Call Now to Set Up Tutoring:

(888) 888-0446

Linear Algebra : Matrices

Study concepts, example questions & explanations for Linear Algebra

Create An Account

All Linear Algebra Resources

4 Diagnostic Tests 108 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #21 : Matrices

and are both matrices with two rows and five columns.

Which of the following are defined?

(a) A + B

(b) A - B

(c)

(d)

Possible Answers:

(a) and (b) only

All four of (a), (b), (c), and (d)

(a), (b) and (d) only

(a), (b) and (c) only

Correct answer:

(a) and (b) only

Explanation:

Two matrices can be added if and only if they have the same number of rows and the same number of columns. This is true of and , so A + B and A - B are defined.

For the product of two matrices to be defined, the number of columns in the first matrix must be equal to the number of rows in the second. cannot be defined, since has five columns and has two rows. cannot be defined for the same reason.

The correct choice is (a) and (b) only.

Report an Error

Example Question #21 : Matrix Matrix Product

is a matrix with three rows and six columns. is a matrix with six rows and three columns.

Which of the following are defined?

(a) A + B

(b) A - B

(c)

(d)

Possible Answers:

(a), (b) and (c) only

(a), (b) and (d) only

(a) and (b) only

All four of (a), (b), (c), and (d)

Correct answer:

Explanation:

Two matrices can be added if and only if they have the same number of rows and the same number of columns. This is not true of and , so A + B and A - B are undefined.

For the product of two matrices to be defined, the number of columns in the first matrix must be equal to the number of rows in the second. has six columns and has six rows, so can be defined. has three columns and has three rows, so can be defined.

Report an Error

Example Question #22 : Matrices

Give the values of , , and so that the matrix

$\begin{bmatrix} 0.4& 0.3& 0.7\\ 0.4& 0.7 & 0.1 \\ a& b& c \end{bmatrix}$

is a stochastic matrix (Markov chain).

Possible Answers:

a = 0, b= 0.4, c = 0.6

a = 0.8, b= 1, c = 0.8

a= 0.2 , b= 0, c= 0.2

The matrix cannot be made into a stochastic matrix regardless of the values of the variables.

a = 0, b=- 0.4, c = 0.6

Correct answer:

a= 0.2 , b= 0, c= 0.2

Explanation:

A stochastic matrix is a matrix of probabilities in which the entry $p_{m,n}$ is the probability that, given the fact that a given system is in a state , the system will be in state next. As such, the elements in each column of , being the probabilities that the system will change from a given state to each other state, respectively, must add up to 1.

Therefore,

a+ 0.4 + 0.4 = 1

a + 0.8 = 1

a = 0.2

b + 0.3 + 0.7 = 1

b+ 1 = 1

b= 0

c + 0.7 + 0.1 = 1

c+ 0.8 = 1

c = 0.2

The correct choice is that a= 0.2 , b= 0, c= 0.2 .

Report an Error

Example Question #22 : Matrix Matrix Product

$A = \begin{bmatrix} 1&0 &0 \\ 0& 1& 0 \end{bmatrix}$ and $B = \begin{bmatrix} 1& 0\\ 0& 1\\ 0& 0 \end{bmatrix}$ .

True or false: AB = BA .

Possible Answers:

False

True

Correct answer:

False

Explanation:

It is not necessary to find and in order to prove that the statement is false.

For the product of two matrices to be defined, the number of columns in the first matrix must be equal to the number of rows in the second. The result of multiplying two matrices when the first matrix has rows and columns and the second has rows and columns is a matrix with rows and columns.

Therefore, since has two rows and three columns, and has three rows and two columns, it follows that has two rows and two columns, and has three rows and three columns. Since and have different dimensions, they cannot be equal.

Report an Error

Example Question #24 : Matrices

Give the values of , , and so that the matrix

$\begin{bmatrix} 0.4& 0.5&0.6 \\ 0.8& 0.4 & 0.1 \\ a & b& c \end{bmatrix}$

is an example of a stochastic matrix (Markov chain).

Possible Answers:

a= - 0.2, b= 0.1, c = 0.3

a= 0.4, b= 0.1, c = 0.5

a= 1.2, b= 0.9, c = 0.7

a= 0.2, b= 0.1, c = 0.3

The matrix cannot be made into a stochastic matrix regardless of the values of the variables.

