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 #23 : Vector Vector Product

$\textbf{u} = (1, x, x^{2}, x^{3}, x^{4})$ .

$\textbf{u} \cdot \textbf{v}$ is equal to the fifth-degree Maclaurin series for $e^{2x}$ for:

Possible Answers:

$\textbf{v}=\left ( 1, 2, 2 , \frac{4}{3}, \frac{2}{3}, \frac{4}{15} \right )$

$\textbf{v}=\left ( 1, 2, 1, \frac{1}{3}, \frac{1}{12}, \frac{1}{60} \right )$

None of the other choices gives the correct response.

$\textbf{v}=\left ( \frac{1}{60} , \frac{1}{12} , \frac{1}{3}, 1, 2, 1\right )$

$\textbf{v}=\left ( \frac{4}{15} , \frac{2}{3} , \frac{4}{3} , 2, 2, 1 \right )$

Correct answer:

$\textbf{v}=\left ( 1, 2, 2 , \frac{4}{3}, \frac{2}{3}, \frac{4}{15} \right )$

Explanation:

The th-degree Maclaurin series for a function f(x) is the polynomial

$\frac{f(0)}{0!} + \frac{f'(0)}{1!}x+ \frac{f''(0)}{2!}x^{2}+ ... + \frac{f^{(n)}(0)}{n!}x^{n}$

If $\textbf{v} = (v_{0}, v_{1}, v_{2}, v_{3}, v_{4}, v_{5})$ ,

then

$\textbf{u} \cdot \textbf{v} = v_{0}+ v_{1}x+ v_{2}x^{2}+ ...+v_{5}x^{5}$ .

Therefore, we want $\textbf{v}$ to be the vector of Maclaurin coefficients by ascending order of degree.

The fifth-degree Maclaurin series for $e^{x}$ is

$e^{x}=1+x+ \frac{1}{2}x^{2}+ \frac{1}{6}x^{3}+ \frac{1}{24}x^{4}+ \frac{1}{120}x^{5}$

The Maclaurin series for $e^{2x}$ can be derived from this by replacing with :

$e^{2x}=1+2x+ \frac{1}{2}(2x)^{2}+ \frac{1}{6}(2x)^{3}+ \frac{1}{24}(2x)^{4}+ \frac{1}{120}(2x)^{5}$

$e^{2x}=1+2x+ \frac{1}{2}(4x^{2})+ \frac{1}{6}(8x^{3})+ \frac{1}{24}(16x^{4})+ \frac{1}{120}(32x^{5})$

$e^{2x}=1+2x+ 2 x^{2}+ \frac{4}{3}x^{3}+ \frac{2}{3}x^{4}+ \frac{4}{15} x^{5}$

Therefore,

$\textbf{v}=\left ( 1, 2, 2 , \frac{4}{3}, \frac{2}{3}, \frac{4}{15} \right )$

Report an Error

Example Question #22 : Vector Vector Product

$\textbf{u} = (-3, 2, 4 ,1 )$

$\textbf{v} = (-2, 0, 1, 4)$

Which of the following is undefined, $\textbf{u} \cdot \textbf{v}$ or $\textbf{u} \times \textbf{v}$ ?

Possible Answers:

Both

$\textbf{u} \cdot \textbf{v}$

$\textbf{u} \times \textbf{v}$

Neither

Correct answer:

$\textbf{u} \times \textbf{v}$

Explanation:

$\textbf{u} \cdot \textbf{v}$ , the dot product of the vectors, is a defined quantity if and only if both vectors are elements of the same vector space. Each has four entries, so both are in $\mathbb{R}^{4}$ . Consequently, $\textbf{u} \cdot \textbf{v}$ is defined.

$\textbf{u} \times \textbf{v}$ , the cross product of the vectors, is a defined vector if and only if both vectors are elements in $\mathbb{R}^{3}$ . As previously mentioned, they are in $\mathbb{R}^{4}$ , so $\textbf{u} \times \textbf{v}$ is undefined.

Report an Error

Example Question #23 : Vector Vector Product

$\textbf{u} = (3, 7, -2)$

$\textbf{v} = (1, -5 )$

Which of the following is undefined, $\textbf{u} \cdot \textbf{v}$ or $\textbf{u} \times \textbf{v}$ ?

Possible Answers:

$\textbf{u} \times \textbf{v}$

Both

Neither

$\textbf{u} \cdot \textbf{v}$

Correct answer:

Both

Explanation:

$\textbf{u} \cdot \textbf{v}$ , the dot product of the vectors, is a defined quantity if and only if both vectors are elements of the same vector space. $\textbf{u}$ has three entries, so $\textbf{u} \in \mathbb{R}^{3}$ ; $\textbf{v}$ has two entries, so $\textbf{v} \in \mathbb{R}^{2}$ . The two are in different vector spaces, so $\textbf{u} \cdot \textbf{v}$ is undefined.

$\textbf{u} \times \textbf{v}$ , the cross product of the vectors, is a defined vector if and only if both vectors are elements in $\mathbb{R}^{3}$ . As previously mentioned, $\textbf{v} \in \mathbb{R}^{2}$ , so $\textbf{u} \times \textbf{v}$ is undefined.

Report an Error

Example Question #831 : Linear Algebra

$\textbf{u} = (3, 7, -2)$

$\textbf{v} = (1, -5, 1 )$

Evaluate $\textbf{u} \times \textbf{v}$

Possible Answers:

(-3, 5, -22)

(-3, -5, -22)

(3, 5,22)

( 17, 5, 8)

(-17, 5, -8)

Correct answer:

(-3, -5, -22)

Explanation:

One way to determine the cross-product of two vectors is to set up a matrix with the first row $\textbf{i}, \textbf{j}, \textbf{k}$ , where these are the unit vectors (1,0,0), (0,1,0),(0,0,1) , respectively, and with the entries of the vectors as the other two rows:

$\begin{vmatrix} \textbf{i} & \textbf{j} & \textbf{k} \\ 3 & 7 & -2 \\ 1 & -5 & 1 \end{vmatrix}$

We can evaluate this as we would evaluate a determinant of a matrix with real entries. Take the products of the upper-left-to-lower-right diagonals, and subtract the products of the lower-left-to-upper-right diagonals:

Cross product

$\textbf{u} \times \textbf{v} = 7\textbf{i}+ ( - 2\textbf{j})+ (-15\textbf{k}) - 7\textbf{k}- 10\textbf{i} - 3 \textbf{j}$

$\textbf{u} \times \textbf{v} = -3\textbf{i}- 5\textbf{j}-22\textbf{k}$

$\textbf{u} \times \textbf{v} = -3(1,0,0 )- 5(0,1,0)-22 (0,0,1)$

$\textbf{u} \times \textbf{v} =(-3, -5, -22)$

Report an Error

Example Question #841 : Linear Algebra

$\textbf{u} = (\ln 2 , 1, -1 )$

$\textbf{v} = (2, 1, - 2)$

If $\textbf{u} \cdot \textbf{v} = \ln a$ , then evaluate .

Possible Answers:

a = 64

$a = 4e^{3}$

$a = \frac{4}{e}$

a = 4e

a = 12

Correct answer:

$a = 4e^{3}$

Explanation:

The dot product $\textbf{u} \cdot \textbf{v}$ is equal to the sum of the products of the numbers in corresponding positions, so

$\textbf{u} \cdot \textbf{v} = (\ln 2) \cdot 2 + 1 \cdot 1 + (-1 )(-2)$

$\textbf{u} \cdot \textbf{v} = 2\ln 2+ 1 + 2$

$\textbf{u} \cdot \textbf{v} = 3+ 2\ln 2$

Applying the properties of logarithms:

$\textbf{u} \cdot \textbf{v} = 3+ \ln 2^{2}$

$\textbf{u} \cdot \textbf{v} = 3+ \ln 4$

$\textbf{u} \cdot \textbf{v} = \ln e^{3}+ \ln 4$

$\textbf{u} \cdot \textbf{v} = \ln 4 e^{3}$

Therefore, $a = 4e^{3}$ .

Report an Error

Example Question #24 : Vector Vector Product

$\textbf{u} = (1, x, x^{2}, x ^{3}, x^{4}, x^{5})$

$\textbf{v} = (y^{5}, y^{3} , y, y^{4}, 1, y^{2})$

The expression $\textbf{u} \cdot \textbf{v}$ yields a polynomial of what degree?

Possible Answers:

None of the other choices gives a correct response.

Correct answer:

Explanation:

The dot product $\textbf{u} \cdot \textbf{v}$ is the sum of the products of entries in corresponding positions, so

$\textbf{u} \cdot \textbf{v} = 1 (y^{5}) + x (y^{3})+ x^{2}(y )+ x^{3}(y^{4})+ x^{4} (1)+ x^{5 }(y^{2})$

$\textbf{u} \cdot \textbf{v} = y^{5} + x y^{3}+ x^{2}y + x^{3}y^{4}+ x^{4} + x^{5 }y^{2}$

The degree of a term of a polynomial is the sum of the exponents of its variables; the individual terms have degrees 5, 4, 3, 7, 4, and 7, in that order. the degree of the polynomial is the highest of these, which is 7.

Report an Error

Example Question #152 : Matrices

A triangle has two sides of length and ; their included angle has measure $\theta$ . The measure of the third side can be obtained from the expression

$\sqrt{\textbf{u} \cdot \textbf{v}}$ ,

where $\textbf{u}= (a , b, 2ab)$ and:

Possible Answers:

$\textbf{v} = (b, a, -\cos \theta)$

$\textbf{v} = (a, b, -\sin \theta)$

$\textbf{v} = (b, a, -\sin \theta)$

$\textbf{v} = (a, b, \cos \theta)$

$\textbf{v} = (a, b, -\cos \theta)$

Correct answer:

$\textbf{v} = (a, b, -\cos \theta)$

Explanation:

Given the lengths and of two sides of a triangle, and the measure of their included angle, $\theta$ , the length of the third side of a triangle can be calculated using the Law of Cosines, which states that

$c^{2} = a^{2}+ b^{2} - 2ab\cos \theta$ .

The dot product $\textbf{u} \cdot \textbf{v}$ is equal to the sum of the products of their corresponding entries, and since $c= \sqrt{\textbf{u} \cdot \textbf{v}}$ , we can substitute $\textbf{u} \cdot \textbf{v}$ for $c^{2}$ :

$\textbf{u} \cdot \textbf{v} = a^{2}+ b^{2} - 2ab\cos \theta$

$\textbf{u} \cdot \textbf{v} = a \cdot a + b \cdot b + 2ab \cdot ( - \cos \theta)$

$\textbf{u} = (a , b , 2ab )$ ; it follows that $\textbf{v} = (a, b, -\cos \theta)$ .

Report an Error

Example Question #28 : Vector Vector Product

and are differentiable functions.

$A = \begin{bmatrix} f &g\end{bmatrix}$

$AB = \begin{bmatrix} \frac{\mathrm{d} }{\mathrm{d} x} \left ( \frac{f}{g} \right )\end{bmatrix}$

Which value of makes this statement true?

Possible Answers:

$B = \begin{bmatrix} - \frac{1}{g^{2}} \cdot \frac{\mathrm{d} g}{\mathrm{d} x} \\ \\ \frac{1}{g^{2}} \cdot \frac{\mathrm{d} f}{\mathrm{d} x} \end{bmatrix}$

$B = \begin{bmatrix} \frac{1}{g^{2}} \cdot \frac{\mathrm{d} g}{\mathrm{d} x} \\ \\- \frac{1}{g^{2}} \cdot \frac{\mathrm{d} f}{\mathrm{d} x} \end{bmatrix}$

$B = \begin{bmatrix} \frac{1}{g^{2}} \cdot \frac{\mathrm{d} g}{\mathrm{d} x} \\ \\ \frac{1}{g^{2}} \cdot \frac{\mathrm{d} f}{\mathrm{d} x} \end{bmatrix}$

$B = \begin{bmatrix} \frac{1}{g^{2}} \cdot \frac{\mathrm{d} f}{\mathrm{d} x} \\ \\ \frac{1}{g^{2}} \cdot \frac{\mathrm{d} g}{\mathrm{d} x} \end{bmatrix}$

$B = \begin{bmatrix} - \frac{1}{g^{2}} \cdot \frac{\mathrm{d} f}{\mathrm{d} x} \\ \\ \frac{1}{g^{2}} \cdot \frac{\mathrm{d} g}{\mathrm{d} x} \end{bmatrix}$

Correct answer:

$B = \begin{bmatrix} - \frac{1}{g^{2}} \cdot \frac{\mathrm{d} g}{\mathrm{d} x} \\ \\ \frac{1}{g^{2}} \cdot \frac{\mathrm{d} f}{\mathrm{d} x} \end{bmatrix}$

Explanation:

Recall the quotient rule of differentiation:

$\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{f}{g} \right ) =\frac{ \frac{\mathrm{d}f }{\mathrm{d} x} \cdot g-f \cdot \frac{\mathrm{d}g }{\mathrm{d} x} }{g^{2}}$

This can be rewritten as

$\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{f}{g} \right ) =\frac{-f \cdot \frac{\mathrm{d}g }{\mathrm{d} x}+ \frac{\mathrm{d}f }{\mathrm{d} x} \cdot g }{g^{2}}$

$\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{f}{g} \right ) =\frac{-f \cdot \frac{\mathrm{d}g }{\mathrm{d} x} }{g^{2}}+\frac{ \frac{\mathrm{d}f }{\mathrm{d} x} \cdot g }{g^{2}}$

$\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{f}{g} \right ) =f \left (- \frac{ \frac{\mathrm{d}g }{\mathrm{d} x} }{g^{2}} \right )+g\left ( \frac{ \frac{\mathrm{d}f }{\mathrm{d} x} }{g^{2}} \right )$

$\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{f}{g} \right ) =f \left (- \frac{1 }{g^{2}} \right ) \frac{\mathrm{d}g }{\mathrm{d} x} +g\left ( \frac{1}{g^{2}} \right ) \frac{\mathrm{d}f }{\mathrm{d} x}$

If $A = \begin{bmatrix} f &g\end{bmatrix}$ and $B = \begin{bmatrix} a \\ b \end{bmatrix}$ ,

then multiply corresponding elements and add the products to get the sole element in :

$AB = \begin{bmatrix} f \cdot a + g \cdot b \end{bmatrix}$

Since we want

$AB = \begin{bmatrix} \frac{\mathrm{d} }{\mathrm{d} x} \left ( \frac{f}{g} \right )\end{bmatrix}=\begin{bmatrix} f \left (- \frac{1 }{g^{2}} \right ) \frac{\mathrm{d}g }{\mathrm{d} x} +g\left ( \frac{1}{g^{2}} \right ) \frac{\mathrm{d}f }{\mathrm{d} x} \end{bmatrix}$ ,

It follows that, of the given choices, $a= \left (- \frac{1 }{g^{2}} \right ) \frac{\mathrm{d}g }{\mathrm{d} x}$ and $b= \left ( \frac{1}{g^{2}} \right ) \frac{\mathrm{d}f }{\mathrm{d} x}$ , and

$B = \begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} - \frac{1}{g^{2}} \cdot \frac{\mathrm{d} g}{\mathrm{d} x} \\ \\ \frac{1}{g^{2}} \cdot \frac{\mathrm{d} f}{\mathrm{d} x} \end{bmatrix}$ .

Report an Error

Example Question #31 : Vector Vector Product

$\textbf{u} = (- 3, 4, 1, 0)$

$\textbf{v} = (0, 1, -2, 4)$

Calculate the angle (nearest degree) between $\textbf{u}$ and $\textbf{v}$ .

Possible Answers:

$85^{\circ }$

$175^{\circ }$

$105^{\circ }$

$5^{\circ }$

The angle is undefined, since the vectors are in $\mathbb{R}^{4}$ .

Correct answer:

$85^{\circ }$

Explanation:

$\textbf{u} = (- 3, 4, 1, 0)$

$\textbf{v} = (0, 1, -2, 4)$

The angle $\; \theta$ between vectors $\textbf{u}$ and $\textbf{v}$ can be calculated using the formula

$\cos \theta = \frac{\textbf{u} \cdot \textbf{v}}{\left \| \textbf{u} \right \|\left \| \textbf{v} \right \|}$ .

$\textbf{u} \cdot \textbf{v}$ , the dot product, is the sum of the products of corresponding entries:

$\textbf{u} \cdot \textbf{v} = -3(0)+4(1)+1(-2)+0(4) = 0 + 4 + (-2)+ 0 = 2$

$\left \| \textbf{u} \right \|$ , the norm of $\textbf{u}$ , is the square root of the sum of the squares of its entries; $\left \| \textbf{v} \right \|$ is defined similarly:

$\left \| \textbf{u} \right \|= \sqrt{ (- 3)^{2}+ 4^{2} + 1^{2} +0^{2}} = \sqrt{9+16+1+0} = \sqrt{26}$

$\left \| \textbf{v} \right \|= \sqrt{ 0^{2}+ 1^{2} +(-2)^{2} +4^{2}} = \sqrt{0+1+4+16} = \sqrt{21}$

$\cos \theta = \frac{2}{ \sqrt{26} \cdot \sqrt{21}} \approx 0.0856$

$\theta \approx \cos ^{-1}0.0856 \approx 85^{\circ }$

Report an Error

Example Question #31 : Vector Vector Product

, , and give the length, width, and height of a rectangular prism.

$\textbf{u} = (L, W, H )$ and $\textbf{v} = (W, H, L )$ .

True or false: $\textbf{u} \cdot \textbf{v}$ gives the surface area of the prism.

Possible Answers:

True

False

Correct answer:

False

Explanation:

The dot product $\textbf{u} \cdot \textbf{v}$ can be calculated by adding the products of the elements in corresponding locations, so

$\textbf{u} \cdot \textbf{v} = LW + WH + LH$ .

The surface area of the prism, , can be found by using the formula:

$A = 2 (LW + WH + LH) = 2( \textbf{u} \cdot \textbf{v})$

Equivalently, $\textbf{u} \cdot \textbf{v}$ gives half the surface area of the prism. The statement is false.

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

ISEE Tutors in Denver, ACT Tutors in Chicago, Algebra Tutors in Denver, SAT Tutors in Washington DC, Physics Tutors in Atlanta, Math Tutors in San Francisco-Bay Area, Biology Tutors in Seattle, English Tutors in Houston, LSAT Tutors in Dallas Fort Worth, SSAT Tutors in San Diego

Popular Courses & Classes

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

Popular Test Prep

GMAT Test Prep in Boston, SAT Test Prep in Dallas Fort Worth, GRE Test Prep in Atlanta, GMAT Test Prep in Miami, SAT Test Prep in Atlanta, ACT Test Prep in San Francisco-Bay Area, LSAT Test Prep in San Francisco-Bay Area, ACT Test Prep in Atlanta, ISEE Test Prep in Miami, SAT Test Prep in Los Angeles

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 #23 : Vector Vector Product

Example Question #22 : Vector Vector Product

Example Question #23 : Vector Vector Product

Example Question #831 : Linear Algebra

Example Question #841 : Linear Algebra

Example Question #24 : Vector Vector Product

Example Question #152 : Matrices

Example Question #28 : Vector Vector Product

Example Question #31 : Vector Vector Product

Example Question #31 : 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: