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 #121 : Matrices

$\begin{align*}&\text{Find the matrix product }\\&\begin{bmatrix}9&8\\13&2\end{bmatrix}\times\begin{bmatrix}20\\-7\end{bmatrix}\end{align*}$

Possible Answers:

$\begin{bmatrix}220\\342\end{bmatrix}$

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

$\begin{bmatrix}124\\246\end{bmatrix}$

$\begin{bmatrix}173\\295\end{bmatrix}$

Correct answer:

$\begin{bmatrix}124\\246\end{bmatrix}$

Explanation:

$\begin{align*}&\text{In order to multiply two matrices, }A\times b\text{, the respective dimensions of each}\\&\text{must be of the form }m\times n\text{ and }n\times p\text{ to create an}\\&m\times p\text{ matrix. Note that order does matter:}\\&(A\times b \neq b \times A)\\&\text{Since our matrices have dimensions: }2\times2\text{ and }2\times1\\&\text{The resultant matrix has dimensions: }2\times1\\&\begin{bmatrix}9&8\\13&2\end{bmatrix}\times\begin{bmatrix}20\\-7\end{bmatrix}\\&\begin{bmatrix}(9)(20)+(8)(-7)\\(13)(20)+(2)(-7)\end{bmatrix}\\&\begin{bmatrix}124\\246\end{bmatrix}\end{align*}$

Report an Error

Example Question #13 : Matrix Vector Product

Let be a matrix and $\bf{x}$ be a vector defined by:

$A = \begin{bmatrix} 2 & 6 & 0 \\ 7 & -1 & 9 \end{bmatrix}$

$\bf{x} = \begin{bmatrix} 2\\ 4 \\-3 \end{bmatrix}$

Find the product $A\bf{x}$ .

Possible Answers:

$A\bf{x} = \begin{bmatrix} 28 & -17 \end{bmatrix}$

$A\bf{x} = \begin{bmatrix} 28 \\ -17 \end{bmatrix}$

$A\bf{x} = \begin{bmatrix} 32 & -17 \end{bmatrix}$

$A\bf{x} = \begin{bmatrix} 24 \\ -7 \end{bmatrix}$

The product does not exist because the dimensions do not match.

Correct answer:

$A\bf{x} = \begin{bmatrix} 28 \\ -17 \end{bmatrix}$

Explanation:

First we check the dimensions. The matrix has 3 columns and the vector $\bf{x}$ has three rows. The dimensions match and the product exists.

$A\bf{x} = \begin{bmatrix} 2 & 6 & 0 \\ 7 & -1 & 9 \end{bmatrix} \cdot\begin{bmatrix} 2\\ 4 \\-3 \end{bmatrix}$

Now we take the dot product of rows in the matrix and the vector $\bf{x}$ .

$\begin{bmatrix} 2 & 6 & 0 \\ 7 & -1 & 9 \end{bmatrix} \cdot\begin{bmatrix} 2\\ 4 \\-3 \end{bmatrix} = \begin{bmatrix} 2\cdot 2 + 6\cdot 4 + 0\cdot -3 \\ 7\cdot 2 + -1\cdot 4 + 9\cdot -3 \end{bmatrix} = \begin{bmatrix} 28 \\ -17 \end{bmatrix}$

Report an Error

Example Question #809 : Linear Algebra

Let be a matrix and $\bf{x}$ be a vector defined as

$A = \begin{bmatrix} 9 & 0 & 3 \\ 7 & 2 & 9 \\ 5 & -9 & -4 \end{bmatrix}$

$\bf{x} = \begin{bmatrix} -9\\ 3 \\ -2 \end{bmatrix}$

Find the product $A\bf{x}$ .

Possible Answers:

The product does not exist because the dimensions do not match.

$A\bf{x}=\begin{bmatrix} 2 & -3 & 9 \end{bmatrix}$

$A\bf{x}=\begin{bmatrix}-75\\ -87 \\ 64 \end{bmatrix}$

$A\bf{x}=\begin{bmatrix}-87 & -75 & -64 \end{bmatrix}$

$A\bf{x}=\begin{bmatrix}-87\\ -75 \\ -64 \end{bmatrix}$

Correct answer:

$A\bf{x}=\begin{bmatrix}-87\\ -75 \\ -64 \end{bmatrix}$

Explanation:

First we check that the dimensions match. Matrix has 3 columns and vector $\bf{x}$ has three rows. The dimensions match and the product exists.

$A\bf{x}= \begin{bmatrix} 9 & 0 & 3 \\ 7 & 2 & 9 \\ 5 & -9 & -4 \end{bmatrix} \cdot \begin{bmatrix} -9\\ 3 \\ -2 \end{bmatrix} =\begin{bmatrix} (9)(-9) + (0)(3) + (3)(-2)\\ (7)(-9)+(2)(3)+(9)(-2)\\ (5)(-9)+(-9)(3)+(-4)(-2) \end{bmatrix} =\begin{bmatrix} -87\\ -75\\ -64 \end{bmatrix}$

Report an Error

Example Question #21 : Matrix Vector Product

Let be a matrix and $\bf{x}$ be a vector defined by

$A = \begin{bmatrix} 0 & 8 & 3 & 6\\ 9 & 0 & 8 & -4 \end{bmatrix}$

$\bf{x} = \begin{bmatrix} 5 \\ 8 \\ -2 \\ 7\end{bmatrix}$

Find the product $A\bf{x}$ .

Possible Answers:

$A\bf{x} = \begin{bmatrix} 100 & 99 \end{bmatrix}$

$A\bf{x} = \begin{bmatrix} 100 & 1 \end{bmatrix}$

$A\bf{x} = \begin{bmatrix} 100 \\ 1 \end{bmatrix}$

$A\bf{x} = \begin{bmatrix} 10 \\ 10 \end{bmatrix}$

The product does not exist because the dimensions do not match.

Correct answer:

$A\bf{x} = \begin{bmatrix} 100 \\ 1 \end{bmatrix}$

Explanation:

Matrix has 4 columns and vector $\bf{x}$ has 4 rows. The dimensions match and the product exists.

$A\bf{x} = \begin{bmatrix} 0 & 8 & 3 & 6\\ 9 & 0 & 8 & -4 \end{bmatrix} \cdot \begin{bmatrix} 5 \\ 8 \\ -2 \\ 7\end{bmatrix} = \begin{bmatrix} (0)(5)+(8)(8)+(3)(-2)+(6)(7) \\ (9)(5)+(0)(8)+(8)(-2)+(-4)(7)\end{bmatrix}$

$A\bf{x}= \begin{bmatrix} 100 \\ 1 \end{bmatrix}$

Report an Error

Example Question #22 : Matrix Vector Product

Let be a matrix and $\bf{x}$ be a vector defined by

$A = \begin{bmatrix} -7 & 3 & 0 \\ -9 & 0 & 2 \\ 0 & 5 & -9 \end{bmatrix}$

$\bf{x} = \begin{bmatrix} -5 \\ 4 \\ 8 \\ 3 \\ 8\end{bmatrix}$

Find the product $A\bf{x}$ .

Possible Answers:

$A\bf{x} = \begin{bmatrix} 47 \\ 61 \\ -52\end{bmatrix}$

The product does not exist because the dimensions do not match.

$A\bf{x} = \begin{bmatrix} 47 & 61 & -52\end{bmatrix}$

$A\bf{x} = \begin{bmatrix} 35 \\ 0 \\ -72\end{bmatrix}$

$A\bf{x} = \begin{bmatrix} 42 \\ 9 \\ -7\end{bmatrix}$

Correct answer:

The product does not exist because the dimensions do not match.

Explanation:

The matrix has 3 columns and the vector $\bf {x}$ has 5 rows. The dimensions do not match and the product does not exist.

Report an Error

Example Question #123 : Matrices

Rewrite the system of equations:

x+2y+z=3

-x+5y+7z=10

2x+3y+6=2

into a matrix vector product:

Av=b

where is a 3x3 matrix and v,b are vectors in $\mathbb{R}^3$ .

Possible Answers:

$\begin{pmatrix} 1& 2 &3 \\ -1&5 &7 \\ 2&3 &6\end{pmatrix}\begin{pmatrix} x\\ y\\z \end{pmatrix}=\begin{pmatrix} 10\\ 3\\2 \end{pmatrix}$

$\begin{pmatrix} -1&5 &7 \\ 1& 2 &3 \\ 2&3 &6\end{pmatrix}\begin{pmatrix} x\\ y\\z \end{pmatrix}=\begin{pmatrix} 3\\ 10\\2 \end{pmatrix}$

