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 #81 : Matrices

is an idempotent matrix.

True or false: It follows that $A^{T}$ is an idempotent matrix.

Possible Answers:

False

True

Correct answer:

True

Explanation:

is an idempotent matrix if $A^{2}= A$ .

For any A , B for which is defined, $(AB)^{T} = B^{T}A^{T}$ . Setting B = A , if is idempotent, this becomes

$(AA)^{T} = A^{T}A^{T}$

$(A^{2})^{T} =( A^{T} )^{2}$

$A^{T} =( A^{T} )^{2}$

It follows that $A^{T}$ is idempotent.

Report an Error

Example Question #82 : Matrices

is an involutory matrix.

True or false: It follows that $A^{T}$ is an involutory matrix.

Possible Answers:

False

True

Correct answer:

True

Explanation:

is an involutory matrix if $A^{2}= I$ .

For any A , B for which is defined, $(AB)^{T} = B^{T}A^{T}$ . Setting B = A , if is involutory, this becomes

$(AA)^{T} = A^{T}A^{T}$

$(A^{2})^{T} = (A^{T}) ^{2}$

$I^{T} = (A^{T}) ^{2}$

$I = (A^{T}) ^{2}$

Thus, $A^{T}$ is involutory, making the statement true.

Report an Error

Example Question #83 : Matrices

is a nilpotent matrix.

True or false: It follows that $A^{T}$ is a nilpotent matrix.

Possible Answers:

True

False

Correct answer:

True

Explanation:

A matrix is nilpotent if there exists $k \in \mathbb{N}$ so that $A^{k} = O$ , the zero matrix.

Suppose is nilpotent.

Case 1: If A = O , which is trivially nilpotent, then $A^{T} = O$ , which is trivially nilpotent.

Case 2: Suppose $A^{2} = O$ . For any A , B for which is defined, $(AB)^{T} = B^{T}A^{T}$ . Setting B = A , if is nilpotent, this becomes

$(AA)^{T} = A^{T}A^{T}$

$O ^{T} =( A^{T} )^{2}$

$O =( A^{T} )^{2}$

It follows that $A^{T}$ is nilpotent. This reasoning can easily be extended to the case k >2 , since $(A_{1}A_{2}...A_{k})^{T} = A_{n}^{T}...A_{2}^{T}A_{1}^{T}$ .

Report an Error

Example Question #84 : Matrices

It is recommended that you use a calculator with matrix capability for this problem.

A five year study of political affiliation reveals that most voters stay loyal to their political party. However, some changes of affiliation are noted over the course of a typical year, as follows:

3% of Republicans become Democrats; 5% of Republicans become independents; 1% of Republicans become Libertarians.

4% of Democrats become Republicans; 3% of Democrats become independents; 1% of Democrats become Libertarians.

17% of Independents become Republicans; 19% of independents become Democrats; 8% become Libertarians.

1% of Libertarians become Republicans; 1% of Libertarians become Democrats; 1% of Libertarians become independents.

Which group of voters has the same affiliation after the end of a forty-year period with greatest probability: Democrats Republicans, Independents, or Libertarians?

Possible Answers:

Libertarians

Republicans

Democrats

Independents

Correct answer:

Libertarians

Explanation:

Form a stochastic (probability) matrix for this system, where the columns represent current political affiliation (Republicans, Democrats, independents, and Libertarians, in that order), and the rows represent affiliation after one year (same order). This matrix

$P=\begin{bmatrix} 0.91 & 0.04 & 0.17 &0.01 \\ 0.03 & 0.92 & 0.19 &0.01 \\ 0.05 & 0.03 &0.56&0.01 \\ 0.01 & 0.01 & 0.08 & 0.97 \end{bmatrix}$

Raise this matrix to the power of 40 to obtain the stochastic matrix with the probabilities that political affiliations will change over 40 years. Observe the entries in the main diagonal, which represent voters who have the same party affiliation at the beginning and the end of the forty-year period:

$P^{40} = \begin{bmatrix}\textbf{0.3147} & 0.3120 & 0.3014 & 0.2395 \\ 0.3199 & \textbf{0.3290} & 0.3123 & 0.2478\\ 0.0648 & 0.0650 & \textbf{0.0631} & 0.0538\\ 0.3006 & 0.2940 & 0.3232 & \textbf{{\color{Red} 0.4589}} \end{bmatrix}$

The greatest number appears in the fourth entry - the voters who were Libertarians at the beginning and end of the forty-year period. The correct choice is Libertarians.

Report an Error

Example Question #84 : Matrices

True or false: $A^ {3}= I$ if and only if A= I .

Possible Answers:

True

False

Correct answer:

False

Explanation:

One direction of the biconditional holds: if A = I , then $A^{3} = I$ . But the other does not, as proved by counterexample.

We give an example of a nonidentity matrix whose cube is equal to the identity. Let

$A = \begin{bmatrix} \cos \frac{2 \pi}{3} + i \sin \frac{2 \pi}{3} & 0 \\ \\ 0 & \cos \frac{2 \pi}{3} + i \sin \frac{2 \pi}{3} \end{bmatrix}$

This matrix is diagonal having its only nonzero entries on its main diagonal; its cube can be taken by cubing each diagonal element:

$A^{3} = \begin{bmatrix}\left ( \cos \frac{2 \pi}{3} + i \sin \frac{2 \pi}{3} \right ) ^{3}& 0 \\ \\ 0 &\left ( \cos \frac{2 \pi}{3} + i \sin \frac{2 \pi}{3} \right ) ^{3} \end{bmatrix}$

By DeMoivre's Theorem,

$\left (\cos \theta + i\sin \theta \right )^{n} = \cos (n \theta) + i\sin (n \theta )$ ;

setting $\theta = \frac{2 \pi}{3} , n = 3$ ,

$\left ( \cos \frac{2 \pi}{3} + i \sin \frac{2 \pi}{3} \right ) ^{3}$

$= \cos \left ( 3 \cdot \frac{2 \pi}{3} \right )+ i \sin \left ( 3 \cdot \frac{2 \pi}{3} \right )$

$= \cos 2 \pi + i \sin 2 \pi$

and

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

Report an Error

Example Question #85 : Matrices

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

Calculate $A ^{101}$ .

Possible Answers:

$\begin{bmatrix} 0 & 0 & i \\ 0 &1 & 0 \\- i & 0 & 0 \end{bmatrix}$

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

$\begin{bmatrix} 0 & 0 &- i \\ 0 &1 & 0 \\ i & 0 & 0 \end{bmatrix}$

$\begin{bmatrix} i & 0 & 0 \\ 0 &1 & 0 \\0 & 0 & -i \end{bmatrix}$

Correct answer:

$\begin{bmatrix} 0 & 0 & i \\ 0 &1 & 0 \\- i & 0 & 0 \end{bmatrix}$

Explanation:

This task isn't as difficult as it seems if we search for a pattern.

First, find the product $A^{2}$ . This can be found by multiplying rows of by columns of - adding the products of corresponding entries, as follows:

$A^{2} = \begin{bmatrix} 0 & 0 & i \\ 0 &1 & 0 \\- i & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & i \\ 0 &1 & 0 \\- i & 0 & 0 \end{bmatrix}$

$= \begin{bmatrix} 0(0)+0(0) + i(-i) & 0(0)+0(1)+ i(0) & 0(i)+0(0)+ i(0) \\ 0(0)+1(0) +0(-i) & 0(0)+1(1)+0(0) & 0(i)+1(0)+0i(0) \\-i(0)+0(0) + 0(-i) & -i(0)+0(1)+0(0) & -i(i)+0(0)+ 0(0) \end{bmatrix}$

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

Thus,

$A ^{101} = A^{100}A =( A^{2 })^{50} A = I^{50}A = IA = A$

The correct response is $\begin{bmatrix} 0 & 0 & i \\ 0 &1 & 0 \\- i & 0 & 0 \end{bmatrix}$ .

Report an Error

Example Question #86 : Matrices

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

Calculate $A^{8}$ .

Possible Answers:

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

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

$\begin{bmatrix} 0 & 0& 0& 0 \\ 0 & 0& 0 & 0 \\ 0 & 0& 0 &0 \\ 0 & 0& 0 & 0 \end{bmatrix}$

$\begin{bmatrix} 128 & 0 & 0 & 128 \\ 0 & 128 & 128 & 0 \\ 0 & 128 & 128 & 0 \\ 128 & 0 & 0 & 128 \end{bmatrix}$

$\begin{bmatrix} 256& 0 & 0 & 256\\ 0 & 256& 256& 0 \\ 0 & 256& 256& 0 \\ 256& 0 & 0 & 256\end{bmatrix}$

Correct answer:

$\begin{bmatrix} 128 & 0 & 0 & 128 \\ 0 & 128 & 128 & 0 \\ 0 & 128 & 128 & 0 \\ 128 & 0 & 0 & 128 \end{bmatrix}$

Explanation:

The question is actually much easier than it looks if you note that

$A^{8} =\left [ (A^{2} )^{2} \right ]^{2}$ .

Calculate $A^{2}$ by multiplying rows by columns, adding the products of corresponding entries:

$A^{2}= \begin{bmatrix} 1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 1 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 1 & 0 & 0 & 1 \end{bmatrix}$

$A^{2}= \begin{bmatrix} 1(1)+0(0)+0(0) + 1(1)& 1(0)+0(1)+0(1) + 1(0) & 1(0)+0(1)+0(1) + 1(0) & 1(1)+0(0)+0(0) + 1(1) \\ 0(1) + 1(0)+1(0)+0(1) & 0(0) + 1(1)+1(1)+0(0) & 0(0) + 1(1)+1(1)+0(0) & 0(1) + 1(0)+1(0)+0(1) \\ 0(1) + 1(0)+1(0)+0(1) & 0(0) + 1(1)+1(1)+0(0) & 0(0) + 1(1)+1(1)+0(0) & 0(1) + 1(0)+1(0)+0(1)\\ 1(1)+0(0)+0(0) + 1(1) & 1(0)+0(1)+0(1) + 1(0) & 1(0)+0(1)+0(1) + 1(0) & 1(1)+0(0)+0(0) + 1(1) \end{bmatrix}$

$A^{2}= \begin{bmatrix} 2 & 0 & 0 &2 \\ 0 & 2&2 & 0 \\ 0 & 2 & 2 & 0 \\2 & 0 & 0 &2 \end{bmatrix}$

= 2A

$A^{8} =\left [ (A^{2} )^{2} \right ]^{2}$

From which we get

$A^{8} = \left [ (2A )^{2} \right ]^{2}$

$=(4A^{2})^{2}$

$=[ 4 (2A) ] ^{2}$

$=(8A)^{2}$

$= 64A^{2}$

= 128A

$= \begin{bmatrix} 128 & 0 & 0 & 128 \\ 0 & 128 & 128 & 0 \\ 0 & 128 & 128 & 0 \\ 128 & 0 & 0 & 128 \end{bmatrix}$

Report an Error

Example Question #88 : Matrix Matrix Product

It is recommended that you use a calculator with matrix capability for this problem.

Two married couples - Mr. and Mrs. Bernoulli and Mr. and Mrs. Pascal - are playing a game with a pair of fair six-sided dice. They call the game 'King and Queen" and agree to the following rules:

Mr. Bernoulli starts out as "King".

Whoever is "King" or "Queen" rolls the dice. If doubles are rolled, the "King"/"Queen" stays as such. If a 7 is rolled, the new "King"/"Queen" is the spouse of the old "King"/"Queen". If a 3, 4, 5, or 6, other than doubles, is rolled, the new "King"/"Queen" is the same-sex member of the other couple. If an 8, 9, 10, or 11, other than doubles, is rolled, the new "King"/"Queen" is the opposite-sex member of the other couple.

Play is to end after eight rolls. Who is the most likely to end up the "King" or "Queen"?

Possible Answers:

Mr. Pascal

Mr. Bernoulli

Mrs. Pascal

Mrs. Bernoulli

The probabilities are virtually equal.

Correct answer:

The probabilities are virtually equal.

Explanation:

If two fair six-sided dice are rolled, the probability of rolling doubles is $\frac{1}{6}$ , making this the probability that the current "King" or 'Queen" remain as such. The probability of rolling a 7 is also $\frac{1}{6}$ , making this the probability that the new "King" or "Queen" will be the spouse of the current one. The probability that a 3, 4, 5, or 6, other than doubles, will be rolled is $\frac{1}{3}$ , making the probability that the new "King" or "Queen" will be the same-sex member of the other couple. The probability that a 8, 9, 10, or 11, other than doubles, will be rolled is $\frac{1}{3}$ , making the probability that the new "King" or "Queen" will be the opposite-sex member of the other couple. These probabilities do not change for any player.

Construct a stochastic (probability) matrix, where columns represent the initial "King" or "Queen", and rows represent the "King" or "Queen" after the first turn. let the columns/rows represent Mr. Bernoulli, Mrs. Bernoulli, Mr. Pascal, and Mrs. Pascal, in that order:

$P= \begin{bmatrix} \frac{1}{6} & \frac{1}{6} & \frac{1}{3} & \frac{1}{3} \\ \\ \frac{1}{6} & \frac{1}{6} & \frac{1}{3} & \frac{1}{3} \\ \\ \frac{1}{3} & \frac{1}{3} &\frac{1}{6} & \frac{1}{6} \\ \\ \frac{1}{3} & \frac{1}{3} &\frac{1}{6} & \frac{1}{6} \end{bmatrix}$

The stochastic matrix representing the probabilities after eight turns will be the eighth power of this. If we round to the nearest four decimal digits, the matrix is very close to:

$P^{8} = \begin{bmatrix} 0.2500 & 0.2500 & 0.2500 & 0.2500 \\ 0.2500 & 0.2500 & 0.2500 & 0.2500 \\ 0.2500 & 0.2500 & 0.2500 & 0.2500 \\ 0.2500 & 0.2500 & 0.2500 & 0.2500 \end{bmatrix}$

This makes each person's chances of ending up "King" or "Queen" virtually equiprobable no matter who actually starts that way.

Report an Error

Example Question #87 : Matrices

It is recommended that you use a calculator with matrix capability for this problem.

A five year study of political affiliation reveals that most voters stay loyal to their political party. However, some changes of affiliation are noted over the course of a typical year, as follows:

3% of Republicans become Democrats; 5% of Republicans become independents; 1% of Republicans become Libertarians.

4% of Democrats become Republicans; 3% of Democrats become independents; 1% of Democrats become Libertarians.

17% of Independents become Republicans; 19% of independents become Democrats; 8% become Libertarians.

1% of Libertarians become Republicans; 1% of Libertarians become Democrats; 1% of Libertarians become independents.

Suppose a voter starts out as an independent. In forty years, which is he most likely to be - a Democrat, a Republican, an independent, or a Libertarian?

Possible Answers:

Libertarian

Democrat

Independent

Republican

Correct answer:

Libertarian

Explanation:

$P=\begin{bmatrix} 0.91 & 0.04 & 0.17 &0.01 \\ 0.03 & 0.92 & 0.19 &0.01 \\ 0.05 & 0.03 &0.56&0.01 \\ 0.01 & 0.01 & 0.08 & 0.97 \end{bmatrix}$

Raise this matrix to the power of 40 to obtain the stochastic matrix with the probabilities that political affiliations will change over 40 years. Observe the probabilities in the third (independent) column:

$P^{40} = \begin{bmatrix}0.3147 & 0.3120 & \textbf{0.3014} & 0.2395 \\ 0.3199 & 0.3290 & \textbf{0.3123} & 0.2478\\ 0.0648 & 0.0650 & \textbf{0.0631} & 0.0538\\ 0.3006 & 0.2940 & \textbf{{\color{Red} 0.3232}} & 0.4589 \end{bmatrix}$

The greatest value appears in the fourth - Libertarian - row. Libertarian is the correct choice.

Report an Error

Example Question #88 : Matrices

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

Calculate $A^{8}$ .

Possible Answers:

$\begin{bmatrix} -128 & 0 & 0 & 128 \\ 0 & -128 & 128 & 0 \\ 0 & -128 & 128 & 0 \\ -128 & 0 & 0 & 128 \end{bmatrix}$

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

$\begin{bmatrix} 0 & 0& 0& 0 \\ 0 & 0& 0 & 0 \\ 0 & 0& 0 &0 \\ 0 & 0& 0 & 0 \end{bmatrix}$

$\begin{bmatrix} 128 & 0 & 0 & 128 \\ 0 & 128 & 128 & 0 \\ 0 & 128 & 128 & 0 \\ 128 & 0 & 0 & 128 \end{bmatrix}$

$\begin{bmatrix} 128 & 0 & 0 & -128 \\ 0 & 128 & -128 & 0 \\ 0 & 128 & -128 & 0 \\ 128 & 0 & 0 & -128 \end{bmatrix}$

Correct answer:

$\begin{bmatrix} 0 & 0& 0& 0 \\ 0 & 0& 0 & 0 \\ 0 & 0& 0 &0 \\ 0 & 0& 0 & 0 \end{bmatrix}$

Explanation:

The question is actually much easier than it looks if you note that

$A^{8} =\left [ (A^{2} )^{2} \right ]^{2}$ .

Calculate $A^{2}$ by multiplying rows by columns, adding the products of corresponding entries:

$A^{2}= \begin{bmatrix} 1 & 0 & 0 & -1 \\ 0 & 1 & -1 & 0 \\ 0 & 1 & -1 & 0 \\ 1 & 0 & 0 & -1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 & -1 \\ 0 & 1 & -1 & 0 \\ 0 & 1 & -1 & 0 \\ 1 & 0 & 0 & -1 \end{bmatrix}$ $= \begin{bmatrix} 1(1)+ 0(0) + 0(0)+ (-1)(1) & 1(0)+ 0(1) + 0(1)+ (-1)(0) &1(0)+ 0(-1) + 0(-1)+ (-1)(0) & 1(-1)+ 0(0) + 0(0)+ (-1)(-1) \\ 0(1)+ 1(0) + (-1)(0)+0(1) & 0(0)+ 1(1) + (-1)(1)+ 0(0) & 0(0)+ 1(-1) + (-1)(-1)+ 0(0)& 0(-1)+ 1(0) + (-1)(0)+0(-1) \\ 0(1)+ 1(0) + (-1)(0)+0(1) & 0(0)+ 1(1) + (-1)(1)+ 0(0) & 0(0)+ 1(-1) + (-1)(-1)+ 0(0)& 0(-1)+ 1(0) + (-1)(0)+0(-1) \\ 1(1)+ 0(0) + 0(0)+ (-1)(1) & 1(0)+ 0(1) + 0(1)+ (-1)(0) &1(0)+ 0(-1) + 0(-1)+ (-1)(0) & 1(-1)+ 0(0) + 0(0)+ (-1)(-1) \end{bmatrix}$

$=\begin{bmatrix} 0 & 0& 0& 0 \\ 0 & 0& 0 & 0 \\ 0 & 0& 0 &0 \\ 0 & 0& 0 & 0 \end{bmatrix}$ ,

the zero matrix.

is nilpotent with index 2, so any higher power of is also equal to the zero matrix.

$A^{8} =\begin{bmatrix} 0 & 0& 0& 0 \\ 0 & 0& 0 & 0 \\ 0 & 0& 0 &0 \\ 0 & 0& 0 & 0 \end{bmatrix}$ .

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

Biology Tutors in Dallas Fort Worth, Statistics Tutors in Atlanta, Calculus Tutors in Chicago, ISEE Tutors in Seattle, French Tutors in Dallas Fort Worth, Reading Tutors in San Francisco-Bay Area, SAT Tutors in Phoenix, GMAT Tutors in Atlanta, Spanish Tutors in New York City, SAT Tutors in Houston

Popular Courses & Classes

GRE Courses & Classes in Atlanta, Spanish Courses & Classes in Miami, GRE Courses & Classes in Denver, MCAT Courses & Classes in Los Angeles, ISEE Courses & Classes in Houston, LSAT Courses & Classes in Atlanta, SSAT Courses & Classes in Phoenix, GRE Courses & Classes in San Diego, ISEE Courses & Classes in Atlanta, GRE Courses & Classes in San Francisco-Bay Area

Popular Test Prep

ACT Test Prep in Phoenix, GMAT Test Prep in San Francisco-Bay Area, LSAT Test Prep in Miami, MCAT Test Prep in Philadelphia, LSAT Test Prep in San Diego, SSAT Test Prep in Houston, GMAT Test Prep in Chicago, LSAT Test Prep in Philadelphia, MCAT Test Prep in Los Angeles, ACT Test Prep in Chicago

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 #81 : Matrices

Example Question #82 : Matrices

Example Question #83 : Matrices

Example Question #84 : Matrices

Example Question #84 : Matrices

Example Question #85 : Matrices

Example Question #86 : Matrices

Example Question #88 : Matrix Matrix Product

Example Question #87 : Matrices

Example Question #88 : Matrices

All Linear Algebra Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: