Math 1180 Final Review I

Math 1180: Final Review I

You should not assume that the final exam will resemble this review sheet. Theproblems below were chosen because they seemed interesting or because they illustratedan important concept. They are not meant to be typical exam problems.

1. Determine whether the vector ~u is in the span of the vectors ~u1, ~u2, ~u3:

~u = 1− x + 2x2, ~u1 = −1− x, ~u2 = 1 + 2x + 3x2, ~u3 = 2 + x + x2.

Answer: Yes.

2. Let A be a 4× 4 matrix with detA = 2. Find det(AB−1) where

B =

5 6 7 81 2 3 40 0 2 30 0 4 7

Answer: 1/4.

3. Find the matrix of the linear transformation in the x-y plane defined as: thecounterclockwise rotation about the origin through 30◦, followed by a reflectionin the line y = −x.

Answer:1

2

(−1

√3

−√

3 1

).

4. Let

A =

1 0 2 32 1 4 11 1 2 −2

.

Find the bases for row(A), col(A), null(A), null(AT ), and rank(A), nullity(A).Answer: row(A) = span{(1 0 2 3), (0 1 0 −5)}, col(A) = span{(1 2 1)T , (0 1 1)T},null(A) = span{(−2, 0, 1, 0)T , (−3, 1, 0, 1)T}, null(AT ) = span{(1,−1, 1)T} (Notethat the answer is not unique). rank(A) = nullity(A) = 2.

5. Let A,B be two matrices of the same size. Prove that if A is similar to B, thenAT is similar to BT .

6. Determine whether the matrix A is diagonalizable, and if so, find an invertiblematrix P and a diagonal matrix D such that P−1AP = D.

A =

1 0 10 1 11 1 0

.

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Math 1180 Final Review I

Is A orthogonally diagonalizable? If so, find a transformation matrix Q and adiagonal matrix D̃ such that QTAQ = D̃.

Answer: D =

1 0 00 −1 00 0 2

, P =

−1 1 11 1 10 −2 1

, Q =

−1/√

2 1/√

6 1/√

3

1/√

2 1/√

6 1/√

3

0 −2/√

6 1/√

3

.

(Note that the answer is not unique)

7. Given the subspace W and a vector ~v:

W = span

1−1−11

,

0332

,

3241

, ~v =

1111

.

(1) Use the Gram-Schmidt process to obtain an orthogonal basis ~u1, ~u2, ~u3 ofW .

(2) Find the orthogonal projection of ~v onto the subspace W .

Answer:

(1) ~u1 =

1−1−11

, ~u2 =

1223

, ~u3 = 12

5−13−3

.

(2) 199

8979115105

.

8. Determine whether T is a linear transformation.

(1) T : P2 → P4 defined by T (p(x)) = x2p(x).Answer: Yes.

(2) T : F → F defined by T (f(x)) = f(x2).Answer: Yes.

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Math 1180 Final Review II

Math 1180: Final Review II

You should not assume that the final exam will resemble this review sheet. Theproblems below were chosen because they seemed interesting or because they illustratedan important concept. They are not meant to be typical exam problems.

1. Given two matrices

A =

1 −1 10 2 20 0 −1

, B =

2 −4 21 −1 01 3 2

.

Find detA, detB, det(A2B−1).Answer: −2, 12, 1/3.

2. Determine whether the vector ~v is in the span of the vectors ~u1, ~u2, ~u3:

~v =

123

, ~u1 =

135

, ~u2 =

301

, ~u3 =

5611

.Answer: No.

3. Find the matrix of the linear transformation in the x-y plane defined as: thecounterclockwise rotation about the origin through 60◦, followed by the orthog-onal projection onto the y-axis.

Answer:1

2

(0 0√3 1

).

4. Given W = span

3

0−1

,−4

38

.

(1) Find an orthonormal basis of W .

(2) Find the projection of the vector

120

onto W .

Answer:

(1) 1√10

3−10

, 17

236

.

(2) 1490

601240333

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Math 1180 Final Review II

5. Let A =

(1 2−2 −3

).

(1) Find the characteristic polynomial of A.

(2) Check if A is diagonalizable. If so, find the transformation matrix P .

Answer: λ2 + 2λ+ 1; Not possible since the geometric multiplicity of −1 is 1.

6. Let A,B be two n× n matrices with eigenvalues λ and µ respectively. Is λ + µnecessarily an eigenvalue of A+B?

7. Given A =

1 0 2 32 1 4 50 1 0 −1

.

(1) Find the bases for row(A), col(A), null(A), and null(AT ).

(2) Find the rank and nullity of A.

Answer: row(A) : (1 0 2 3), (0 1 0 − 1); col(A) :

120

,0

11

; null(A) :−3101

,−2010

; null(AT ) :

1−11

. rank(A) = nullity(A) = 2.

8. Are the following sets vector spaces? If so, find the dimension of them and givea basis.

(1) V = {p ∈ P3 : p(0) = 0}.Answer: Yes. dimV = 3.

(2) V = {A ∈M2×2 : Tr(A) = 0}.Answer: Yes. dimV = 3.

(3) V = {A ∈M2×2 : det(A) ≥ 0}.Answer: No.

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Math 1180 Final Review III

Math 1180: Final Review III

You should not assume that the final exam will resemble this review sheet. Theproblems below were chosen because they seemed interesting or because they illustratedan important concept. They are not meant to be typical exam problems.

1. Let A =

0 2 −1−1 3 −1−2 4 −1

.

(1) Find the eigenvalues and the basis for each eigenspace of A.

(2) Determine whether A is diagonalizable. If yes, find the transformationmatrix P and the corresponding diagonal matrix D.

Answer:

(1) λ = 0, E0 = span

1

12

; λ = 1, E1 = span

2

10

,−1

01

.

(2) Yes. P =

1/2 2 −11/2 1 01 0 1

, D =

0 0 00 1 00 0 1

.

2. Let W be a subspace of Rn. Prove that ~w is in W if and only if projW (~w) = ~w.

3. (1) Determine whether the vector ~v is a linear combination of the vectors~u1, ~u2, ~u3, ~u4:

~v = −2x2 + 5x+ 17, ~u1 = 2x2 + x+ 1, ~u2 = 11x2 + 6x+ 7,

~u3 = −x2 + 2x+ 7, ~u4 = 2x2 − 2.

(2) Solve the linear system2x1 + 11x2 − x3 + 2x4 = −2x1 + 6x2 + 2x3 = 5x1 + 7x2 + 7x3 − 2x4 = 17.

Answer:

(1) Yes.

(2)

x1x2x3x4

=

−671200

+

−12

201

t+

28−510

s.

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Math 1180 Final Review III

4. Given W = span

12−12

,

20−23

.

(1) Find an orthogonal basis of W .

(2) Find the projection of the vector

100−2

onto W .

(3) Find a basis for the orthogonal complement of W .

Answer:

(1)

12−12

,

1−2−11

.

(2) 170

−31−2231−52

.

(3)

−6−104

,

1010

.

5. Find det

1 0 2 −12 3 1 11 0 2 02 1 1 2

.

Answer: 6.

6. Given A =

2 1 2 1 10 2 −4 0 14 3 2 1 −13 4 −2 1 1

.

(1) Find the bases of row(A), col(A), null(A) and null(AT ).

(2) Find the rank and nullity of A.

Answer: row(A) : (1 3 − 4 0 0), (0 1 − 2 − 1 − 3), (0 0 0 2 7); col(A) :2043

,

1234

,

1011

; null(A) :

−22100

,

3−10−72

; null(AT ) :

1−1−21

. rank(A) =

2

Math 1180 Final Review III

3, nullity(A) = 2.

7. Find the inverse of 1 0 2 10 −1 0 02 0 2 00 0 0 −1

or show that it does not exist.

Answer:

−1 0 1 −10 −1 0 01 0 −1

21

0 0 0 −1

.

8. Let T : P2 → P2 be a linear transformation for which

T (1) = 3− x, T (x) = 4x− 2x2, T (2 + x2) = 1 + 2x2.

Find T (3− x+ 2x2).Answer: −1− 3x+ 6x2.

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Math 1180 Final Review IV

Math 1180: Final Review IV

You should not assume that the final exam will resemble this review sheet. Theproblems below were chosen because they seemed interesting or because they illustratedan important concept. They are not meant to be typical exam problems.

1. Let A =

4 2 22 1 1−8 −4 −4

. Show that A is diagonalizable. Then find A2012.

Answer: A2012 =

4 2 22 1 1−8 −4 −4

2. A linear transformation T from R2 to R2 is defined as the rotation clockwise by

2π/3, followed by a stretch by a factor of 4 in the x-component. Find the matrixof T . Is T invertible? If so, find its matrix.

Answer:1

2

(−4 4

√3

−√

3 −1

). The inverse matrix is

1

8

(−1 −4

√3√

3 −4

).

3. Solve the linear system using Cramer’s method.x− z = 52y − x = −15z − 2x+ y = 2.

Answer:

xyz

=

−11−4−16

.

4. Find an orthogonal basis of R4 containing

1111

.

Answer (not unique):

1111

,

1−100

,−1−13−1

,

110−2

.

5. Let W be a line in R3 with parametric equation

x = t, y = 2t, z = −t.

Find a basis for W⊥.Answer: (Not unique) (−2, 1, 0)T , (1, 0, 1)T .

1

Math 1180 Final Review IV

6. Find det

13

1 2 −1 0 10 0 2 1 00 0 −1 0 00 0 0 2 10 −3 0 4 10

.

Answer: 1/81.

7. Given A =

−1 3 4 0 −20 1 −3 −1 2−3 7 18 2 −10

.

(1) Find the bases of row(A), col(A), null(A), and null(AT ).

(2) Find the rank and nullity of A.

Answer: row(A) : (−1 3 4 0−2), (0 1−3−1 2); col(A) :

−10−3

,3

17

; null(A) :133100

,

31010

,−8−2001

, null(AT ) :

−321

. rank(A) = 2, nullity(A) = 3.

8. Let {~v1, · · · , ~vk} be a basis for a vector space V and let T : V → V be a lineartransformation. Prove that if T (~v1) = ~v1, · · · , T (~vk) = ~vk, then T is the identitytransformation on V .

2

Math 1180 Final Review V

Math 1180: Final Review V

You should not assume that the final exam will resemble this review sheet. Theproblems below were chosen because they seemed interesting or because they illustratedan important concept. They are not meant to be typical exam problems.

1. Given the matrix

A =

1 1 2 1 −12 1 −3 0 11 2 9 3 −4

.

(1) Find the bases for row(A), col(A), null(A), null(AT ), and determine the rankand nullity of A.

(2) Find a basis for the orthogonal complement W⊥ of W , where W is a sub-space spanned by

~v1 =

1121−1

, ~v2 =

21−301

, ~v3 =

1293−4

.

Answer:

(1) row(A) : [1 0 − 5 − 1 2], [0 1 7 2 − 3]; col(A) :

121

,1

12

; null(A) :5−7100

,

1−2010

,−23001

; null(AT ) :

−311

. rank(A) = 2, nullity(A) = 3.

(2)

−5−7100

,

1−2010

,−23001

.

2. A linear transformation T from R2 to R2 is defined as the rotation counterclock-wise by 60◦, followed by a projection onto the x-axis, then a stretching by afactor of 3. Find the matrix of T .

Answer: 3

(12−√32

0 0

).

1

Math 1180 Final Review V

3. Determine whether

(−3 301 4

)is in the span of

(1 21 0

),

(0 30 1

), and

(2 11 1

).

Answer: Yes.

4. (1) Is the matrix A =

1 0 12 3 11 2 0

invertible? If so, find the inverse.

Answer: A−1 =

2 −2 3−1 1 −1−1 2 −3

.

(2) Solve the linear system x+ z = 32x+ 3y + z = −3x+ 2y = 1.

Answer: x = 15, y = −7, z = −12.

5. Given A =

0 1 −14 0 −24 2 −4

.

(1) Find all eigenvalues and the basis for each eigenspace of A.

Answer: λ = −2, E−2 = span

0

11

,1

02

, λ = 0, E0 = span

1

22

.

(2) Determine whether A is diagonalizable. If it is, find an invertible matrix Pand a diagonal matrix D such that P−1AP = D. (You don’t have to findP−1)

Answer: Yes. P =

0 1 11 0 21 2 2

, D =

−2 0 00 −2 00 0 −2

.

6. The subspace W is spanned by

~v1 =

11−11

, ~v2 =

−1−210

.(1) Construct an orthogonal basis for W .

Answer:

11−11

,

0−101

.

2

Math 1180 Final Review V

(2) Find the orthogonal projection of ~v =

10−5−2

onto W .

Answer:

12−10

.

(3) Find the component of ~v orthogonal to W .

Answer:

0−2−4−2

.7. Determine whether W is a subspace of V .

(1) V = M2×2, W =

{(a bb 3a

)}.

Answer: Yes.

(2) V = F , W = {f ∈ F : f(−x) = f(x)}.Answer: Yes.

(3) V = D, W = {f ∈ D : f ′(x) ≤ 0}.Answer: No.

(4) V = F , W = {f ∈ F : f(1) = 1}.Answer: No.

8. Let T : V → V be a linear transformation such that T ◦ T = T . Prove that ~vand T (~v) are linearly dependent if and only if T (~v) = ~v or T (~v) = ~0.

3