$\begin{pmatrix} 1& 2 &3 \\ -1&5 &7 \\ 2&3 &6\end{pmatrix}\begin{pmatrix} x\\ y\\z \end{pmatrix}=\begin{pmatrix} 3\\ 10\\2 \end{pmatrix}$

$\begin{pmatrix} 1& -1 &2 \\ 2&5 &3 \\ 3&7 &6\end{pmatrix}\begin{pmatrix} x\\ y\\z \end{pmatrix}=\begin{pmatrix} 3\\ 10\\2 \end{pmatrix}$

Correct answer:

$\begin{pmatrix} 1& 2 &3 \\ -1&5 &7 \\ 2&3 &6\end{pmatrix}\begin{pmatrix} x\\ y\\z \end{pmatrix}=\begin{pmatrix} 3\\ 10\\2 \end{pmatrix}$

Explanation:

To write

x+2y+z=3

-x+5y+7z=10

2x+3y+6=2

into matrix vector form, we recall that matrix multiplication with a vector is done such that the first element in the resulting vector is the dot product of the first row of with the vector v=(x,y,z)^T , the second element is the dot product of the second row with , and so on. The first row is thus (1,2,3) , the second row is (-1,5,7) , and the third row is (2,3,6) . So the left side of the equality is

$\begin{pmatrix} 1& 2 &3 \\ -1&5 &7 \\ 2&3 &6\end{pmatrix}\begin{pmatrix} x\\ y\\z \end{pmatrix}$

The right side is the vector (3,10,2)^T , so the final answer is

$\begin{pmatrix} 1& 2 &3 \\ -1&5 &7 \\ 2&3 &6\end{pmatrix}\begin{pmatrix} x\\ y\\z \end{pmatrix}=\begin{pmatrix} 3\\ 10\\2 \end{pmatrix}$

which is equivalent to

x+2y+z=3

-x+5y+7z=10

2x+3y+6=2

Report an Error

Example Question #812 : Linear Algebra

Let $A = \begin{bmatrix} 2\sqrt{2} & \sqrt{3} \\ - \sqrt{2} & 2\sqrt{3} \end{bmatrix}$ and $B = \begin{bmatrix} \sqrt{2} \\ \sqrt{3} \end{bmatrix}$ .

Find .

Possible Answers:

$\begin{bmatrix}7 \\ 4 \end{bmatrix}$

$\begin{bmatrix} 4 & \sqrt{6} \\ - \sqrt{6} &6\end{bmatrix}$

is not defined.

$\begin{bmatrix} 4 - \sqrt{6} \\ 6+ \sqrt{6} \end{bmatrix}$

$\begin{bmatrix} 4 & 3 \\ -2 & 6 \end{bmatrix}$

Correct answer:

$\begin{bmatrix}7 \\ 4 \end{bmatrix}$

Explanation:

First, it must be established that is defined. This is the case if and only if has as many columns as has rows. Since has two columns and has two rows, is defined.

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,

$AB = \begin{bmatrix} 2\sqrt{2} & \sqrt{3} \\ - \sqrt{2} & 2\sqrt{3} \end{bmatrix} \begin{bmatrix} \sqrt{2} \\ \sqrt{3} \end{bmatrix}$

$= \begin{bmatrix} 2\sqrt{2} (\sqrt{2})+ \sqrt{3} (\sqrt{3} )\\ - \sqrt{2} (\sqrt{2})+ 2\sqrt{3}(\sqrt{3} ) \end{bmatrix}$

$= \begin{bmatrix}7 \\ 4 \end{bmatrix}$

Report an Error

Example Question #23 : Matrix Vector Product

Multiply $\begin{bmatrix} 5 &3 &-1 \\ 0&6 &3 \\1 &-2 &1 \end{bmatrix} \cdot \begin{bmatrix} 1\\5 \\-2 \end{bmatrix}$

Possible Answers:

$\begin{bmatrix} 7\\29 \\16 \end{bmatrix}$

$\begin{bmatrix} 22\\24 \\-11 \end{bmatrix}$

$\begin{bmatrix} 3\\ 37\\12 \end{bmatrix}$

$\begin{bmatrix} 18\\ 36\\ -8\end{bmatrix}$

Correct answer:

$\begin{bmatrix} 22\\24 \\-11 \end{bmatrix}$

Explanation:

To multiply, add:

$1 \cdot \begin{bmatrix} 5\\ 0\\ 1\end{bmatrix} + 5 \cdot \begin{bmatrix} 3\\ 6\\ -2\end{bmatrix} + -2 \cdot \begin{bmatrix} -1\\3 \\1 \end{bmatrix}$

$\begin{bmatrix} 5\\ 0\\ 1\end{bmatrix} + \begin{bmatrix} 15\\ 30\\ -10\end{bmatrix} + \begin{bmatrix} 2\\ -6\\ -2\end{bmatrix} = \begin{bmatrix} 22\\ 24\\ -11\end{bmatrix}$

Report an Error

Example Question #1 : Vector Vector Product

Compute $A\cdot B$ , where

$A=\begin{bmatrix} 2\\ 3\\ 4\\ \end{bmatrix}$

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

Possible Answers:

$A\cdot B =\begin{bmatrix} 2 & 6& 10\\ 3& 9& 15 \\ 4& 12& 20 \end{bmatrix}$

Not possible

$A\cdot B =\begin{bmatrix} 2 & 9& 10\\ 3& 6& 15 \\ 4& 20& 12 \end{bmatrix}$

Correct answer:

$A\cdot B =\begin{bmatrix} 2 & 6& 10\\ 3& 9& 15 \\ 4& 12& 20 \end{bmatrix}$

Explanation:

Before we compute the product of , and , we need to check if it is possible to take the product. We will check the dimensions. is $3\times1$ , and is $1\times3$ , so the dimensions of the resulting matrix will be $3\times3$ . Now let's compute it.

$A\cdot B=\begin{bmatrix} 2\cdot1 & 2\cdot3 & 2\cdot 5\\ 3 \cdot 1& 3\cdot3 & 3 \cdot 5 \\ 4 \cdot 1 & 4\cdot3 & 4\cdot 5 \end{bmatrix} =\begin{bmatrix} 2 & 6& 10\\ 3& 9& 15 \\ 4& 12& 20 \end{bmatrix}$

Report an Error

Example Question #2 : Vector Vector Product

Find the vector-vector product of the following vectors.

$A=[1 \ 2]$

$B=\begin{bmatrix} 3\\ 5 \end{bmatrix}$

Possible Answers:

It's not possible to multiply these vectors

Correct answer:

Explanation:

$A\cdot B=1\cdot3+2\cdot5=3+10=13$

Report an Error

View Linear Algebra Tutors

Megan
Certified Tutor

University of Wisconsin-Madison, Bachelor of Science, Chemical Engineering.

View Linear Algebra Tutors

Yoonsik
Certified Tutor

Seoul National University, Bachelor of Science, Physics. University of Pennsylvania, Doctor of Philosophy, Atomic and Molecul...

View Linear Algebra Tutors

Keivan
Certified Tutor

University of Calgary, Doctorate (PhD), Mathematical Finance.

All Linear Algebra Resources

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

Popular Subjects

GMAT Tutors in Washington DC, SAT Tutors in Phoenix, ACT Tutors in San Diego, Physics Tutors in Seattle, MCAT Tutors in Denver, Math Tutors in Denver, Biology Tutors in Chicago, SSAT Tutors in San Diego, GRE Tutors in Atlanta, Chemistry Tutors in Seattle

Popular Courses & Classes

LSAT Courses & Classes in New York City, GRE Courses & Classes in Miami, GMAT Courses & Classes in Miami, ISEE Courses & Classes in New York City, SAT Courses & Classes in Denver, MCAT Courses & Classes in Los Angeles, SSAT Courses & Classes in New York City, SSAT Courses & Classes in Dallas Fort Worth, SSAT Courses & Classes in Denver, GRE Courses & Classes in Seattle

Popular Test Prep

SAT Test Prep in Seattle, LSAT Test Prep in Los Angeles, GMAT Test Prep in Seattle, SSAT Test Prep in Houston, ACT Test Prep in San Francisco-Bay Area, ACT Test Prep in Seattle, ISEE Test Prep in Atlanta, LSAT Test Prep in Washington DC, LSAT Test Prep in Dallas Fort Worth, MCAT Test Prep in Philadelphia

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 #121 : Matrices

Example Question #13 : Matrix Vector Product

Example Question #809 : Linear Algebra

Example Question #21 : Matrix Vector Product

Example Question #22 : Matrix Vector Product

Example Question #123 : Matrices

Example Question #812 : Linear Algebra

Example Question #23 : Matrix Vector Product

Example Question #1 : Vector Vector Product

Example Question #2 : Vector Vector Product

All Linear Algebra Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: