Berkeley Math 54 Sample Quizzes

Sample Quiz 1, MATH 54, Fall 2014

Name (Last, First):

Student ID:

1. Assume the following assertions are true:

There are courses at Berkeley.In every Berkeley course, there is a student who understands everything.

Circle each of the following assertions which must also then be true:

a. There is a Berkeley course in which all students understand something.

b. There is a Berkeley course in which all students understand nothing.

c. There is no Berkeley course in which each student understands nothing.

d. There is no Berkeley course in which each student does not understand something.

e. There is a Berkeley course in which there is a student who understands everything.

c, d, and e must be true.

2. For what numbers a, b, c is the following matrix in row echelon form (REF) or reduced rowechelon form (RREF)? 0 a 1 b 00 0 0 c 0

0 0 0 0 a

REF: a = 0 and any b, c; or a, c 6= 0 and any b.

RREF: a, c = 1 and b = 0; or a = 0, b = 0, c = 1; or a, c = 0 and any b.

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Quiz 2, MATH 54, Fall 2014

Name (Last, First):

Student ID:

1. Find the general solution of the linear system corresponding to the following augmentedmatrix.[1 2 1 42 4 5 6

]Solution. First, convert the matrix into REF by applying the row operations.[

1 2 1 42 4 5 6

][1 2 1 40 0 7 14

](Replacement)

where the corresponding system of linear equations is

x1 2x2 x3 = 47x3 = 14

Therefore, x3 = 2 and x1 2x2 = 2.If we pick x2 as a free variable, the general solution reads

x1 = 2x2 + 2

x2 = free variable

x3 = 2

1


2. Determine if b is a linear combination of the vectors formed from the columns of the matrix A.

A =

1 4 20 3 52 8 4

,b = 373

Solution. Note that b is a linear combination of the column vectors of A if and only if Ax = b

has a solution.Calculate a REF of the augmented matrix [A|b] :

[A|b] =

1 4 2 30 3 5 72 8 4 3

1 4 2 30 3 5 70 0 0 3

The last row (0 0 0 | 3) indicates that the solution does not exist. Hence, b is NOT a linear

combination of the column vectors of A.

2


Name (Last, First):

Student ID:

1. Determine if the columns of the matrix form a linearly independent set.

2

40 2 31 3 61 1 0

3

5

Solution 1. The set is linearly dependent. This is because

3

2

4011

3

5+ 3

2

4231

3

5 2

2

4360

3

5 =

2

4000

3

5

so that the three columns form a linearly dependent set.

Solution 2. Lets take the associated augmented matrix for the homogeneous equation.

2

40 2 3 01 3 6 01 1 0 0

3

5

In order to make the matrix to be in Row Echelon Form, we need the below row reduction steps.

2

40 2 3 01 3 6 01 1 0 0

3

5

2

41 3 6 00 2 3 01 1 0 0

3

5

2

41 3 6 00 2 3 00 4 6 0

3

5

2

41 3 6 00 2 3 00 0 0 0

3

5

The first is obtained by interchanging 1st and 2nd rows.The second : Changing 3rd row into 1st row + 3rd row.The last : Changing 3rd row into 3rd row + (2) 2nd row.

Hence, we can find a solution z =free from 3rd row. y = 32z from 2nd row. x = y from 1stand 2nd row. Find one example of x, y, and z by setting z = 2, we get the weights x, y, and z(not all zeros) that make the linear combination xx1 + yx2 + zx3 become zero.

1

David Nadler


2. Let T (x, y) = (2x+ y, x). Show that T is a one-to-one linear transformation. Does T map R2onto R2?

Solution. First of all,

T ((x1, y1)) + T ((x2, y2)) = (2x1 + y1, x1) + (2x2 + y2, x2)

= (2(x1 + x2) + (y1 + y2), x1 + x2) = T ((x1 + x2, y1 + y2))

Also, T (c(x, y)) = (2cx+ cy, cx) = c(2x+ y, x) = cT ((x, y)). Hence, T is a linear transformation.Now, in order to prove that T is one-to-one, (because T is a linear transformation and by

Theorem 11 (Chapter 1.9)) we only need to show that

If T (x, y) = (0, 0) then x = 0 and y = 0.

Suppose that T (x, y) = (0, 0), then it implies that 2x+ y = 0, x = 0. So, obviously, you get x = 0and y = 0. Henceforth, T is a one-to-one linear transformation.

For the last question, the answer is YES. To get this answer, we need the argument below.

For an arbitrary element in R2, say (z, w), if we define x = w, y = z 2w thenT (x, y) = (2x+ y, x) = (2w + z 2w,w) = (z, w).

Therefore, T maps R2 onto R2.

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Math 54 Name (Print):Practice Quiz #4

1. Suppose A =

[1 23 4

]is the standard matrix for an invertible linear function T : R2 R2 and

B =

[1 0 43 4 0

]is the standard matrix for a linear function S : R3 R2. Find the standard

matrix for T T T1 S.

Solution: First ntoice that T and T1 cancel. That is, T T T1 S = T I S = T S.Since the standard matrix for a composition of two linear functions is given by the product of

their standard matrices, the matrix for T S is[1 23 4

] [1 0 43 4 0

]=

[7 8 415 16 12

]

Math 54 Practice Quiz #4 - Page 2 of 2

2. Let A =

1 2 4 10 1 1 00 0 0 0

. Find a basis for Col(A) and a basis for Nul(A).Solution: A is already in echelon form, so we dont need to do too much work. The first twocolumns of A are the pivot columns of A, so they form a basis for Col(A).

To find a basis for Nul(A), we want to parameterize the solution space of Ax = 0. Thisequation corresponds to the system of equations x1 2x2 + 4x3 + x4 = 0, x2 + x3 = 0, and0 = 0. So the general solution is given by x2 = x3 and x1 = 6x3 x4 with x3 and x4 free.

This gives us

x1x2x3x4

=6x3 x4x3x3x4

= x36110

+ x41001

. So6110

,1001

form a basis forNul(A).


Name (Last, First):

Student ID:

(1) Let A be a 6 8 matrix. Suppose that the null space of A has a basis consisting of 3 vectors.Do the columns of A span R6? Justify your answer.Solution: According to the Rank Theorem (Section 2.7, Theorem 13), we have rank(A) +dimNul(A) = 8. We are given that dimNul(A) = 3. Thus rank(A) = 5. By Section 2.6Theorem 12, the pivot columns of A form a basis for the column space of A, so there are5 pivot columns. Thus there are 5 pivot rows. This means there is a row which does notcontain a pivot; thus the columns of A do not span R6.

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(2) Compute the determinant of the following matrix.3 1 0 24 9 2 00 1 0 31 2 0 2

Solution: We use Section 3.1, Theorem 1 to compute the determinant:

det

3 1 0 24 9 2 00 1 0 31 2 0 2

(1)= 0 (1)1+3 det 4 9 00 1 31 2 2

+ 2 (1)2+3 det 3 1 20 1 31 2 2

+ 0 (1)3+3 det

3 1 24 9 01 2 2

+ 0 (1)4+3 det3 1 24 9 00 1 3

= 2 (1)2+3 det

3 1 20 1 31 2 2

(2)= 2 (1)2+3

(3 (1)1+1 det

[1 32 2

]+ 0 (1)2+1 det

[1 22 2

]

+ (1) (1)3+1 det[1 21 3

])

= 2 (1)2+3(3 (1)1+1(8) + (1) (1)3+1(5)

)= 58 .

In the step marked (1), we use the cofactor expansion across the 3rd column; in the stepmarked (2), we use the cofactor expansion across the 1st column.

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Name (Last, First):

Student ID:

1. Is the set W of 2 2 symmetric matrices a subspace of the vector space V of all 2 2 matrices?

(Recall that a matrix A is symmetric if and only if AT = A. Equivalently, a symmetric 2 2

matrix is of the form

[a bb c

].)

There are two ways of solving this problem, choose your favorite one!

Solution 1: All we need to show that the zero-matrix is in W , and that W is closed underaddition and scalar multiplication.

Zero-vector: The 2 2 zero-matrix O =[0 00 0

]satisfies OT = O, therefore O is in W

Closed under addition: Suppose A and B are in W . Then AT = A and BT = B. But then,by properties of transpose:

(A + B)T = AT + BT = A + B

And therefore A + B is symmetric. Since A and B were arbitrary in W , it follows that W isclosed under addition

Closed under scalar multiplication: Suppose A in in W and c is a real number, thenAT = 0, and so, by properties of transposes:

(cA)T = c(AT

)= cA

And therefore cA is symmetric. Since A and c were arbitrary, it follows that W is closed undermultiplication.

Therefore, W is a subspace of V .

Solution 2:

Notice that every matrix

[a bb c

]can be written as:[

a bb c

]=

[a 00 0

]+

[0 bb 0

]+

[0 00 c

]= a

[1 00 0

]+ b

[0 11 0

]+ c

[0 00 1

]Therefore, it follows that:

W = Span

{[1 00 0

],

[0 11 0

],

[0 00 1

]}But since the span of any number of vectors in V is a subspace of V , it follows that W is a

subspace of V .

1


2. Let B ={[

14

],

[23

]}be a basis of R2.

a. Calculate the change-of-coordinates matrix PB from B to the standard basis of R2.

PB =

[1 24 3

]

b. Use part a. to calculate [x]B given x =

[16

]We know that:

x = PB [x]B

And therefore:

[x]B = (PB)1 x =

[1 24 3

]1 [16

]=

1

5

[3 24 1

] [16

]=

1

5

[1510

]=

[32

]

2


Name (Last, First):

Student ID:

1) Find the dimension of the subspace H inside of R4 given by all vectors of the form2a+ 4b+ c+ 5da 7b 4c+ 7da+ b+ c 4da b 3d

where a, b, c, d are any real numbers.

Solution:

2a+ 4b+ c+ 5da 7b 4c+ 7da+ b+ c 4da b 3d

= a

2111

+ b

4711

+ c

1410

+ d

5743

.

So H = Span{

2111

,

4711

,

1410

,

5743

}.

Thus dim(H) equals the rank of the matrix A =

2 4 1 51 7 4 71 1 1 41 1 0 3

.

A row-reduces to

1 0 0.5 3.50 1 0.5 0.50 0 0 00 0 0 0

, so dim(H) = rank(A) = 2.

1


2) If A is a 9 6 matrix, what is the largest possible dimension of the row space of A? What is thelargest possible dimension of the null space Nul(A)?

Solution: The number of pivots in A cannot be more than the number of rows or the numberof columns of A, so A has at most 6 pivots. Since dim(Row(A)) = rank(A) = dim(Col(A)), thelargest possible value for dim(Row(A)) is also 6.

By the Rank Theorem, rank(A)+dim(Nul(A)) = 6, so the largest possible value for dim(Nul(A))is 6, which occurs when rank(A) = 0.

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Sample Quiz 8 Solution, MATH 54, Fall 2014

Name (Last, First):

Student ID:

1) Consider the matrix

A =

4 1 12 1 02 0 1

Find a basis for the eigenspace with eigenvalue = 1.

Solution: The eigenspace of A corresponding to an eigenvalue is the null space of the matrixA I. For = 1, we have

A I = A I =

3 1 12 0 02 0 0

A vector x =

x1x2x3

is a solution to (A I)x = 0 if and only if3x1 + x2 + x3 = 0 2x1 = 0

so if and only ifx1 = 0 x3 = x2

Thus the general solution is

x =

x1x2x3

= 0x2x2

= x2 011

Thus the eigenspace is one-dimensional with basis

v1 =

011

1


2) Find the characteristic equation and the eigenvalues of the matrix

A =

[1 14 1

]

Solution:

det(A I) = det[1 14 1

]= (1 )2 4 = (1 2)(1 + 2) = (1 )(3 )

Thus the characteristic equation is

(1 )(3 ) = 0

and the eigenvalues are 1 and 3.

2

Sample Quiz 9 solution, MATH 54, Fall 2014

Name (Last, First):

Student ID:

1. Consider the matrix

A =

(5 513 3

).

Use a change of basis to represent A as a rotation and scaling transformation. In other words, finda real matrix

C =

(a bb a

)and an invertible real matrix P such that A = PCP1.

Solution. The characteristic equation is

0 = det(A I) = det(

5 513 3

)= (5 )(3 ) + 65 = 2 + 2+ 50.

This has two complex zeros = 1

1 50 = 1 7i.

We can choose the eigenvalue = 1 7i and find an associated eigenvector:

Nul(A I) = Nul(

4 + 7i 513 4 + 7i

)= Nul

(4 + 7i 5

0 0

)(we didnt really row reduce here; the fact that was an eigenvalue tells us that there must be a

row of zeros in the REF). We can choose the eigenvector v =

(5

4 + 7i

), since

(4 + 7i 5

0 0

)(5

4 + 7i

)=

(5(4 + 7i) + 5(4 + 7i)

0

)= 0.

Then Theorem 9 tells us that A = PCP1, where

P =(Re[v] Im[v]

)=

(5 04 7

)and

C =

(a bb a

)=

(1 77 1

), (where a ib = = 1 7i).

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Sample Quiz 9 solution, MATH 54, Fall 2014

2. Inside of R4, consider the vectors

v1 =

0111

, v2 =

1011

, v3 =

1101

.Find all vectors that are simultaneously orthogonal to v1, v2, and v3 with respect to the dot product.

Solution. First, let us find a basis of the null space

Nul

vT1vT2vT3

= Nul0 1 1 11 0 1 1

1 1 0 1

= Nul1 0 1 11 1 0 1

0 1 1 1

= Nul

1 0 1 10 1 1 00 1 1 1

= Nul

1 0 1 10 1 1 00 0 2 1

A basis is given by

1112

, and all scalings of this vector are exactly those vectors that aresimultaneously orthogonal to v1, v2 and v3.

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Name (Last, First):

Student ID:

1. Find the set of all x in R2 minimizing Ax b where

A =

2 14 12 1

, b = 31

5

.Solution. This is equivalent to solving the least-squares problem for Ax = b, which we can do bysolving the normal equations ATAx = ATb. We can compute

ATA =

[24 00 3

], ATb =

[129

].

Then ATAx = ATb has the unique solution

x =

[1/23

],

which is the unique value of x minimizing Ax b.

1


2. Let H be the subspace of R4 given by

H = span

1110

,0211

,1131

.

Find an orthonormal basis for H.

Solution. Let v1, v2, v3 be the three vectors given above that span H. Well use the Gram-Schmidtprocess:

u1 = v1 =

1110

,

u2 = v2 v2 u1u1 u1

u1 =

1101

,

u3 = v3 v3 u1u1 u1

u1 v3 u2u2 u2

u2 =

1120

.Then {u1, u2, u3} is an orthogonal basis for H. To produce an orthonormal basis, we normalizethese vectors, yielding the set

13

1110

, 131101

, 16

1120

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Name (Last, First):

Student ID:

1. Find an orthogonal matrix P and diagonal matrix D such that A = PDP1 where

A =

[2 33 2

]

Solution. The characteristic equation is det

[2 3

3 2

]= 2 4 5 = 0

Therefore eigenvalues are 1 = 5, 2 = 1

The eigenvector for 1 = 5 will be nonzero vectors in Nul

[3 33 3

], we can pick

[11

]The eigenvector for 2 = 1 will be nonzero vectors in Nul

[3 33 3

], we can pick

[11

]We know that for a symmetric matrix, eigenvectors with different eigenvalues are orthogonal to

each other, and so to orthogonally diagonalize the matrix, we only have to normalize the eigenvectorsto unit vectors. [

11

]

[ 2222

] [11

]

[ 22

22

],

So P =

[ 22

22

22

22

], D =

[5 00 1

].

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2. Solve the initial value problem

y + 6y + 5y = 0, y(0) = 1, y(0) = 1

Solution. The auxiliary equation is 2 + 6 + 5 = 0, the solutions are 1,5, since we have twodifferent real solutions, the general solution to the equation will be C1e

x + C2e5x. Plug in the

initial condition, we have:C1 + C2 = 1

C1 5C2 = 1

Solve the equation above, we have C1 =32 , C2 =

12 , so the solution is

32ex 12e

5x

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Name (Last, First):

Student ID:

1. Find a particular solution to the following differential equation.

y 2y 3y = 3tet

Solution. We first solve the correpsonding homogeneous differential equation:

y 2y 3y = 0

The auxiliary equation isr2 2r 3 = 0

which has two distinct zeroes 3 and 1. Since 1 is not a zero of the auxiliary equation, yp = (at+b)etwill work for some real numbers a and b. We plug in yp to the differential equation:

3tet = yp 2yp 3yp= (4at 4b)et

This implies that a = 34 and b = 0. Therefore,

yp = 3

4tet

is a particular solution to the given differentail equation.

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2. Find a general solution to the following differential equation.

y + 3y + 2y = t + et

Solution. We first solve the correpsonding homogeneous differential equation:

y + 3y + 2y = 0

The auxiliary equation isr2 + 3r + 2 = 0

which has two distinct zeroes 2 and 1. We then get a general solution yh of the homogeneousequation:

yh = c1e2t + c2e

t

where c1 and c2 are arbitray real numbers. To find a particular solution for the given differentialequation, we break it up into two parts; we find a particular solution for

y + 3y + 2y = t

andy + 3y + 2y = et

respectively. And then we add them to obtain a particular solution for the given differentialequation. For the former, 0 is not a zero of the auxiliary equation and so yp1 = at+ b will work forsome real numbers a and b. Plug it in to the former:

t = yp1 + 3yp1 + 2yp1

= 2at + 3a + 2b

This implies that a = 12 and b = 34 . So,

yp1 =1

2t 3

4

is a particular solution to the former.For the latter, 1 is not a zero of the auxiliary equation and so yp1 = ce

t will work for some realnumber c. Plug it in to the latter:

et = yp2 + 3yp2 + 2yp2

= 6cet

This implies that c = 16 . So,

yp2 =1

6et

is a particular solution to the latter. Therefore,

yp := yp1 + yp2 =1

2t 3

4+

1

6et

is a particular solution to the original differentail equation. All in all,

yh + yp = c1e2t + c2e

t +1

2t 3

4+

1

6et

is a general solution for the differential equation wherer c1 and c2 are arbitrary real numbers.

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Name (Last, First):

Student ID:

1. Find a general solution to the homogeneous equation:(d

dt 5)3( d2

dt2+ 4

)y = 0

Solution. The auxiliary equation becomes

(r 5)3(r2 + 4) = 0

with the roots r = 5 (triple)

r = 2i (single)

r = 2i (single)

For the triple root r = 5, we obtain three linearly independent solutions

y1 = e5t, y2 = te

5t, y3 = t2e5t.

The complex roots r = 2i correspond to

y4 = cos 2t, y5 = sin 2t

Hence, a general solution y consists of all possible linear combinations of those five linearly inde-pendent solutions:

y = C1e5t + C2te

5t + C3t2e5t + C4 cos 2t + C5 sin 2t

where C1, , C5 are arbitrary.

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2. Let

x1 =

[ sin tcos t

], x2 =

[cos tsin t

].

Determine if {x1,x2} form a fundamental solution set of the system:

x =

[0 11 0

]x

Solution. First, show that x1 and x1 are solutions to the system by checking

x1 =

[ sin tcos t

]=

[ cos t sin t

]=

[0 11 0

] [ sin tcos t

]=

[0 11 0

]x1

x2 =

[cos tsin t

]=

[ sin tcos t

]=

[0 11 0

] [cos tsin t

]=

[0 11 0

]x2

Since x1 and x2 are solutions to a homogeneous linear system, it suffices to check if the Wron-skian is equal to zero at some point t0 in order to determine the linear independence of {x1,x2}.Then the Wronskian reads

W [x1,x2](t) = det

[ sin t cos tcos t sin t

]Choose any point t0, say t0 = 0, so that

W [x1,x2](0) = det

[ sin 0 cos 0cos 0 sin 0

]= det

[0 11 0

]= 1 6= 0

Therefore, the Wronskian is always nonzero at any t, which implies that the set {x1,x2} is linearlyindependent. Hence, {x1,x2} is a fundamental solution set of the system.

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Berkeley Math 54 Sample Quizzes

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