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 #101 : Matrix Matrix Product

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

Calculate $(AA^{T})^{99}$ .

Possible Answers:

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

$(AA^{T})^{99}$ is not defined.

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

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

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

Correct answer:

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

Explanation:

$A^{T}$ , the transpose of , can be found by transposing rows with columns.

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

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

Matrices are multiplied by multiplying each row of the first matrix by each column of the second - that is, by adding the products of the entries in corresponding positions. Thus,

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

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

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

This is a square matrix, so it can be raised to a power. To raise a diagonal matrix to a power, simply raise each number in the main diagonal to that power:

$(AA^{T})^{99} = \begin{bmatrix} (-1)^{99} & 0 \\ 0 & (-1)^{99} \end{bmatrix}= \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}$ .

Report an Error

Example Question #102 : Matrix Matrix Product

$A=\begin{bmatrix} a & 1 \\ -1 & b \end{bmatrix}$ and $C = \begin{bmatrix} c & -1 \\ 1 & d \end{bmatrix}$ , where all four variables stand for real quantities.

Which must be true of and regardless of the values of the variables?

Possible Answers:

$AC= -(CA)^{T}$

AC= CA

None of the statements given in the other choices are correct.

AC=- (CA)

$AC= (CA)^{T}$

Correct answer:

None of the statements given in the other choices are correct.

Explanation:

Matrices are multiplied by multiplying each row of the first matrix by each column of the second - that is, by adding the products of the entries in corresponding positions. Thus,

$AC=\begin{bmatrix} a & 1 \\ -1 & b \end{bmatrix}\begin{bmatrix} c & -1 \\ 1 & d \end{bmatrix}$

$=\begin{bmatrix} a (c)+ 1(1)& a (-1)+ 1(d)\\ -1 (c)+ b(1)& -1 (-1)+ b(d)\end{bmatrix}$

$=\begin{bmatrix} ac+1&-a+d \\ b-c& bd+1\end{bmatrix}$

$CA=\begin{bmatrix} c & -1 \\ 1 & d \end{bmatrix}\begin{bmatrix} a & 1 \\ -1 & b \end{bmatrix}$

$=\begin{bmatrix} c (a)+ (-1)(-1)& c(1)+(-1)(b)\\ 1 (a)+ d(-1)& 1(1)+d(b)\end{bmatrix}$

$=\begin{bmatrix} ac+1& -b+c\\ a-d& bd+1\end{bmatrix}$

AC= CA is false in general.

$(CA)^{T}$ , the transpose of , is the result of transposing rows and columns:

$(CA)^{T} =\begin{bmatrix} ac+1& a-d \\ -b+c & bd+1\end{bmatrix}$ . so

$AC= (CA)^{T}$ is false in general.

$-(CA) =\begin{bmatrix} -ac-1& b-c\\ -a-d&- bd-1\end{bmatrix}$ , so

AC=- (CA) is false in general.

$-(CA)^{T} =\begin{bmatrix} -ac-1&-a-d \\ b-c &- bd-1\end{bmatrix}$ , so

$AC= -(CA)^{T}$ is false in general.

Thus, none of the four given statements need be true.

Report an Error

Example Question #103 : Matrix Matrix Product

Let $A = \begin{bmatrix} \cos \frac{\pi }{5} \\ \sin \frac{\pi }{5} \end{bmatrix}$

Find $(AA^{T})^{2}$ .

Possible Answers:

$\begin{bmatrix} \cos ^{2}\frac{\pi }{5} - \sin^{2} \frac{\pi }{5}\end{bmatrix}$

$\begin{bmatrix} \cos ^{2}\frac{\pi }{5} & 1 \\ 1& \sin^{2} \frac{\pi }{5}\end{bmatrix}$

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

$\begin{bmatrix} \cos ^{2}\frac{\pi }{5} & \sin \frac{\pi }{5}\cos \frac{\pi }{5} \\ \sin \frac{\pi }{5}\cos \frac{\pi }{5} & \sin^{2} \frac{\pi }{5}\end{bmatrix}$

$(AA^{T})^{2}$ is undefined.

Correct answer:

$\begin{bmatrix} \cos ^{2}\frac{\pi }{5} & \sin \frac{\pi }{5}\cos \frac{\pi }{5} \\ \sin \frac{\pi }{5}\cos \frac{\pi }{5} & \sin^{2} \frac{\pi }{5}\end{bmatrix}$

Explanation:

is equal to the two-entry column matrix $\begin{bmatrix} \cos \frac{\pi }{5} \\ \sin \frac{\pi }{5} \end{bmatrix}$ , so $A^{T}$ , the transpose, is the row matrix

$A^{T} = \begin{bmatrix} \cos \frac{\pi }{5} & \sin \frac{\pi }{5} \end{bmatrix}$

The product of two matrices is calculated by multiplying rows by columns - adding the corresponding entries in the rows of the first matrix by the columns of the second - so

$AA^{T}= \begin{bmatrix} \cos \frac{\pi }{5} \\ \sin \frac{\pi }{5} \end{bmatrix}\begin{bmatrix} \cos \frac{\pi }{5} & \sin \frac{\pi }{5} \end{bmatrix}$

$= \begin{bmatrix} \cos \frac{\pi }{5}(\cos \frac{\pi }{5}) & \sin \frac{\pi }{5}(\cos \frac{\pi }{5}) \\ \cos \frac{\pi }{5}(\sin \frac{\pi }{5}) & \sin \frac{\pi }{5}(\sin \frac{\pi }{5}) \end{bmatrix}$

$= \begin{bmatrix} \cos ^{2}\frac{\pi }{5} & \sin \frac{\pi }{5}\cos \frac{\pi }{5} \\ \sin \frac{\pi }{5}\cos \frac{\pi }{5} & \sin^{2} \frac{\pi }{5}\end{bmatrix}$

Report an Error

Example Question #101 : Matrices

Multiply: $\begin{bmatrix} 2 & 0& -5\\ -1& 3& 1\end{bmatrix} \cdot \begin{bmatrix} 2\\ 2\\ 1\end{bmatrix}$

Possible Answers:

$\begin{bmatrix} 9\\8 \end{bmatrix}$

$\begin{bmatrix} -1\\5 \end{bmatrix}$

$\begin{bmatrix} -1\\8 \end{bmatrix}$

$\begin{bmatrix} 9\\5 \end{bmatrix}$

$\begin{bmatrix} -1\\ -3\end{bmatrix}$

Correct answer:

$\begin{bmatrix} -1\\5 \end{bmatrix}$

Explanation:

To multiply, add:

$2 \begin{bmatrix} 2\\ -1\end{bmatrix} + 2 \begin{bmatrix} 0\\3 \end{bmatrix} + 1 \begin{bmatrix} -5\\ 1\end{bmatrix}$

$= \begin{bmatrix} 4\\-2 \end{bmatrix} + \begin{bmatrix} 0\\ 6\end{bmatrix} + \begin{bmatrix} -5\\ 1\end{bmatrix} = \begin{bmatrix} -1\\ 5\end{bmatrix}$

Report an Error

Example Question #2 : Matrix Vector Product

Compute AB.

$A=\begin{bmatrix} 1&3 \\2 &5 \end{bmatrix}$

$B = \begin{bmatrix} 4& 1& 3\\ 6& -2&0 \end{bmatrix}$

Possible Answers:

$AB = \begin{bmatrix} 4& 3&3 \\ 12&-10 &0 \end{bmatrix}$

$AB = \begin{bmatrix} -14&7 & 3\\ -22& 12&6 \end{bmatrix}$

None of the other answers.

$AB = \begin{bmatrix} 22& -5&3 \\ 38&-8 &6 \end{bmatrix}$

$AB = \begin{bmatrix} 5&4 &3 \\ 8&3 &0 \end{bmatrix}$

Correct answer:

$AB = \begin{bmatrix} 22& -5&3 \\ 38&-8 &6 \end{bmatrix}$

Explanation:

Because the number of columns in matrix A and the number of rows in matrix B are equal, we know that product AB does in fact exist. Matrix AB should have the same number of rows as A and the same number of columns as B. In this case, AB is a 2x3 matrix:

$AB=\begin{bmatrix} (1\cdot 4+3\cdot 6) & (1\cdot 1+3\cdot -2) &(1\cdot 3+3\cdot 0) \\ (2\cdot 4+5\cdot 6) & (2\cdot 1+5\cdot -2) &(2\cdot 3+5\cdot 0) \end{bmatrix}$

Report an Error

Example Question #1 : Matrix Vector Product

Compute AB

$A=\begin{bmatrix} 5\end{bmatrix}$

$B = \begin{bmatrix} -5&8 &12 &7 \end{bmatrix}$

Possible Answers:

$AB = \begin{bmatrix} -25\\40 \\60 \\35 \end{bmatrix}$

$AB = \begin{bmatrix} 0&13 &17 &12 \end{bmatrix}$

$AB= \begin{bmatrix} -25&40 &60 &35 \end{bmatrix}$

$AB = \begin{bmatrix} 0\\ 13\\17 \\12 \end{bmatrix}$

None of the other answers

Correct answer:

$AB= \begin{bmatrix} -25&40 &60 &35 \end{bmatrix}$

Explanation:

$AB = \begin{bmatrix} 5\cdot -5& 5\cdot 8&5\cdot 12 &5\cdot 7 \end{bmatrix}$

Report an Error

Example Question #3 : Matrix Vector Product

Calculate $A\bf{x}$ , given

$A=\begin{bmatrix} 3& 4\\ 0&-2 \\ 1&3 \end{bmatrix}$ , $\bf{x}=\begin{bmatrix} 6\\ 2 \end{bmatrix}$

Possible Answers:

$\begin{bmatrix} 24\\ -4\\ 6 \end{bmatrix}$

$\begin{bmatrix} 26\\ -4\\ 12 \end{bmatrix}$

$\begin{bmatrix} 8\\ 0\\ 6 \end{bmatrix}$

$\begin{bmatrix} 18\\ 0\\ 6 \end{bmatrix}$

Correct answer:

$\begin{bmatrix} 26\\ -4\\ 12 \end{bmatrix}$

Explanation:

By definition,

$A\bf{x}=\begin{bmatrix} 3&4 \\ 0&-2 \\ 1& 3 \end{bmatrix} \begin{bmatrix} 6\\ 2 \end{bmatrix}= \begin{bmatrix} 3(6)+4(2)\\ 0(6)-2(2)\\ 1(6)+3(2) \end{bmatrix}= \begin{bmatrix} 26\\ -4\\ 12 \end{bmatrix}$ .

Report an Error

Example Question #1 : Matrix Vector Product

Calculate , given

$A = \begin{bmatrix} 1&4 \end{bmatrix}$

$B=\begin{bmatrix} 3\\ 4 \end{bmatrix}$

Possible Answers:

$AB= \begin{bmatrix} 3& 4\\ 1& 4 \end{bmatrix}$

AB= 16

$AB = \begin{bmatrix} 1& 4\\ 3& 4 \end{bmatrix}$

AB = 19

Correct answer:

AB = 19

Explanation:

By definition,

$AB = \begin{bmatrix} 1& 4 \end{bmatrix} \begin{bmatrix} 3\\ 4 \end{bmatrix} = [3(1)+4(4)]= [19]=19$ . A matrix with only one entry is simply a scalar.

Report an Error

Example Question #5 : Matrix Vector Product

Calculate , given

c = 3

$A= \begin{bmatrix} 3&1 \\ 5&-2 \end{bmatrix}$ .

Possible Answers:

cA = 21

$cA = \begin{bmatrix} 9&3 \\ 15& -6 \end{bmatrix}$

$cA = \begin{bmatrix} 9&1 \\ 5&-6 \end{bmatrix}$

Can not be determined, is not a vector.

Correct answer:

$cA = \begin{bmatrix} 9&3 \\ 15& -6 \end{bmatrix}$

Explanation:

By definition,

$cA = 3\begin{bmatrix} 3&1 \\ 5& -2 \end{bmatrix}= \begin{bmatrix} 3(3)&3(1) \\ 3(5)&3(-2) \end{bmatrix}= \begin{bmatrix} 9&3 \\ 15&-6 \end{bmatrix}$ .

Report an Error

Example Question #1 : Matrix Vector Product

$\begin{align*}&\text{Find the product }A\times B \\&\text{Where }A=\begin{bmatrix}3&10\\3&-10\end{bmatrix}\text{ and }B=\begin{bmatrix}2&-8\\11&-19\end{bmatrix}\end{align*}$

Possible Answers:

$\begin{bmatrix}85&-245\\-135&135\end{bmatrix}$

$\text{The multiplication cannot be performed.}$

$\begin{bmatrix}116&-214\\-104&166\end{bmatrix}$

$\begin{bmatrix}182&-148\\-38&232\end{bmatrix}$

Correct answer:

$\begin{bmatrix}116&-214\\-104&166\end{bmatrix}$

Explanation:

$\begin{align*}&\text{In order to multiply two matrices, }A\times b\text{, the respective dimensions of each}\\&\text{must be of the form }m\times n\text{ and }n\times p\text{ to create an}\\&m\times p\text{ matrix. Note that order does matter:}\\&(A\times b \neq b \times A)\\&\text{Since A has dimensions: }2\times2\\&\text{and B has dimensions: }2\times2\\&\text{The resultant matrix has dimensions: }2\times2\\&\begin{bmatrix}3&10\\3&-10\end{bmatrix}\times\begin{bmatrix}2&-8\\11&-19\end{bmatrix}\\&\begin{bmatrix}(3)(2)+(10)(11)&(3)(-8)+(10)(-19)\\(3)(2)+(-10)(11)&(3)(-8)+(-10)(-19)\end{bmatrix}\\&\begin{bmatrix}116&-214\\-104&166\end{bmatrix}\end{align*}$

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

Sanford
Certified Tutor

Westminster College, Bachelor of Science, Social Studies Teacher Education.

View Linear Algebra Tutors

Raha
Certified Tutor

Amirkabir University of Technology, Bachelor of Science, Chemical Engineering. Wayne State University, Master of Science, Che...

All Linear Algebra Resources

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

Popular Subjects

Biology Tutors in Houston, Computer Science Tutors in New York City, Math Tutors in Phoenix, SAT Tutors in Houston, Statistics Tutors in New York City, French Tutors in Philadelphia, French Tutors in San Diego, GRE Tutors in Phoenix, Computer Science Tutors in Denver, GMAT Tutors in Dallas Fort Worth

Popular Courses & Classes

Spanish Courses & Classes in San Francisco-Bay Area, ACT Courses & Classes in Chicago, ACT Courses & Classes in Denver, MCAT Courses & Classes in Dallas Fort Worth, LSAT Courses & Classes in Phoenix, Spanish Courses & Classes in Seattle, GMAT Courses & Classes in Boston, GRE Courses & Classes in Miami, MCAT Courses & Classes in Houston, ACT Courses & Classes in Miami

Popular Test Prep

SAT Test Prep in Washington DC, GRE Test Prep in Dallas Fort Worth, ACT Test Prep in Denver, ACT Test Prep in Chicago, LSAT Test Prep in Los Angeles, MCAT Test Prep in Phoenix, ISEE Test Prep in Houston, LSAT Test Prep in Atlanta, LSAT Test Prep in Miami, SAT Test Prep in Seattle

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 #101 : Matrix Matrix Product

Example Question #102 : Matrix Matrix Product

Example Question #103 : Matrix Matrix Product

Example Question #101 : Matrices

Example Question #2 : Matrix Vector Product

Example Question #1 : Matrix Vector Product

Example Question #3 : Matrix Vector Product

Example Question #1 : Matrix Vector Product

Example Question #5 : Matrix Vector Product

Example Question #1 : Matrix Vector Product

All Linear Algebra Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: