1
0
2
3
0
6
−3 1
0
−3 1
0
0 1
6
→
→
→
6 −2 0
6 −2 0
6 −2 0
0 −1 −6
0 −1 −6
0 −1 −6
1 0
2
1 0
2
1 0
0 1
6
0 1
6
0 1
→
→
0 −2 −12
0 0
0
0 0
0 0
0 −1 −6
0 −1 −6
2
6
0
0
In rref(M ), the pivot columns are the first and second columns,
which means C(M ) is given by the span of the first and second
columns of M.
70
3
0
−3
C(M ) = Span(
, 1 )
6
−2
−1
0
If we augment rref(M ) with the zero vector, we get the system of
equations
x1 + 2x3 = 0
x2 + 6x3 = 0
Solve the system for the pivot variables in terms of the free variable.
x1 = − 2x3
x2 = − 6x3
Then the solution to the system is
x1
−2
x2 = x3 −6
[1]
x3
which means the null space is given as
−2
N(M ) = Span( −6 )
[1]
Find M T,
71
3 −3 6
0
M = 0 1 −2 −1
6 0
0 −6
T
then put M T into reduced row-echelon form.
1 −1 2
0
1 −1 2
0
1 0 0 −1
0 1 −2 −1 → 0 1 −2 −1 → 0 1 −2 −1 →
6 0
0 −6
0 6 −12 −6
0 6 −12 −6
1 0 0 −1
0 1 −2 −1
[0 0 0
0]
In rref(M T ), the pivot columns are the first and second columns,
which means C(M T ) is given by the span of the first and second
columns of M T.
3
−3
C(M T ) = Span( 0 , 1 )
6 [0]
If we augment rref(M T ) with the zero vector, we get the system of
equations
x1 − x4 = 0
x2 − 2x3 − x4 = 0
Solve the system for the pivot variables in terms of the free
variables.
x1 = x4
72
x2 = 2x3 + x4
Then the solution to the system is
x1
x2
x3 = x3
x4
0
2
+ x4
1
0
1
1
0
1
which means the left null space is given as
0
1
1
2
N(M T ) = Span(
,
0 )
1
0
1
Because M is an m × n = 4 × 3 matrix, the row space and null space
are defined in ℝn = ℝ3, and the column space and left null space are
defined in ℝm = ℝ4.
The dimension of the column space and row space is the rank of M,
r = 2. The dimension of the null space is n − r = 3 − 2 = 1, and the
dimension of the left null space is m − r = 4 − 2 = 2.
In summary, the four fundamental subspaces are
3
0
−3
C(M ) = Span(
, 1 )
6
−2
−1
0
ℝ4
Dim = 2
73
10.
−2
N(M ) = Span( −6 )
[1]
ℝ3
Dim = 1
3
−3
C(M T ) = Span( 0 , 1 )
6 [0]
ℝ3
Dim = 2
0
1
1
2
N(M T ) = Span(
,
0 )
1
0
1
ℝ4
Dim = 2
Put each line into slope-intercept form,
y=−
1
x−2
3
y =x+4
y=1
then graph all three in the same plane.
74
While there are three points where some of the lines intersect,
there’s no single point where all three lines intersect, which means
there’s no solution to A x ⃗ = b ,⃗ which means there’s no vector
x ⃗ = (x, y) that satisfies the equation A x ⃗ = b ,⃗
1 3 x
−6
−1 1 [y] = 4
[ 0 1]
[1]
In other words, b ⃗ = (−6,4,1) is not in the column space of A. The next
best thing we can do is find the least squares solution. By building
the matrix equation, we’ve already found
1 3
A = −1 1
[ 0 1]
Now we’ll find A T.
75
AT =
1 −1 0
[3 1 1]
Then A T A is
1 3
1
−1
0
AT A =
[3 1 1] [−1 1]
0 1
AT A =
1(1) − 1(−1) + 0(0) 1(3) − 1(1) + 0(1)
[3(1) + 1(−1) + 1(0) 3(3) + 1(1) + 1(1)]
AT A =
1+1+0 3−1+0
[3 − 1 + 0 9 + 1 + 1]
AT A =
2 2
[2 11]
And A T b ⃗ is
−6
1
−1
0
AT b ⃗ =
[3 1 1] [ 4 ]
1
AT b ⃗ =
1(−6) − 1(4) + 0(1)
[3(−6) + 1(4) + 1(1)]
AT b ⃗ =
−6 − 4 + 0
[−18 + 4 + 1]
AT b ⃗ =
−10
[−13]
Then we get
76
A T A x *⃗ = A T b ⃗
−10
2 2
x *⃗ =
[2 11]
[−13]
⃗ we’ll put the augmented matrix into reduced rowThen to find x *,
echelon form.
2 2 | −10
1 1 | −5
1 1 | −5
→
→
→
[2 11 | −13] [2 11 | −13] [0 9 | −3]
1 1 | −5
1
[0 1 | − 3 ]
→
1 0 | − 14
3
0 1 |
− 13
Then the least squares solution is given by the augmented matrix as
x *⃗ =
11.
14
1
− ,−
( 3
3)
Define v 1⃗ = (2,0, − 2), v 2⃗ = (0, − 2,4), and v 3⃗ = (−2, − 2,3).
V = Span( v 1⃗ , v 2⃗ , v 3⃗ )
The length of v 1⃗ is
| | v 1⃗ | | =
22 + 02 + (−2)2 =
4+0+4 =
8=2 2
Then if u 1⃗ is the normalized version of v 1⃗ , we can say
77
2
u 1⃗ =
0
2 2 [−2]
1
So we can say that V is spanned by u 1⃗ , v 2⃗ , and v 3⃗ .
V1 = Span( u 1⃗ , v 2⃗ , v 3⃗ )
We’ll name w 2⃗ as the vector that connects ProjV1 v 2⃗ to v 2⃗ .
w 2⃗ = v 2⃗ − ProjV1 v 2⃗
w 2⃗ = v 2⃗ − ( v 2⃗ ⋅ u 1⃗ ) u1 ⃗
Plug in the values we already have.
0
0
2
2
1
1
w 2⃗ = −2 − ( −2 ⋅
0
0
[4]
[ 4 ] 2 2 [−2]) 2 2 [−2]
0
0
2
2
1
w 2⃗ = −2 − ( −2 ⋅ 0 ) 0
[ 4 ] 8 [ 4 ] [−2] [−2]
0
2
1
w 2⃗ = −2 − ((0)(2) + (−2)(0) + (4)(−2)) 0
[4] 8
[−2]
0
2
1
w 2⃗ = −2 − (0 + 0 − 8) 0
[4] 8
[−2]
0
2
1
w 2⃗ = −2 − (−8) 0
[4] 8
[−2]
78
0
2
w 2⃗ = −2 + 0
[ 4 ] [−2]
2
w 2⃗ = −2
[2]
So w 2⃗ is orthogonal to u 1⃗ , but it hasn’t been normalized, so let’s
normalize it. The length of w 2⃗ is
| | w 2⃗ | | =
22 + (−2)2 + 22
| | w 2⃗ | | =
4+4+4
| | w 2⃗ | | =
12
| | w 2⃗ | | = 2 3
Then the normalized version of w 2⃗ is u 2⃗ .
2
u 2⃗ =
−2
[
2 3 2]
1
So we can say that V is spanned by u 1⃗ , u 2⃗ , and v 3⃗ . Then the vector
w 3⃗ is given by
w 3⃗ = v 3⃗ − ProjV1 v 3⃗ − ProjV2 v 3⃗
w 3⃗ = v 3⃗ − ( v 3⃗ ⋅ u 1⃗ ) u1 ⃗ − ( v 3⃗ ⋅ u 2⃗ )u2⃗
Plug in the values we already have.
79
−2
−2
2
2
1
1
w 3⃗ = −2 − ( −2 ⋅
0 )
0
[3]
[ 3 ] 2 2 [−2] 2 2 [−2]
−2
2
2
1
1
− ( −2 ⋅
−2 )
−2
[3] 2 3[2] 2 3[2]
−2
2
2
−2
2
2
1 −2
1
w 3⃗ = −2 − ( −2 ⋅ 0 ) 0 − ( −2 ⋅ −2 ) −2
[ 3 ] 8 [ 3 ] [−2] [−2] 12 [ 3 ] [ 2 ] [ 2 ]
−2
2
1
w 3⃗ = −2 − (−2(2) − 2(0) + 3(−2)) 0
[3] 8
[−2]
2
1
− (−2(2) − 2(−2) + 3(2)) −2
12
[2]
−2
2
2
1
1
w 3⃗ = −2 − (−4 + 0 − 6) 0 − (−4 + 4 + 6) −2
[3] 8
[−2] 12
[2]
−2
5 2
1 2
w 3⃗ = −2 +
0 −
−2
[ 3 ] 4 [−2] 2 [ 2 ]
5
2
−2
w 3⃗ = −2 + 0
[3]
− 52
1
− −1
[1]
80
w 3⃗ =
w 3⃗ =
5
2
−2 +
−1
−2 + 0 + 1
3−
5
2
−1
− 12
−1
− 12
So w 3⃗ is orthogonal to u 2⃗ , but it hasn’t been normalized, so let’s
normalize it. The length of w 3⃗ is
| | w 3⃗ | | =
1
1
−
+ (−1)2 + −
( 2)
( 2)
| | w 3⃗ | | =
1
1
+1+
4
4
| | w 3⃗ | | =
3
2
2
2
Then the normalized version of w 3⃗ is u 3⃗ .
u 3⃗ =
2
3
− 12
−1
− 12
Therefore, we can say that u 1⃗ , u 2⃗ , and u 3⃗ form an orthonormal
basis for V.
81
2
2
1
V3 = Span(
0 ,
−2 ,
[
]
[
2 2 −2 2 3 2 ]
1
1
V3 = Span(
12.
2
−
−
1
0
1
3
, −
2
1
3
, −
1
3
−
2
3
1
−2
−1 )
− 12
1
6
2
3
)
1
6
Starting with
−1 4 −2
A= 0 2 0
[ 0 0 −1]
we’ll first find the determinant | λIn − A | .
1 0 0
−1 4 −2
λ 0 1 0 − 0 2 0
[0 0 1] [ 0 0 −1]
λ 0 0
−1 4 −2
0 λ 0 − 0 2 0
[ 0 0 −1]
0 0 λ
λ + 1 −4
2
0
λ−2
0
0
0
λ+1
82
Then let’s work along the first column, since it includes two 0 values,
to find the determinant of this resulting matrix.
(λ + 1)
λ−2
0
−4
2
−4 2
−0
+0
0 λ+1
0
λ+1
λ−2 0
The last two determinants cancel, leaving us with just
(λ + 1)
λ−2
0
0
λ+1
(λ + 1)[(λ − 2)(λ + 1) − (0)(0)]
(λ + 1)2(λ − 2)
Remember that we’re trying to satisfy | λIn − A | = 0, so we can set
this characteristic polynomial equal to 0 to get the characteristic
equation, and then we’ll solve for λ.
(λ + 1)2(λ − 2) = 0
λ = − 1 or λ = 2
With these three eigenvalues, we’ll have three eigenspaces, given
by Eλ = N(λIn − A). Given
Eλ = N
λ + 1 −4
2
0
λ−2
0
0
0
λ+1
we get
83
−1 + 1
−4
2
0
−1 − 2
0
0
0
−1 + 1
E−1 = N
0 −4 2
E−1 = N
0 −3 0
([0 0 0])
and
2 + 1 −4
2
0
2−2
0
0
0
2+1
E2 = N
3 −4 2
E2 = N
0 0 0
([0 0 3])
Therefore, the eigenvectors in the eigenspace E−1 will satisfy
0 −4 2
0
0 −3 0 v ⃗ = 0
[0 0 0]
[0]
0 −4 2 | 0
0 1
0 −3 0 | 0 → 0 −3
0 0 0 | 0
0 0
0 1 − 12
0 0
0 0
1
0
− 12
0
0
| 0
0 1 − 12
| 0 → 0 0 − 32
| 0
0 0 0
| 0
| 0 →
| 0
0 1 0 | 0
| 0 → 0 0 1 | 0
0 0 0 | 0
| 0
| 0
84
This gives
v2 = 0
v3 = 0
So we can say
v1
1
v2 = v1 0
[0]
v3
Which means that E−1 is defined by
1
E−1 = Span( 0 )
[0]
And the eigenvectors in the eigenspace E2 will satisfy
3 −4 2
0
0 0 0 v⃗= 0
[0 0 3]
[0]
3 −4 2 | 0
1 − 43
0 0 0 | 0 → 0 0
0 0 3 | 0
0 0
1 − 43
0
0
0
0
2
3
| 0
2
3
| 0
1 − 43
0 | 0 → 0
3 | 0
0
0
0
2
3
| 0
3 | 0 →
0 | 0
1 − 43 0 | 0
1 | 0 → 0
0 | 0
0
0
0
1 | 0
0 | 0
This gives
85
v1 −
4
v2 = 0
3
v3 = 0
or
v1 =
4
v2
3
v3 = 0
So we can say
4
v1
3
v2 = v2
1
v3
0
Which means that E2 is defined by
4
3
E2 = Span( 1 )
0
Let’s put these results together. For the eigenvalues λ = − 1 and
λ = 2, respectively, we got
4
1
3
E−1 = Span( 0 ) and E2 = Span( 1 )
[0]
0
86
Each of these spans represents a line in ℝ3. So for any vector v ⃗
along any of these lines, when we apply the transformation T to the
vector v ,⃗ T( v ⃗ ) will be a vector along the same line, it might just be
scaled up or scaled down. Specifically,
• since λ = 2 in the eigenspace E2, any vector v ⃗ in E2, under the
transformation T, will be scaled by 2, meaning that
T( v ⃗ ) = λ v ⃗ = 2 v ,⃗ and
• since λ = − 1 in the eigenspace E−1, any vector v ⃗ in E−1, under
the transformation T, will be scaled by −1, meaning that
T( v ⃗ ) = λ v ⃗ = − v .⃗
87
88
Linear Algebra Final Exam
This exam is comprehensive over the entire course and includes 12
questions. You have 60 minutes to complete the exam.
The exam is worth 100 points. The 8 multiple choice questions are worth 5
points each (40 points total) and the 4 free response questions are worth
15 points each (60 points total).
Mark your multiple choice answers on this cover page. For the free
response questions, show your work and make sure to circle your final
answer.
1. (5 pts)
A
B
C
D
E
2. (5 pts)
A
B
C
D
E
3. (5 pts)
A
B
C
D
E
4. (5 pts)
A
B
C
D
E
5. (5 pts)
A
B
C
D
E
6. (5 pts)
A
B
C
D
E
7. (5 pts)
A
B
C
D
E
8. (5 pts)
A
B
C
D
E
89
1. (5 pts) What is the reduced row-echelon form of the matrix?
1 −3 2
0
−2 0
1
1
A=
0
2 −1 0
3
1
2 −2
A
1
0
rref(A) =
0
0
0
1
0
0
0 0
0 0
1 1/2
0 1
B
1
0
rref(A) =
0
0
0 0
0
1 −1 0
0 1 1/2
0 0
0
C
1
0
rref(A) =
0
0
0 0 −3
1 −2 1
0 1
0
0 0
0
D
1
0
rref(A) =
0
0
0
1
0
0
0
0
1
0
0
0
0
1
E
1
0
rref(A) =
0
0
0
1
0
0
0
2
1
0
3
0
0
0
90
2. (5 pts) What is the matrix product?
−2 1 2 5
AB = 0 −1 4 2
1 −3 1 −1
1
1
2
−5 0
2
2 −3 −1
2
0 −1
A
7 −8 −9
AB = 17 −12 −8
16 −2 −4
B
−8 7 −9
AB = −12 17 −8
−2 16 −4
C
−9 −8 7
AB = −8 −12 17
−4 −2 16
D
17 −12 −8
AB = 7 −8 −9
16 −2 −4
E
7 −8 −9
AB = 16 −2 −4
17 −12 −8
91
3. (5 pts) Find the vector sum 2 a ⃗ − 3 b ⃗ + 5 c ⃗ − d ,⃗ if a ⃗ = (2,6, − 1),
b ⃗ = (−3,1,1), c ⃗ = (0,5, − 2), and d ⃗ = (1,4, − 4).
A
(−12, − 30,11)
D
(14,38, − 19)
B
(6, − 36,5)
E
(−6,36, − 5)
C
(12,30, − 11)
4. (5 pts) What is the length of x ⃗ = (4, − 2,1,0)?
A
| | x ⃗| | =
13
D
| | x ⃗| | = −
B
| | x ⃗| | =
21
E
| | x ⃗| | =
C
| | x ⃗| | = −
21
9
13
92
5. (5 pts) Find the equation of a plane with normal vector n ⃗ = (2, − 6,1) that
passes through (5,2, − 3).
A
2x − 6y + z = − 5
D
2x − 6y + z = − 1
B
2x − 6y + z = 5
E
2x + 6y + z = − 5
C
2x − 6y + z = 1
6. (5 pts) Transform x ⃗ = (4, − 1) with T ∘ S, if S : ℝ2 → ℝ2 and T : ℝ2 → ℝ2.
−2x1 + x2
S( x ⃗ ) =
[ 3x2 ]
T( x ⃗ ) =
x1 − 4x2
[ −4x2 ]
A
(T ∘ S)( x ⃗ ) = (−12, − 3)
D
(T ∘ S)( x ⃗ ) = (12,3)
B
(T ∘ S)( x ⃗ ) = (−3, − 12)
E
(T ∘ S)( x ⃗ ) = (3,12)
C
(T ∘ S)( x ⃗ ) = (−3,12)
93
7. (5 pts) Find the transformation in the alternate basis, [T( x ⃗ )]B of the
vector x ⃗ = (6, − 1).
T( x ⃗ ) =
1 −2
x⃗
[4 −4]
−3
2
B = Span(
,
[−1] [ 0 ])
A
28
[T( x ⃗ )]B =
64
[− 3 ]
−28
[
]
B
[T( x ⃗ )]B =
C
−28
[T( x ⃗ )]B =
64
[− 3 ]
64
3
D
E
64
3
[T( x ⃗ )]B =
[28]
[T( x ⃗ )]B =
− 64
3
[−28]
94
8. (5 pts) Use the Gram-Schmidt process to change the basis of V into an
orthonormal basis.
2
0
−1
V = Span( −1 , 0 , 1 )
[ 1 ] [−3] [ 1 ]
A
B
C
D
E
0
−1 , −
[
2 1]
V = Span(
1
V = Span(
1
V = Span(
1
4
3
1
−3 ,
2 )
[
]
[
34 −3
17 2]
1
4
0
1
−3 , −
−1 ,
[
]
[
34 −3]
2 1
3
2 )
[
17 2]
1
4
0
3
1
1
−3 ,
−1 ,
2 )
[
]
[
]
[
34 −3
17 2]
2 1
4
0
3
1
1
−3 ,
−1 ,
2 )
[
]
[
]
[
34 −3
17 2]
2 1
V = Span( −
1
V = Span( −
1
0
−1 , −
[
2 1]
4
3
1
−3 ,
2 )
[
]
[
34 −3
17 2]
1
95
9. (15 pts) Solve A x ⃗ = b ,⃗ using values b1 = − 2, b2 = 1, and b3 = 1.
1
1
A= 1
2
[−2 −3]
10. (15 pts) Find the four fundamental subspaces of A, including their
spaces and dimensions.
1 −2 0 1
A = −1 0
2 4
[0
1 −1 2]
96
11. (15 pts) Find the orthogonal complement of V.
−1
0
2
0
−2
1
V = Span(
,
,
1
0
−2 )
1
0
2
12. (15 pts) Find the Eigenvalues and Eigenvectors of the matrix, then
describe what’s happening in the Eigenbases.
3 6 −8
A= 0 0 6
0 0 2
97
Linear Algebra Final Exam Answer Key
1. (5 pts)
A
B
C
D
E
2. (5 pts)
A
B
C
D
E
3. (5 pts)
A
B
C
D
E
4. (5 pts)
A
B
C
D
E
5. (5 pts)
A
B
C
D
E
6. (5 pts)
A
B
C
D
E
7. (5 pts)
A
B
C
D
E
8. (5 pts)
A
B
C
D
E
99
9. (15 pts) x ⃗ =
−5
[3]
1
−2
1
10. (15 pts) C(A) = Span( −1 , 0 , 4 )
[ 0 ] [ 1 ] [2]
ℝ3
Dim = 3
2
1
N(A) = Span(
1 )
0
ℝ4
Dim = 1
1
−1
0
−2
0
C(A T ) = Span(
,
, 1 )
0
2
−1
1
4
2
ℝ4
Dim = 3
0
N(A ) = 0
[0]
ℝ3
Dim = 0
T
2
11. (15 pts) V ⊥ = Span(
10
3
5
3
)
1
100
−10
1
−2
12. (15 pts) E0 = Span( 1 ), E2 = Span( 3 ), E3 = Span( 0 )
[0]
[ 1 ]
[0]
• Since λ = 0 in the eigenspace E0, any vector v ⃗ in E0,
under the transformation T, will be scaled down to the
zero vector, meaning that T( v ⃗ ) = λ v ⃗ = 0 v ⃗ = O.⃗
• Since λ = 2 in the eigenspace E2, any vector v ⃗ in E2,
under the transformation T, will be scaled by 2, meaning
that T( v ⃗ ) = λ v ⃗ = 2 v .⃗
• Since λ = 3 in the eigenspace E3, any vector v ⃗ in E3,
under the transformation T, will be scaled by 3, meaning
that T( v ⃗ ) = λ v ⃗ = 3 v .⃗
101
Linear Algebra Final Exam Solutions
1.
D. To put A into reduced row-echelon form, start by working on the
first column.
1 −3 2
0
1 −3 2
0
1 −3 2
0
−2 0
1
1
0 −6 5
1
0 −6 5
1
→
→
0
2 −1 0
0 2 −1 0
0 2 −1 0
3
1
2 −2
3 1
2 −2
0 10 −4 −2
Find the pivot entry in the second column, then zero out the rest of
the second column.
1 −3
0
2
5
1
0
1
−6 −6
0 2 −1 0
0 10 −4 −2
1
0
− 12
− 56
− 12
− 16
→ 0 1
0 2 −1 0
0 10 −4 −2
→
1
0
0
1
0
0
− 12 − 12
− 56 − 16
2
3
1
3
0 10 −4 −2
1 0 − 12 − 12
0 1 − 56 − 16
0 0
0 0
2
3
13
3
1
3
1
−3
Find the pivot entry in the third column, then zero out the rest of the
third column.
102
1
1
1 0 −2 −2
0 1 − 56 − 16
0 0
1
0 0
13
3
1
2
− 13
→
1 0 0 − 14
0 1 0
0 0 1
0 0 0
1
0
0
0
2.
0
1
0
0
0
0
1
0
1
4
1
2
5
2
1
1 0
0
0 1 − 56 − 16
0 1
0
0 0
1
0 0
13
3
1 0
0
−4
0 0
1
0 0
13
3
1
2
− 13
→
1 0 0 − 14
→
0 1 0
0 0 1
0 0 0
1
4
1
2
→
1
1 0 0 0
0 1 0 14
0 0 1
1
2
0 0 0 1
− 14
1
4
1
2
− 13
1 0 0 0
0 1 0 0
→
0 0 1 12
0 0 0 1
0
0
0
1
A. Multiply each row of A by the first column of B.
−2 1 2 5
AB = 0 −1 4 2
1 −3 1 −1
1
1
2
−5 0
2
2 −3 −1
2
0 −1
−2(1) + 1(−5) + 2(2) + 5(2) . . .
AB = 0(1) − 1(−5) + 4(2) + 2(2) . . .
1(1) − 3(−5) + 1(2) − 1(2) . . .
...
...
...
103
−2 − 5 + 4 + 10 . . .
AB =
0+5+8+4
...
1 + 15 + 2 − 2 . . .
7 ...
AB = 17 . . .
[16 . . .
...
...
...
...
...
...]
Multiply each row of A by the second column of B.
7 −2(1) + 1(0) + 2(−3) + 5(0) . . .
AB = 17 0(1) − 1(0) + 4(−3) + 2(0) . . .
16 1(1) − 3(0) + 1(−3) − 1(0) . . .
7 −2 + 0 − 6 + 0 . . .
AB = 17 0 − 0 − 12 + 0 . . .
16 1 − 0 − 3 − 0 . . .
7 −8 . . .
AB = 17 −12 . . .
16 −2 . . .
Multiply each row of A by the third column of B.
7 −8 −2(2) + 1(2) + 2(−1) + 5(−1)
AB = 17 −12 0(2) − 1(2) + 4(−1) + 2(−1)
16 −2 1(2) − 3(2) + 1(−1) − 1(−1)
7 −8 −4 + 2 − 2 − 5
AB = 17 −12 0 − 2 − 4 − 2
16 −2 2 − 6 − 1 + 1
104
7 −8 −9
AB = 17 −12 −8
16 −2 −4
3.
C. The vector sum is
2 a ⃗− 3 b ⃗ + 5 c ⃗− d ⃗
2(2,6, − 1) − 3(−3,1,1) + 5(0,5, − 2) − (1,4, − 4)
Apply the scalars to each vector.
(4,12, − 2) + (9, − 3, − 3) + (0,25, − 10) + (−1, − 4,4)
Add each of the vector components.
(4 + 9 + 0 − 1,12 − 3 + 25 − 4, − 2 − 3 − 10 + 4)
(12,30, − 11)
4.
B. The length of x ⃗ = (4, − 2,1,0) is given by
| | x ⃗| | =
x12 + x22 + x32 + x42
| | x ⃗| | =
42 + (−2)2 + 12 + 02
| | x ⃗| | =
16 + 4 + 1 + 0
| | x ⃗| | =
21
105
5.
A. The equation of a plane with normal vector n ⃗ = (a, b, c) = (2, − 6,1),
which passes through (x0, y0, z0) = (5,2, − 3), is given by
a(x − x0) + b(y − y0) + c(z − z0) = 0
2(x − 5) − 6(y − 2) + 1(z − (−3)) = 0
Simplify to get the equation of the plane into standard form.
2x − 10 − 6y + 12 + z + 3 = 0
2x − 6y + z + 5 = 0
2x − 6y + z = − 5
6.
E. Given x ⃗ = (4, − 1) and
S( x ⃗ ) =
−2x1 + x2
[ 3x2 ]
T( x ⃗ ) =
x1 − 4x2
[ −4x2 ]
start by using S to transform the standard basis vectors.
S
−2(1) + 0
1
−2
=
=
([0]) [ 3(0) ] [ 0 ]
−2(0) + 1
0
1
S
=
=
([1]) [ 3(1) ] [3]
106
So the transformation S can be written as the matrix-vector product
S( x ⃗ ) =
−2 1
x⃗
[ 0 3]
Use T to transform the standard basis vectors.
T
1 − 4(0)
1
1
=
=
([0]) [ −4(0) ] [0]
T
0 − 4(1)
0
−4
=
=
([1]) [ −4(1) ] [−4]
So the transformation T can be written as the matrix-vector product
1 −4
T( y ⃗ ) =
y⃗
[0 −4]
If we call the matrix from S
A=
−2 1
[ 0 3]
and we call the matrix from T
B=
1 −4
[0 −4]
then the composition of the transformations is
T(S( x ⃗ )) = BA x ⃗
T(S( x ⃗ )) =
1 −4 −2 1
x⃗
[0 −4] [ 0 3]
107
T(S( x ⃗ )) =
1(−2) − 4(0) 1(1) − 4(3)
x⃗
[0(−2) − 4(0) 0(1) − 4(3)]
T(S( x ⃗ )) =
−2 −11
x⃗
[ 0 −12]
To transform x ⃗ = (4, − 1), multiply this transformation matrix by
x ⃗ = (4, − 1).
7.
T
−2 −11
4
4
=
( ([−1])) [ 0 −12] [−1]
T
−2(4) − 11(−1)
4
=
( ([−1])) [ 0(4) − 12(−1) ]
T
3
4
=
( ([−1])) [12]
S
S
S
C. Transform x ⃗ = (6, − 1) with the transformation T.
T( x ⃗ ) =
1 −2
x⃗
[4 −4]
T
6
6
1 −2
=
([−1]) [4 −4] [−1]
T
1(6) − 2(−1)
6
=
([−1]) [4(6) − 4(−1)]
108
T
8
6
=
([−1]) [28]
Now find C −1.
[C | I ] =
2 −3 | 1 0
[−1 0 | 0 1]
[C | I ] =
[−1
[C | I ] =
[C | I ] =
[C | I ] =
1
− 32
| 0 1]
0
1 − 32
|
1 − 32
|
0
|
1 0 |
1
2
1
2
|
0 − 32
1
1
2
|
1
2
− 13
0
0
0
1
0
− 23
−1
1
2
[0 1 | − 3 − 3 ]
To convert to the alternate basis, we need to multiply the
transformed vector by C −1.
0 −1
8
6
C T
=
1
2
([−1]) [− 3 − 3 ] [28]
−1
0(8) − 1(28)
6
C T
=
([−1]) [− 1 (8) − 2 (28)]
3
3
−1
109
−28
6
C T
=
([−1]) [− 64
3 ]
−1
This is the vector x ⃗ = (6, − 1) after the transformation T, and
converted into the alternate basis. So
−28
[T( x ⃗ )]B =
64
[− 3 ]
8.
C. The vectors v 1⃗ = (0, − 1,1), v 2⃗ = (2,0, − 3), and v 3⃗ = (−1,1,1) form
the basis for V.
V = Span( v 1⃗ , v 2⃗ , v 3⃗ )
Normalize v 1⃗ . The length of v 1⃗ is
| | v 1⃗ | | =
02 + (−1)2 + 12 =
0+1+1 =
2
Then the normalized version of v 1⃗ is
u 1⃗ =
0
−1
[
2 1]
1
Replace v 2⃗ with a vector that’s both orthogonal to u 1⃗ , and normal.
Name w 2⃗ as the vector that connects ProjV1 v 2⃗ to v 2⃗ .
w 2⃗ = v 2⃗ − ProjV1 v 2⃗
110
w 2⃗ = v 2⃗ − ( v 2⃗ ⋅ u 1⃗ ) u1 ⃗
2
2
0
0
1
1
w 2⃗ = 0 − ( 0 ⋅
−1 )
−1
[−3]
[−3]
[
]
[
2 1
2 1]
2
0
1
w 2⃗ = 0 − (2(0) + 0(−1) − 3(1)) −1
[−3] 2
[1]
2
3 0
w 2⃗ = 0 +
−1
[−3] 2 [ 1 ]
2
w 2⃗ = 0
[−3]
0
3
−
+
2
3
2
2
3
−
w 2⃗ =
2
− 32
Normalize w 2⃗ . The length of w 2⃗ is
| | w 2⃗ | | =
3
3
22 + −
+ −
( 2) ( 2)
| | w 2⃗ | | =
4+
| | w 2⃗ | | =
34
4
2
2
9 9
+
4 4
111
| | w 2⃗ | | =
34
2
Then the normalized version of w 2⃗ is u 2⃗ :
u 2⃗ =
2
34
1
u 2⃗ = ⋅
2
u 2⃗ =
2
− 32
− 32
4
−3
34 [−3]
2
4
−3
34 [−3]
1
Replace v 3⃗ with a vector that’s both orthogonal to u 1⃗ and u 2⃗ , and
normal. Name w 3⃗ as the vector that connects ProjV2 v 3⃗ to v 3⃗ .
w 3⃗ = v 3⃗ − ProjV2 v 3⃗
w 3⃗ = v 3⃗ − [( v 3⃗ ⋅ u 1⃗ ) u1 ⃗ + ( v 3⃗ ⋅ u 2⃗ )u2⃗]
0
0
−1
1
1
w 3⃗ = v 3⃗ − ( 1 ⋅
−1 )
−1 + ( v 3⃗ ⋅ u 2⃗ )u2⃗
[ [1]
]
2[1]
2[1]
0
1
w 3⃗ = v 3⃗ −
(−1(0) + 1(−1) + 1(1)) −1 + ( v 3⃗ ⋅ u 2⃗ )u2⃗
[2
[1]
]
112
0
1
w 3⃗ = v 3⃗ −
(0) −1 + ( v 3⃗ ⋅ u 2⃗ )u2⃗
[2 [ 1 ]
]
−1
−1
w 3⃗ = 1 − 0 + ( 1 ⋅
[1] [
[1]
4
4
1
−3 )
−3
34 [−3]
34 [−3]]
1
4
4
−1
−1
1
w 3⃗ = 1 − ( 1 ⋅ −3 ) −3
[ 1 ] 34 [ 1 ] [−3] [−3]
4
−1
1
w 3⃗ = 1 − (−1(4) + 1(−3) + 1(−3)) −3
[ 1 ] 34
[−3]
4
−1
5
w 3⃗ = 1 +
−3
[ 1 ] 17 [−3]
−1
w 3⃗ = 1 +
[1]
w 3⃗ =
20
17
− 15
17
− 15
17
3
17
2
17
2
17
1 3
w 3⃗ =
2
17 [ ]
2
113
Normalize w 3⃗ . The length of w 3⃗ is
| | w 3⃗ | | =
3
2
2
+
+
( 17 ) ( 17 ) ( 17 )
| | w 3⃗ | | =
9
4
4
+
+
289 289 289
| | w 3⃗ | | =
17
289
| | w 3⃗ | | =
17
2
2
2
17
Then the normalized version of w 3⃗ is u 3⃗ :
u 3⃗ =
17
17
3
17
2
17
2
17
1
17 3
u 3⃗ =
⋅
2
17
[
17 2]
u 3⃗ =
3
2
[
17 2]
1
The vectors u 1⃗ , u 2⃗ , and u 3⃗ form an orthonormal basis for V.
114
V = Span(
9.
4
0
3
1
1
−3 ,
−1 ,
2 )
2 [ 1 ] 34 [−3] 17 [2]
1
Put the matrix A into reduced row-echelon form.
1 0
1 0
1
1
1 1
1
1
1 → 0 1 → 0 1 → 0 1
1
2 → 0
[−2 −3] [−2 −3] [0 −1] [0 −1] [0 0]
To find the complementary solution, augment rref(A) with the zero
vector to get a system of equations.
x1 = 0
x2 = 0
Then the only vector that satisfies the null space is
x1
0
=
[x2] [0]
The complementary solution is
x n⃗ =
0
[0]
To find the particular solution, augment A with b ⃗ = (b1, b2, b3), then
put it in reduced row-echelon form.
115
1
1 | b1
1
1 |
b1
1 1 |
b1
1
2 | b2 → 0
1 | b2 − b1 → 0 1 | b2 − b1
−2 −3 | b3
−2 −3 |
b3
0 −1 | 2b1 + b3
1 0 | 2b1 − b2
1 0 |
2b1 − b2
0 1 | b2 − b1 → 0 1 |
b2 − b1
0 −1 | 2b1 + b3
0 0 | b1 + b2 + b3
From the third row, the system is constrained.
b1 + b2 + b3 = 0
b1 = − b2 − b3
Choose values of b1, b2, and b3 that make this equation true, like
b2 = 1, b3 = 1, and b1 = − 2. Then the matrix is
1 0 | 2(−2) − 1
1 0 | −5
0 1 | 1 − (−2) → 0 1 | 3
0 0 | −2 + 1 + 1
0 0 | 0
This gives a system of equations
x1 = − 5
x2 = 3
So the particular solution is
x p⃗ =
−5
[3]
116
The general solution is the sum of the particular and complementary
solutions.
x ⃗ = x p⃗ + x n⃗
x⃗=
0
−5
+
[ 3 ] [0]
Of course, adding the zero vector doesn’t change the value of the
general solution, so the general solution is
x⃗=
10.
−5
[3]
First put A into reduced row-echelon form.
1 −2 0 1
1 −2 0 1
1 −2 0 1
A = −1 0
2 4 → 0 −2 2 5 → 0 1 −1 2
[0
1 −1 2]
0 1 −1 2
0 −2 2 5
1 0 −2 5
1 0 −2 5
1 0 −2 5
→ 0 1 −1 2 → 0 1 −1 2 → 0 1 −1 2
0 0 0 9
0 0 0 1
0 −2 2 5
1 0 −2 0
1 0 −2 0
→ 0 1 −1 2 → 0 1 −1 0
[0 0 0 1] [0 0 0 1]
In rref(A), the pivot columns are the first, second, and fourth
columns, which means C(A) is given by the span of the first, second,
and fourth columns of A.
117
1
−2
1
C(A) = Span( −1 , 0 , 4 )
[ 0 ] [ 1 ] [2]
From rref(A), we get the system of equations
x1 − 2x3 = 0
x2 − x3 = 0
x4 = 0
Solve the system for the pivot variables.
x1 = 2x3
x2 = x3
x4 = 0
Then the solution to the system is
x1
x2
x3 = x3
x4
2
1
1
0
which means the null space is given as
2
1
N(A) = Span(
1 )
0
118
Find A T,
1 −1 0
−2 0
1
AT =
0
2 −1
1
4
2
then put A T into reduced row-echelon form.
1 −1 0
1 −1 0
1 −1 0
0 −2 1
−2 0
1
0 −2 1
AT =
→
→
0 2 −1
0
2 −1
0 2 −1
0 5
2
1
4
2
1 4
2
1 −1 0
1 −1 0
0 1 − 12
0 −2 1
→
→
0 0 0
0 0
0
0 5 2
0 5
2
1 0 − 12
1
→ 0 1 −2
0 0 0
0 0 1
→
1 0 0
0 1 − 12
0 0
0 0
0
1
→
1
0
→
0
0
1 −1 0
0 1 − 12
0
0
0
0
0
1
0
0
0
1 0 − 12
→
9
2
0
1 0
0
0 1
→
0
0 0
1
0 0
0 1 − 12
0 0
0 0
0
9
2
0
0
1
0
In rref(A T ), the pivot columns are the first, second, and third
columns, which means C(A T ) is given by the span of the first,
second, and third columns of A T.
119
1
−1
0
−2
0
C(A T ) = Span(
,
, 1 )
0
2
−1
1
4
2
From rref(A T ), we get the system of equations
x1 = 0
x2 = 0
x3 = 0
Then the solution to the system is
x1
0
x2 = 0
[0]
x3
which means the left null space is given as
0
N(A T ) = 0
[0]
Because A is an m × n = 3 × 4 matrix, the row space and null space
are defined in ℝn = ℝ4, and the column space and left null space are
defined in ℝm = ℝ3.
The dimension of the column space and row space is the rank of A,
r = 3. The dimension of the null space is n − r = 4 − 3 = 1, and the
dimension of the left null space is m − r = 3 − 3 = 0.
120
In summary, the four fundamental subspaces are
1
−2
1
C(A) = Span( −1 , 0 , 4 )
[ 0 ] [ 1 ] [2]
ℝ3
Dim = 3
2
1
N(A) = Span(
1 )
0
ℝ4
Dim = 1
1
−1
0
−2
0
C(A T ) = Span(
,
, 1 )
0
2
−1
1
4
2
ℝ4
Dim = 3
0
N(A ) = 0
[0]
ℝ3
Dim = 0
T
11.
The subspace V is a plane in ℝ4, spanned by the three vectors
v 1⃗ = (2, − 2,1,1), v 2⃗ = (−1,0,0,2), and v 3⃗ = (0,1, − 2,0). Its orthogonal
complement V ⊥ is the set of vectors which are orthogonal to both all
three.
−1
0
2
0
−2
1
V ⊥ = { x ⃗ ∈ ℝ4 | x ⃗ ⋅
= 0 , x ⃗⋅
= 0, x ⃗ ⋅
1
0
−2 }
1
0
2
Let x ⃗ = (x1, x2, x3, x4) to get three equations from these dot products.
121
2x1 − 2x2 + x3 + x4 = 0
−x1 + 2x4 = 0
x2 − 2x3 = 0
Put these equations into an augmented matrix,
2 −2 1 1 | 0
−1 0
0 2 | 0
0
1 −2 0 | 0
then put it into reduced row-echelon form.
−1 0
0 2 | 0
1 0
0 −2 | 0
2 −2 1 1 | 0 → 2 −2 1
1 | 0
0
1 −2 0 | 0
0 1 −2 0 | 0
1 0
0 −2 | 0
1 0
0 −2 | 0
0 −2 1
5 | 0 → 0 1 −2 0 | 0
0 1 −2 0 | 0
0 −2 1
5 | 0
1 0 0 −2 | 0
1 0 0 −2 | 0
0 1 −2 0 | 0 → 0 1 −2 0 | 0
0 0 −3 5 | 0
0 0 1 − 53 | 0
1 0 0
−2
0 1 0 − 10
3
0 0 1 − 53
| 0
| 0
| 0
122
The rref form gives the system of equations
x1 − 2x4 = 0
x2 −
10
x4 = 0
3
x3 −
5
x4 = 0
3
Solve the system for the pivot variables, x1, x2, and x3.
x1 = 2x4
10
x2 =
x
3 4
5
x3 = x4
3
So we could also express the system as
x1
x2
x3 = x4
x4
2
10
3
5
3
1
The orthogonal complement is
2
V ⊥ = Span(
10
3
5
3
)
1
123
12.
Starting with
3 6 −8
A= 0 0 6
0 0 2
first, find the determinant | λIn − A | .
1 0 0
3 6 −8
λ 0 1 0 − 0 0 6
[0 0 1]
0 0 2
λ 0 0
3 6 −8
0 λ 0 − 0 0 6
0 0 λ
0 0 2
λ − 3 −6
8
0
λ
−6
0
0 λ−2
Then let’s work along the first column, since it includes two 0 values,
to find the determinant of this resulting matrix.
(λ − 3)
−6 8
λ −6
−6
8
−0
+0
0 λ−2
0 λ−2
λ −6
The last two determinants cancel, leaving us with just
(λ − 3)
λ −6
0 λ−2
124
(λ − 3)λ(λ − 2) − (−6)(0)
λ(λ − 3)(λ − 2)
Remember that we’re trying to satisfy | λIn − A | = 0, so we can set
this characteristic polynomial equal to 0 to get the characteristic
equation, and then we’ll solve for λ.
λ(λ − 3)(λ − 2) = 0
λ = 0 or λ = 2 or λ = 3
With these three eigenvalues, we’ll have three eigenspaces, given
by Eλ = N(λIn − A). Given
Eλ = N
λ − 3 −6
8
0
λ
−6
0
0 λ−2
we get
E0 = N
0 − 3 −6
8
0
0
−6
0
0 0−2
E0 = N
−3 −6 8
0
0 −6
0
0 −2
and
125
E2 = N
2 − 3 −6
8
0
2
−6
0
0 2−2
E2 = N
−1 −6 8
0
2 −6
0
0
0
E3 = N
3 − 3 −6
8
0
3
−6
0
0 3−2
E3 = N
0 −6 8
0 3 −6
0 0
1
and
Therefore, the eigenvectors in the eigenspace E0 will satisfy
−3 −6 8
0
0 −6
0
0 −2
0
v⃗= 0
[0]
−3 −6 8 | 0
1 2 − 83 | 0
1 2 − 83 | 0
0
0 −6 | 0 → 0 0 −6 | 0 → 0 0 1 | 0
0
0 −2 | 0
0 0 −2 | 0
0 0 −2 | 0
126
1 2 0 | 0
1 2 0 | 0
0 0 1 | 0 → 0 0 1 | 0
0 0 −2 | 0
0 0 0 | 0
This gives
v1 + 2v2 = 0
v3 = 0
or
v1 = − 2v2
v3 = 0
So we can say
v1
−2
v2 = v2 1
[0]
v3
Which means that E0 is defined by
−2
E0 = Span( 1 )
[0]
And the eigenvectors in the eigenspace E2 will satisfy
−1 −6 8
0
2 −6
0
0
0
0
v⃗= 0
[0]
127
−1 −6 8 | 0
1 6 −8 | 0
1 0 10 | 0
0
2 −6 | 0 → 0 2 −6 | 0 → 0 2 −6 | 0
0
0
0 | 0
0 0 0 | 0
0 0 0 | 0
1 0 10 | 0
0 1 −3 | 0
0 0 0 | 0
This gives
v1 + 10v3 = 0
v2 − 3v3 = 0
or
v1 = − 10v3
v2 = 3v3
So we can say
v1
−10
v2 = v3 3
[ 1 ]
v3
Which means that E2 is defined by
−10
E2 = Span( 3 )
[ 1 ]
And the eigenvectors in the eigenspace E3 will satisfy
128
0 −6 8
0 3 −6
0 0
1
0
v⃗= 0
[0]
0 −6 8 | 0
0 1 − 43 | 0
0 1 − 43 | 0
0 3 −6 | 0 → 0 3 −6 | 0 → 0 0 −2 | 0
0 0
1 | 0
0 0 1 | 0
0 0 1 | 0
0 1 − 43
0 0
0 0
1
1
0 1 0 | 0
0 1 0 | 0
| 0 → 0 0 1 | 0 → 0 0 1 | 0
0 0 1 | 0
0 0 0 | 0
| 0
| 0
This gives
v2 = 0
v3 = 0
So we can say
v1
1
v2 = v1 0
[0]
v3
Which means that E3 is defined by
1
E3 = Span( 0 )
[0]
Let’s put these results together. For the eigenvalues λ = 0, λ = 2, and
λ = 3, respectively, we got
129
−10
1
−2
E0 = Span( 1 ), E2 = Span( 3 ), and E3 = Span( 0 )
[0]
[ 1 ]
[0]
Each of these spans represents a line in ℝ3. So for any vector v ⃗
along any of these lines, when you apply the transformation T to the
vector v ,⃗ T( v ⃗ ) will be a vector along the same line, it might just be
scaled up or scaled down. Specifically,
• since λ = 0 in the eigenspace E0, any vector v ⃗ in E0, under the
transformation T, will be scaled down to the zero vector,
meaning that T( v ⃗ ) = λ v ⃗ = 0 v ⃗ = O,⃗
• since λ = 2 in the eigenspace E2, any vector v ⃗ in E2, under the
transformation T, will be scaled by 2, meaning that
T( v ⃗ ) = λ v ⃗ = 2 v ,⃗ and
• since λ = 3 in the eigenspace E3, any vector v ⃗ in E3, under the
transformation T, will be scaled by 3, meaning that
T( v ⃗ ) = λ v ⃗ = 3 v .⃗
130
131