Correct answer:

The matrix cannot be made into a stochastic matrix regardless of the values of the variables.

Explanation:

Therefore,

a + 0.4 + 0.8 = 1

a+ 1.2 = 1

a = -0.2

However, since a stochastic matrix is, by definition, a matrix of probabilities, each element of the matrix must fall in the interval [0,1] . This rule has been violated, so no values of the variables can make the matrix a stochastic matrix.

Report an Error

Example Question #25 : Matrices

Markov 1

Mary and Marty are playing a game whose board comprises two spaces, marked "X' and "Y", as shown above. For each turn, one player rolls a fair six-sided die, then moves to the space indicated based on where the player is already and what the player rolls. For example, if the player is on space "X" and rolls a "2", he moves to space "Y", since one of the numbers on the arrow going from "X" to 'Y" is "2"; similarly, if he is on space "X" and rolls a "6", he will stay where he is.

Give the stochastic matrix which represents the probabilities that, given that a player is on a particular space, he or she will be on each given space two turns later.

Possible Answers:

$\begin{bmatrix} \frac{4}{9} & \frac{1}{6} \\ \\ \frac{5}{9} &\frac{5}{6} \end{bmatrix}$

$\begin{bmatrix} \frac{1}{3}& \frac{1}{3} \\ \\ \frac{2}{3} &\frac{2}{3} \end{bmatrix}$

$\begin{bmatrix} \frac{4}{9} &\frac{4}{9} \\ \\ \frac{5}{9} &\ \frac{5}{9} \end{bmatrix}$

$\begin{bmatrix} \frac{1}{6} & \frac{1}{3} \\ \\ \frac{5}{6} &\frac{2}{3} \end{bmatrix}$

$\begin{bmatrix} \frac{4}{9} & \frac{1}{3} \\ \\ \frac{5}{9} &\frac{2}{3} \end{bmatrix}$

Correct answer:

$\begin{bmatrix} \frac{4}{9} & \frac{1}{3} \\ \\ \frac{5}{9} &\frac{2}{3} \end{bmatrix}$

Explanation:

A stochastic matrix is a matrix of probabilities in which the entry $p_{m,n}$ is the probability that, given the fact that a given system is in a state , the system will be in state next.

We will let State 1 be that the player is on Space "X" and State 2 be that the player is on Space "Y". If the player is on Space "X", the probability that he will still be on that space after one turn is $\frac{1}{6}$ , since one roll out of six will allow him to stay there; similarly, the probability that he will be on Space "Y" is $\frac{5}{6}$ . If a player is on Space "B", the same probabilities are, respectively, $\frac{1}{2}$ and $\frac{1}{2}$ . The stochastic matrix representing these probabilities is

$P = \begin{bmatrix} \frac{1}{6}& \frac{1}{2}\\ \\ \frac{5}{6}& \frac{1}{2} \end{bmatrix}$ .

The stochastic matrix representing the probabilities of being on a particular space after two terms is the square of this matrix $P^{2}$ , which is

$P ^{2}= \begin{bmatrix} \frac{1}{6}& \frac{1}{2}\\ \\ \frac{5}{6}& \frac{1}{2} \end{bmatrix}\begin{bmatrix} \frac{1}{6}& \frac{1}{2}\\ \\ \frac{5}{6}& \frac{1}{2} \end{bmatrix}$

The entry in column , row of the product $PP= P^{2}$ is the product of row of and column of - the sum of the products of the numbers that appear in the corresponding positions of the row and the column. Therefore,

$P ^{2}= \begin{bmatrix} \frac{1}{6} \cdot \frac{1}{6}+ \frac{1}{2} \cdot \frac{5}{6} & \frac{1}{6} \cdot \frac{1}{2} + \frac{1}{2} \cdot \frac{1}{2} \\ \\ \frac{5}{6} \cdot \frac{1}{6}+ \frac{1}{2}\cdot \frac{5}{6} &\frac{5}{6} \cdot \frac{1}{2} + \frac{1}{2} \cdot \frac{1}{2} \end{bmatrix}$

$= \begin{bmatrix} \frac{1}{36}+ \frac{5}{12} & \frac{1}{12} + \frac{1}{4} \\ \\ \frac{5}{36}+ \frac{5}{12} &\frac{5}{12} + \frac{1}{4} \end{bmatrix}$

$= \begin{bmatrix} \frac{4}{9} & \frac{1}{3} \\ \\ \frac{5}{9} &\frac{2}{3} \end{bmatrix}$ ,

the correct choice.

Report an Error

Example Question #23 : Matrix Matrix Product

A B and are both defined products. has four rows and three columns.

Which of the following is true about the number of rows and columns in ?

Possible Answers:

can have any number of rows but must have four columns.

must have three rows but can have any number of columns.

must either have three rows and four columns or four rows and three columns.

must have three rows and four columns.

must have four rows and three columns.

Correct answer:

must have three rows and four columns.

Explanation:

For a matrix product A B to be defined, it must hold that the number of rows in is equal to the number of columns in . has three columns, so must have three rows. For to be defined, by similar reasoning, must have four columns.

Report an Error

Example Question #28 : Matrices

AB = C

where

$B = \begin{bmatrix} 2& 3& -4& 2\\ -1&4 & 1& 0\\ 3& -2& 5&1 \\ 1&1 &2 &3 \end{bmatrix}$

and

$C = \begin{bmatrix} 2& 3& -4& 2\\ 3&-12 & -3& 0\\ 3& -2& 5&1 \\ 1&1 &2 &3 \end{bmatrix}$

Evaluate .

(You may assume that both and have inverses. )

Possible Answers:

None of the other choices gives a correct response.

$A = \begin{bmatrix} -3&0 &0 &0 \\ 0&1 & 0& 0\\ 0&0 &1&0 \\ 0&0 &0 & 1 \end{bmatrix}$

$A = \begin{bmatrix} 1&0 &0 &0 \\ 0&1 & 0& 0\\ 0&0 &-3&0 \\ 0&0 &0 & 1 \end{bmatrix}$

$A = \begin{bmatrix} 1&0 &0 &0 \\ 0&-3 & 0& 0\\ 0&0 &1 &0 \\ 0&0 &0 & 1 \end{bmatrix}$

$A = \begin{bmatrix} 1&0 &0 &0 \\ 0&1 & 0& 0\\ 0&0 &1&0 \\ 0&0 &0 & -3 \end{bmatrix}$

Correct answer:

$A = \begin{bmatrix} 1&0 &0 &0 \\ 0&-3 & 0& 0\\ 0&0 &1 &0 \\ 0&0 &0 & 1 \end{bmatrix}$

Explanation:

While A B =C implies that $A= C B^{-1}$ , it is not necessary to do the calculations in order to find .

differs from only in its second row, in which each element is times the corresponding element in :

$-1 \cdot( -3 )= 3$

$4 \cdot( -3 )= -12$

$1 \cdot( -3 )=- 3$

$0 \cdot( -3 )= 0$

Therefore, is the result of a row operation on , namely,

$-3R 2 \rightarrow R2$

Therefore, is the product of an elementary matrix and ; the elementary matrix for the given row operation is the one in which the operation is performed on the (four-by-four) identity matrix, which is

$A = \begin{bmatrix} 1&0 &0 &0 \\ 0&-3 & 0& 0\\ 0&0 &1 &0 \\ 0&0 &0 & 1 \end{bmatrix}$ .

Report an Error

Example Question #22 : Matrices

AB = C

where

$B = \begin{bmatrix} 2& 3& -4& 2\\ -1&4 & 1& 0\\ 3& -2& 5&1 \\ 1&1 &2 &3 \end{bmatrix}$

and

$C = \begin{bmatrix} 3& -2& 5&1 \\ -1&4 & 1& 0 \\ 2& 3& -4& 2 \\ 1&1 &2 &3 \end{bmatrix}$

Evaluate .

(You may assume that both and have inverses. )

Possible Answers:

$A = \begin{bmatrix} 0&0 & 1&0 \\ 0&1 & 0& 0\\ 1& 0& 0& 0\\ 0& 0& 0& 1 \end{bmatrix}$

$A = \begin{bmatrix} 0&0 & -1&0 \\ 0&1 & 0& 0\\ -1& 0& 0& 0\\ 0& 0& 0& 1 \end{bmatrix}$

$A = \begin{bmatrix} 1&0 & 0&0 \\ 0&0 & 0& 1\\ 0& 0& 1& 0\\ 0& 1& 0& 0 \end{bmatrix}$

$A = \begin{bmatrix}-1&0 & 0&0 \\ 0&1 & 0& 0\\ 0& 0& -1& 0\\ 0& 0& 0& 1 \end{bmatrix}$

$A = \begin{bmatrix} 1&0 & 0&0 \\ 0&0 & 0& -1\\ 0& 0& 1& 0\\ 0&- 1& 0& 0 \end{bmatrix}$

Correct answer:

$A = \begin{bmatrix} 0&0 & 1&0 \\ 0&1 & 0& 0\\ 1& 0& 0& 0\\ 0& 0& 0& 1 \end{bmatrix}$

Explanation:

While A B =C implies that $A= C B^{-1}$ , it is not necessary to do the calculations in order to find .

differs from only in that its first and third rows have reversed positions. Therefore, is the result of a row operation on , namely,

$R1 \leftrightarrow R3$

$A = \begin{bmatrix} 0&0 & 1&0 \\ 0&1 & 0& 0\\ 1& 0& 0& 0\\ 0& 0& 0& 1 \end{bmatrix}$ .

Report an Error

Example Question #28 : Matrices

Let $L = \begin{bmatrix} 1& 4\\ 0& 2 \end{bmatrix}$ , $U = \begin{bmatrix} 2&2 \\ 0& -3 \end{bmatrix}$ , and $A = \begin{bmatrix} 2&-10 \\ 0& -6 \end{bmatrix}$

True or false: is an example of a valid -factorization of .

Possible Answers:

False, because is not the right kind of matrix.

True

False, because is not a factorization of .

Correct answer:

False, because is not the right kind of matrix.

Explanation:

An -factorization is a way of expressing a matrix as a product of two matrices and . For the factorization to be valid:

1) must be a Lower triangular matrix - all elements above its main diagonal (upper left corner to lower right corner) must be "0".

2) must be an Upper triangular matrix - all elements below its main diagonal must be "0".

3) LU = A

can be seen to have a nonzero element above its main diagonal - the 4 in Row 1, Column 2. Consequently, is not a lower triangular matrix. This makes invalid as an -factorization.

Report an Error

View Linear Algebra Tutors

Chelsea
Certified Tutor

The University of Texas at Dallas, Bachelor of Science, Physics. The University of Texas at Dallas, Master of Science, Bioinf...

View Linear Algebra Tutors

Nima
Certified Tutor

Sahand University of Technology, Master of Science, Electrical Engineering. Institut national de la recherche scientifique, D...

View Linear Algebra Tutors

Liban
Certified Tutor

Emory University, Bachelor of Science, Mathematics/Economics.

All Linear Algebra Resources

4 Diagnostic Tests 108 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

GMAT Tutors in Atlanta, English Tutors in Houston, Algebra Tutors in New York City, ISEE Tutors in Chicago, SAT Tutors in San Francisco-Bay Area, Spanish Tutors in Chicago, GRE Tutors in Chicago, Chemistry Tutors in Los Angeles, Statistics Tutors in Boston, Reading Tutors in Miami

Popular Courses & Classes

SAT Courses & Classes in New York City, LSAT Courses & Classes in Atlanta, ACT Courses & Classes in San Diego, ISEE Courses & Classes in San Francisco-Bay Area, SSAT Courses & Classes in Atlanta, GRE Courses & Classes in Los Angeles, GMAT Courses & Classes in Seattle, ACT Courses & Classes in Seattle, ACT Courses & Classes in Los Angeles, SAT Courses & Classes in Houston

Popular Test Prep

GRE Test Prep in Miami, SAT Test Prep in San Diego, MCAT Test Prep in Houston, GRE Test Prep in Denver, ACT Test Prep in Dallas Fort Worth, SSAT Test Prep in Houston, ACT Test Prep in Seattle, GMAT Test Prep in Miami, ISEE Test Prep in Washington DC, GRE Test Prep in Phoenix

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

Linear Algebra : Matrices

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

Example Question #21 : Matrices

Example Question #21 : Matrix Matrix Product

Example Question #22 : Matrices

Example Question #22 : Matrix Matrix Product

Example Question #24 : Matrices

Example Question #25 : Matrices

Example Question #23 : Matrix Matrix Product

Example Question #28 : Matrices

Example Question #22 : Matrices

Example Question #28 : Matrices

All Linear Algebra Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: