Linear Algebra exam

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MAT 242 Test 2 SOLUTIONS, FORM T

1. Let ~v1 =

 

5 1 −5 −1

 , ~v2 =

  −3 −3 −1 −3

 , ~v3 =

  −5 −13 −15 −17

 , and ~v4 =

 

3 −3 −3

 .

a. [10 points] The set {~v1,~v2,~v3,~v4} is linearly dependent. Find a nontrivial linear combination of these vectors that adds up to ~0.

Solution: To do this, you need to find a nontrivial solution to c1~v1 + c2~v2 + c3~v3 + c4~v4 = ~0, whose matrix looks like [ ~v1 ~v2 ~v3 ~v4 | ~0 ], or

 5 −3 −5 3 1 −3 −13 −3 −5 −1 −15 −3 −1 −3 −17 3

∣∣∣∣∣∣∣ 0 0 0 0

  −−−−−−→RREF

 

1 0 2 0 0 1 5 0 0 0 0 1 0 0 0 0

∣∣∣∣∣∣∣ 0 0 0 0

 

The nontrivial solutions are thus given by

c1 = −2t c2 = −5t c3 = t

c4 = 0

where t can be any real number

Choosing t = 1 produces the linear combination −2v1 − 5v2 + v3. Grading: +5 points for setting up the system of linear equations; +5 points for parameterizing

it; +5 points for finding a linear combination. Grading for common mistakes: +7 points (total) for using the column space method instead.

b. [10 points] Is the vector

  −4

3 −6 −5

  in the span of {~v1,~v2,~v3,~v4}? Justify your answer.

Solution: You must determine whether the augmented matrix [~v1 ~v2 ~v3 ~v4|~u] is a system that has at least one solution.

 5 −3 −5 3 1 −3 −13 −3 −5 −1 −15 −3 −1 −3 −17 3

∣∣∣∣∣∣∣ −4

3 −6 −5

  −−−−−−→RREF

 

1 0 2 0 0 1 5 0 0 0 0 1 0 0 0 0

∣∣∣∣∣∣∣ 0 0 0 1

 

Since this system has no solutions, the vector is not in the span.

Grading: +4 points for setting up the matrix, +4 points for the RREF, +3 points for deter- mining how many solutions there were, +4 points for answering NO. Grading for common mistakes: −8 points for using ~0 instead of ~u.

MAT 242 Test 2 SOLUTIONS, FORM T

2. [25 points] For the matrix A below, find a basis for the null space of A, a basis for the row space of A, a basis for the column space of A, the rank of A, and the nullity of A. The reduced row echelon form of A is the matrix R given below.

A =

 

5 −5 10 −15 0 2 4 22 0 −1 5 0 25 −10 2 −1 −2 −11 0 −4

  R =

 

1 0 5 −2 0 0 1 3 1 0 0 0 0 0 1 0 0 0 0 0

 

Solution: To find a basis for the null space, you need to solve the system of linear equations A~x = ~0, or equivalently, R~x = ~0. Parameterizing the solutions to this equation, and writing the answer in expanded form, produces 

 x1 x2 x3 x4 x5

  = α1 ·

  −5 −3

1 0 0

  + α2 ·

 

2 −1

0 1 0

 

so     −5 −3

1 0 0

  ,  

2 −1

0 1 0

   

is a basis for the null space of A. The nullity is the number of vectors in this basis, namely 2. A basis for the row space can be found by taking the nonzero rows of R:

{[ 1, 0, 5, −2, 0 ] , [ 0, 1, 3, 1, 0 ] , [ 0, 0, 0, 0, 1 ]}

A basis for the column space can be found by taking the columns of A which have pivots in them, so

   

5 2 5 −1

  ,   −5

4 0 −2

  ,  

0 −1

2 −4

   

is a basis for the column space of A. Lastly, the rank of A is the number of vectors in a basis for the row space (or column space)

of A, so the rank of A is 3. Grading: +5 points for each of: finding a basis for the null space, a basis for the row space, a

basis for the column space, the nullity, and the rank. Grading for common mistakes: −3 points for forgetting a variable in the parameterization; −3 points for choosing columns of R for the column space of A; −3 points for choosing rows from A for the row space of A; −3 points for choosing the non-pivot columns of A for the null space of A.

MAT 242 Test 2 SOLUTIONS, FORM T

3. [15 points] Find the eigenvalues of A =

  6 −6 180 2 0 −3 2 −9

 .

Solution: Solve for when the determinant of A−λI equals zero:∣∣∣∣∣∣ 6 −λ −6 18

0 2 −λ 0 −3 2 −9 −λ

∣∣∣∣∣∣ = λ3 + λ2 − 6λ = λ (λ + 3) (λ− 2) . The eigenvalues make this product equal to zero, so they are 0, −3, and 2.

Grading: +5 points for A − λI, +5 points for finding the determinant, +5 points for finding the eigenvalues. Grading for common mistakes: +5 points (total) for using row operations on the matrix A, +3 points (total) for finding the eigenvectors of 0.

4. Let B be the (ordered) basis

    12

  ,  −2−5

  ,  −3−9 −1

    and C the basis

   −10 −2

  ,   3−1

  ,   31

   .

a. [10 points] Find the coordinates of

  168

  with respect to the basis C.

Solution:

[~u]C = C̃ −1 ·~u =

 −1 3 30 −1 1 −2 6 7

 −1 ·

  168

  =

 −4−2

 

Grading: +3 points for the formula, +4 points for substitution, +3 points for calculation.

b. [10 points] Find the change-of-basis matrix from C to B.

Solution:

B−1 ·C =

  1 −2 −32 −5 −9

0 0 −1

 −1 ·

 −1 3 30 −1 1 −2 6 7

  =

 −11 35 34−8 25 26

2 −6 −7

 

Grading: +3 points for the formula, +4 points for substitution, +3 points for calculation.

Grading for common mistakes: +7 points (total) for C−1B =

 −19 41 60−4 9 14 −2 4 5

  (backwards), +5 points

(total) for BC−1 =

 −3 −1 12 −1 −2

2 0 −1

 , +5 points (total) for CB−1 =

  1 −1 3−2 1 −4

2 −2 5

 , +3 points (to-

tal) for B−1 =

  5 −2 32 −1 3

0 0 −1

 , +7 points (total) for B−1C~x

  13961060 −282

 , +3 points (total) for any other

vector.

MAT 242 Test 2 SOLUTIONS, FORM T

5. The eigenvalues of the matrix A =

  0 6 60 4 4

0 −2 −2

  are 0 (with multiplicity 2) and 2 (with multiplicity 1).

(You do not need to find these.) Do the following for the matrix A:

a. [10 points] Find a basis for the eigenspace of each eigenvalue.

Solution: The eigenspace of an eigenvalue λ is the null space of A−λI. So, if λ = 0,

A− (0)I =

  0 6 60 4 4

0 −2 −2

  −−−−−−→RREF

  0 1 10 0 0

0 0 0

  ,

whose solutions (with a vector of 0s on the right-hand side) are parameterized by x1x2 x3

  = s ·

  10

  + t ·

  0−1

 

and a basis for the eigenspace of 0 is

    10

  ,   0−1

    .

If λ = 2,

A− (2)I =

 −2 6 60 2 4

0 −2 −4

  −−−−−−→RREF

  1 0 30 1 2

0 0 0

  ,

whose solutions are parameterized by  x1x2 x3

  = s ·

 −3−2

 

and a basis for the eigenspace of 2 is

   −3−2

    . .

Grading: +3 points for A−λI, +3 points for the RREF, +4 points for finding the null space basis. Grading for common mistakes: +5 points (total) for finding the column space of A−lambdaI.

b. [10 points] Is the matrix A diagonalizable? If so, find matrices D and P such that A = PDP−1

and D is a diagonal matrix. If A is not diagonalizable explain carefully why it is not.

Solution: YES. The dimension of each eigenspace equals the (given) multiplicity of each eigen- value. The matrix P can be constructed by gluing the basis vectors together; then the (i, i) entry in D is the eigenvalue that the ith column of P belongs to. One pair of matrices that diagonalizes A is

D =

  0 0 00 0 0

0 0 2

  and P =

  0 1 −3−1 0 −2

1 0 1

 

Grading: +3 points for YES, +7 points for P and D. Full credit — marked +10∗ points — was given for an answer consistent with any mistakes made in part (a).

MAT 242 Test 2 SOLUTIONS, FORM W

1. Let S =

   

4 4 3 2

  ,  

8 8 6 4

  ,  

3 −1 −2 −1

  ,   −7 −3 −1 −1

  ,  

4 0 0 1

   .

a. [10 points] Find a basis for the subspace W spanned by S, and the dimension of W .

Solution: To find a basis, use the column space (CS) approach: Glue the vectors together to get

the matrix Ṽ , find the RREF, and take the original vectors that have pivots in their columns:

Ṽ =

 

4 8 3 −7 4 4 8 −1 −3 0 3 6 −2 −1 0 2 4 −1 −1 1

  −−−−−−→RREF

 

1 2 0 −1 0 0 0 1 −1 0 0 0 0 0 1 0 0 0 0 0

 

A basis for the subspace will consist of the columns of Ṽ which have a pivot in their column, namely   

4 4 3 2

  ,  

3 −1 −2 −1

  ,  

4 0 0 1

    . The dimension of the subspace is the number of vectors in this basis,

which is 3.

Grading: +5 points for Ṽ , +5 points for the RREF, +5 points for the dimension. Grading for common mistakes: +5 points (total) for finding a basis for the null space; −3 points for using the columns of the RREF.

b. [10 points] Is the vector

  −11 −7 −4 −3

  in the span of S? Justify your answer.

Solution: You must determine whether the augmented matrix [~v1 ~v2 ~v3 ~v4|~u] is a system that has at least one solution.

 4 8 3 −7 4 4 8 −1 −3 0 3 6 −2 −1 0 2 4 −1 −1 1

∣∣∣∣∣∣∣ −11 −7 −4 −3

  −−−−−−→RREF

 

1 2 0 −1 0 0 0 1 −1 0 0 0 0 0 1 0 0 0 0 0

∣∣∣∣∣∣∣ −2 −1

0 0

 

Since this system has infinitely many solutions, it has at least one, and the vector is in the span.

Grading: +4 points for setting up the matrix, +4 points for the RREF, +3 points for determin- ing how many solutions there were, +4 points for answering YES. Grading for common mistakes: −8 points for using ~0 instead of ~u.

MAT 242 Test 2 SOLUTIONS, FORM W

A =

  −3 −3 −12 0 −3 −2 −9 −5 −2 5 13 −35 −5 1 −2 −30

  R =

 

1 0 1 5 0 1 3 −5 0 0 0 0 0 0 0 0

 

 x1 x2 x3 x4

  = α1 ·

  −1 −3

1 0

  + α2 ·

  −5

5 0 1

 

so     −1 −3

1 0

  ,   −5

5 0 1

   

is a basis for the null space of A. The nullity is the number of vectors in this basis, namely 2. A basis for the row space can be found by taking the nonzero rows of R:

{[ 1, 0, 1, 5 ] , [ 0, 1, 3, −5 ]}

A basis for the column space can be found by taking the columns of A which have pivots in them, so

    −3 −3 −2 −5

  ,   −3 −2

5 1

   

is a basis for the column space of A. Lastly, the rank of A is the number of vectors in a basis for the row space (or column space)

of A, so the rank of A is 2. Grading: +5 points for each of: finding a basis for the null space, a basis for the row space, a

MAT 242 Test 2 SOLUTIONS, FORM W

3. [15 points] Find the eigenvalues of A =

  0 −3 00 −3 0

0 3 0

 .

Solution: Solve for when the determinant of A−λI equals zero:∣∣∣∣∣∣ −λ −3 0 0 −3 −λ 0 0 3 −λ

∣∣∣∣∣∣ = λ3 + 3λ2 = λ2 (λ + 3) . The eigenvalues make this product equal to zero, so they are 0, 0, and −3.

4. Let B be the (ordered) basis

   −13 −1

  ,  −27 −4

  ,   0−2

    and C the basis

    10

  ,   −3−1 −12

  ,  −10 −2

   .

a. [10 points] Find the coordinates of

  2−19

  with respect to the basis B.

Solution:

[~u]B = B̃ −1 ·~u =

 −1 −2 03 7 −2 −1 −4 3

 −1 ·

  2−19

  =

  4−3

 

Grading: +3 points for the formula, +4 points for substitution, +3 points for calculation.

b. [10 points] If the coordinates of ~u with respect to B are

  6−1 −2

 , what are the coordinates of ~u with

respect to C?

Solution:

[~u]C = C̃ −1 · B̃ · [~u]B =

  1 −3 −10 −1 0

3 −12 −2

 −1 ·

 −1 −2 03 7 −2 −1 −4 3

  ·   6−1 −2

  =

 −90−15 −41

 

Grading: +3 points for the formula, +4 points for substitution, +3 points for calculation.

Grading for common mistakes: +7 points (total) for B−1C[~u] =

  285−148 −91

  (backwards), +5 points

(total) for CB−1[~u] =

  19535

664

 , +5 points (total) for BC−1[~u] =

  617 −47

 , +3 points (total) for

C−1[~u] =

  −81 −17

 , +7 points (total) for the change-of-basis matrix C−1B =

 −17 −42 15−3 −7 2 −7 −19 9

 ,

+3 points (total) for a different change-of-basis matrix.

MAT 242 Test 2 SOLUTIONS, FORM W

5. The eigenvalues of the matrix A =

 −3 0 00 −3 0

0 1 −4

  are −3 (with multiplicity 2) and −4 (with multiplic-

ity 1). (You do not need to find these.) Do the following for the matrix A:

a. [10 points] Find a basis for the eigenspace of each eigenvalue.

Solution: The eigenspace of an eigenvalue λ is the null space of A−λI. So, if λ = −3,

A− (−3)I =

  0 0 00 0 0

0 1 −1

  −−−−−−→RREF

  0 1 −10 0 0

0 0 0

  ,

whose solutions (with a vector of 0s on the right-hand side) are parameterized by x1x2 x3

  = s ·

  10

  + t ·

  01

 

and a basis for the eigenspace of −3 is

    10

  ,   01

    .

If λ = −4,

A− (−4)I =

  1 0 00 1 0

0 1 0

  −−−−−−→RREF

  1 0 00 1 0

0 0 0

  ,

whose solutions are parameterized by  x1x2 x3

  = s ·

  00

 

and a basis for the eigenspace of −4 is

    00

    . .

Grading: +3 points for A−λI, +3 points for the RREF, +4 points for finding the null space basis. Grading for common mistakes: +5 points (total) for finding the column space of A−lambdaI.

b. [10 points] Is the matrix A diagonalizable? If so, find matrices D and P such that A = PDP−1

and D is a diagonal matrix. If A is not diagonalizable explain carefully why it is not.

D =

 −3 0 00 −3 0

0 0 −4

  and P =

  0 1 01 0 0

1 0 1

 

Grading: +3 points for YES, +7 points for P and D. Full credit — marked +10∗ points — was given for an answer consistent with any mistakes made in part (a).

MAT 242 Test 2 SOLUTIONS, FORM O

1. Let S =

   

1 −4

1 0

  ,   −4 16 −4

  ,  

3 1 4 −3

  ,  

0 3 −4 −1

  ,  

2 −11

6 1

   .

a. [10 points] Find a basis for the subspace W spanned by S, and the dimension of W .

Solution: To find a basis, use the column space (CS) approach: Glue the vectors together to get

the matrix Ṽ , find the RREF, and take the original vectors that have pivots in their columns:

Ṽ =

 

1 −4 3 0 2 −4 16 1 3 −11

1 −4 4 −4 6 0 0 −3 −1 1

  −−−−−−→RREF

 

1 −4 0 0 2 0 0 1 0 0 0 0 0 1 −1 0 0 0 0 0

 

A basis for the subspace will consist of the columns of Ṽ which have a pivot in their column, namely   

1 −4

1 0

  ,  

3 1 4 −3

  ,  

0 3 −4 −1

    . The dimension of the subspace is the number of vectors in this basis,

which is 3.

b. [10 points] Is the vector

 

65 −10

6 52

  in the span of S? Justify your answer.

Solution: You must determine whether the augmented matrix [~v1 ~v2 ~v3 ~v4|~u] is a system that has at least one solution.

 1 −4 3 0 2 −4 16 1 3 −11

1 −4 4 −4 6 0 0 −3 −1 1

∣∣∣∣∣∣∣ 65 −10

6 52

  −−−−−−→RREF

 

1 −4 0 0 2 0 0 1 0 0 0 0 0 1 −1 0 0 0 0 0

∣∣∣∣∣∣∣ 0 0 0 1

 

Since this system has no solutions, the vector is not in the span.

MAT 242 Test 2 SOLUTIONS, FORM O

A =

 

0 −4 −5 9 15 −1 −3 1 6 13

2 −4 4 −8 30 0 4 −1 −3 −21 −2 −5 −4 17 15

  R =

 

1 0 0 −4 3 0 1 0 −1 −5 0 0 1 −1 1 0 0 0 0 0 0 0 0 0 0

 

 x1 x2 x3 x4 x5

  = α1 ·

 

4 1 1 1 0

  + α2 ·

  −3

5 −1

0 1

 

so    

4 1 1 1 0

  ,   −3

5 −1

0 1

   

is a basis for the null space of A. The nullity is the number of vectors in this basis, namely 2. A basis for the row space can be found by taking the nonzero rows of R:

{[ 1, 0, 0, −4, 3 ] , [ 0, 1, 0, −1, −5 ] , [ 0, 0, 1, −1, 1 ]}

A basis for the column space can be found by taking the columns of A which have pivots in them, so

   

0 −1

2 0 −2

  ,   −4 −3 −4

4 −5

  ,   −5

1 4 −1 −4

   

is a basis for the column space of A. Lastly, the rank of A is the number of vectors in a basis for the row space (or column space)

of A, so the rank of A is 3. Grading: +5 points for each of: finding a basis for the null space, a basis for the row space, a

MAT 242 Test 2 SOLUTIONS, FORM O

3. [15 points] Find the eigenvalues of A =

  0 −4 00 −2 0

2 0 2

 .

Solution: Solve for when the determinant of A−λI equals zero:∣∣∣∣∣∣ −λ −4 0 0 −2 −λ 0 2 0 2 −λ

∣∣∣∣∣∣ = λ3 − 4λ = λ (λ + 2) (λ− 2) . The eigenvalues make this product equal to zero, so they are 0, −2, and 2.

4. Let B be the (ordered) basis

    10 −3

  ,   0−1 −3

  ,  −2−3 −2

    and C the basis

   −1−3

  ,   12 −1

  ,   27 −1

   .

a. [10 points] Find the coordinates of

 −13−38

  with respect to the basis C.

Solution:

[~u]C = C̃ −1 ·~u =

 −1 1 2−3 2 7

1 −1 −1

 −1 ·

 −13−38

  =

 −3−6 −5

 

Grading: +3 points for the formula, +4 points for substitution, +3 points for calculation.

b. [10 points] If the coordinates of ~u with respect to B are

  0−8

 , what are the coordinates of ~u with

respect to C?

Solution:

[~u]C = C̃ −1 · B̃ · [~u]B =

 −1 1 2−3 2 7

1 −1 −1

 −1 ·

  1 0 −20 −1 −3 −3 −3 −2

  ·   0−8

  =

  519

 

Grading: +3 points for the formula, +4 points for substitution, +3 points for calculation.

Grading for common mistakes: +7 points (total) for B−1C[~u] =

  26−39

  (backwards), +5 points

(total) for CB−1[~u] =

  −67−109

 , +5 points (total) for BC−1[~u] =

  9−12 −62

 , +3 points (total)

for C−1[~u] =

  119

 , +7 points (total) for the change-of-basis matrix C−1B =

 −4 −8 −131 −2 −7 −2 −3 −4

 ,

+3 points (total) for a different change-of-basis matrix.

MAT 242 Test 2 SOLUTIONS, FORM O

5. The eigenvalues of the matrix A =

 −5 −16 −320 −13 −16

0 8 11

  are −5 (with multiplicity 2) and 3 (with multiplic-

ity 1). (You do not need to find these.) Do the following for the matrix A:

a. [10 points] Find a basis for the eigenspace of each eigenvalue.

Solution: The eigenspace of an eigenvalue λ is the null space of A−λI. So, if λ = −5,

A− (−5)I =

  0 −16 −320 −8 −16

0 8 16

  −−−−−−→RREF

  0 1 20 0 0

0 0 0

  ,

whose solutions (with a vector of 0s on the right-hand side) are parameterized by x1x2 x3

  = s ·

  10

  + t ·

  0−2

 

and a basis for the eigenspace of −5 is

    10

  ,   0−2

    .

If λ = 3,

A− (3)I =

 −8 −16 −320 −16 −16

0 8 8

  −−−−−−→RREF

  1 0 20 1 1

0 0 0

  ,

whose solutions are parameterized by  x1x2 x3

  = s ·

 −2−1

 

and a basis for the eigenspace of 3 is

   −2−1

    . .

Grading: +3 points for A−λI, +3 points for the RREF, +4 points for finding the null space basis. Grading for common mistakes: +5 points (total) for finding the column space of A−lambdaI.

b. [10 points] Is the matrix A diagonalizable? If so, find matrices D and P such that A = PDP−1

and D is a diagonal matrix. If A is not diagonalizable explain carefully why it is not.

D =

 −5 0 00 −5 0

0 0 3

  and P =

  0 1 −2−2 0 −1

1 0 1

 

Grading: +3 points for YES, +7 points for P and D. Full credit — marked +10∗ points — was given for an answer consistent with any mistakes made in part (a).