Call Now to Set Up Tutoring:

(888) 888-0446

Linear Algebra : Linear Equations

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 : Linear Equations

$\begin{align*}&\text{Find nontrivial solutions of }\begin{bmatrix}x\\y\\z\end{bmatrix}\\&\text{for the function: }\begin{bmatrix}-14&0&3\\-13&-7&9\\\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\\end{bmatrix}\end{align*}$

$\begin{align*}&x=\frac{(3z)}{14}\\&y=-\frac{(87z)}{98}\end{align*}$

Possible Answers:

$\begin{align*}&x=-\frac{(3z)}{7}\\&y=-\frac{(261z)}{98}\end{align*}$

$\begin{align*}&x=-\frac{(3z)}{56}\\&y=-\frac{(29z)}{196}\end{align*}$

$\begin{align*}&x=-\frac{(12z)}{7}\\&y=-\frac{(348z)}{49}\end{align*}$

Correct answer:

$\begin{align*}&x=-\frac{(3z)}{7}\\&y=-\frac{(261z)}{98}\end{align*}$

Explanation:

$\begin{align*}&\text{When faced with a system of equations like the}\\&\text{one represented by }\\&\begin{bmatrix}-14&0&3\\-13&-7&9\\\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\\end{bmatrix}\\&\text{We can manipulate coefficients to eliminate}\\&\text{variables in each row. That is to say, to reduce it to echelon form.}\\&\text{Reducing our matrix }\begin{bmatrix}-14&0&3\\-13&-7&9\\\end{bmatrix}\\&\text{We find the reduced echelon form: }\begin{bmatrix}1&0&-\frac{3}{14}\\0&1&-\frac{87}{98}\\\end{bmatrix}\\&\text{Since each row is equal to zero, we can therefore relate variables to find}\\&\text{nontrivial solutions:}\\&x=\frac{(3z)}{14}\\&y=\frac{(87z)}{98}\end{align*}$

Report an Error

Example Question #82 : Linear Equations

$\begin{align*}&\text{Find the nontrivial solutions for the system of equations: }\\&\begin{bmatrix}-2&13&12\\-9&-18&-18\\\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\\end{bmatrix}\end{align*}$

Possible Answers:

$\begin{align*}&x=\frac{(10z)}{17}\\&y=\frac{(8z)}{51}\end{align*}$

$\begin{align*}&x=\frac{(2z)}{119}\\&y=\frac{(112z)}{17}\end{align*}$

$\begin{align*}&x=\frac{(6z)}{17}\\&y=\frac{(8z)}{17}\end{align*}$

$\begin{align*}&x=-\frac{(2z)}{17}\\&y=\frac{(16z)}{17}\end{align*}$

Correct answer:

$\begin{align*}&x=-\frac{(2z)}{17}\\&y=\frac{(16z)}{17}\end{align*}$

Explanation:

$\begin{align*}&\text{When faced with a system of equations like the}\\&\text{one represented by }\\&\begin{bmatrix}-2&13&12\\-9&-18&-18\\\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\\end{bmatrix}\\&\text{We can manipulate coefficients to eliminate}\\&\text{variables in each row. That is to say, to reduce it to echelon form.}\\&\text{Reducing our matrix }\begin{bmatrix}-2&13&12\\-9&-18&-18\\\end{bmatrix}\\&\text{We find the reduced echelon form: }\begin{bmatrix}1&0&\frac{2}{17}\\0&1&\frac{16}{17}\\\end{bmatrix}\\&\text{Since each row is equal to zero, we can therefore relate variables to find}\\&\text{nontrivial solutions:}\\&x=-\frac{(2z)}{17}\\&y=-\frac{(16z)}{17}\end{align*}$

Report an Error

Example Question #83 : Linear Equations

$\begin{align*}&\text{Find nonzero solutions for }\begin{bmatrix}x\\y\\z\end{bmatrix}\\&\text{for the system: }\begin{bmatrix}-7&-2&-11\\-19&6&-14\\\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\\end{bmatrix}\end{align*}$

Possible Answers:

$\begin{align*}&x=\frac{(47z)}{80}\\&y=\frac{(333z)}{80}\end{align*}$

$\begin{align*}&x=\frac{(47z)}{200}\\&y=\frac{(111z)}{16}\end{align*}$

$\begin{align*}&x=\frac{(47z)}{280}\\&y=\frac{(111z)}{560}\end{align*}$

$\begin{align*}&x=-\frac{(47z)}{40}\\&y=\frac{(111z)}{80}\end{align*}$

Correct answer:

$\begin{align*}&x=-\frac{(47z)}{40}\\&y=\frac{(111z)}{80}\end{align*}$

Explanation:

$\begin{align*}&\text{When faced with a system of equations like the}\\&\text{one represented by }\\&\begin{bmatrix}-7&-2&-11\\-19&6&-14\\\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\\end{bmatrix}\\&\text{We can manipulate coefficients to eliminate}\\&\text{variables in each row. That is to say, to reduce it to echelon form.}\\&\text{Reducing our matrix }\begin{bmatrix}-7&-2&-11\\-19&6&-14\\\end{bmatrix}\\&\text{We find the reduced echelon form: }\begin{bmatrix}1&0&\frac{47}{40}\\0&1&\frac{111}{80}\\\end{bmatrix}\\&\text{Since each row is equal to zero, we can therefore relate variables to find}\\&\text{nontrivial solutions:}\\&x=-\frac{(47z)}{40}\\&y=-\frac{(111z)}{80}\end{align*}$

Report an Error

Example Question #84 : Linear Equations

$\begin{align*}&\text{Find the nontrivial solutions for the system of equations: }\\&\begin{bmatrix}-1&16&-18\\0&5&10\\\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\\end{bmatrix}\end{align*}$

Possible Answers:

$\begin{align*}&x=200z\\&y=\frac{(2z)}{5}\end{align*}$

$\begin{align*}&x=-50z\\&y=2z\end{align*}$

$\begin{align*}&x=25z\\&y=6z\end{align*}$

$\begin{align*}&x=350z\\&y=\frac{(2z)}{7}\end{align*}$

Correct answer:

$\begin{align*}&x=-50z\\&y=2z\end{align*}$

Explanation:

$\begin{align*}&\text{When faced with a system of equations like the}\\&\text{one represented by }\\&\begin{bmatrix}-1&16&-18\\0&5&10\\\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\\end{bmatrix}\\&\text{We can manipulate coefficients to eliminate}\\&\text{variables in each row. That is to say, to reduce it to echelon form.}\\&\text{Reducing our matrix }\begin{bmatrix}-1&16&-18\\0&5&10\\\end{bmatrix}\\&\text{We find the reduced echelon form: }\begin{bmatrix}1&0&50\\0&1&2\\\end{bmatrix}\\&\text{Since each row is equal to zero, we can therefore relate variables to find}\\&\text{nontrivial solutions:}\\&x=-50z\\&y=-2z\end{align*}$

Report an Error

Example Question #85 : Linear Equations

$\begin{align*}&\text{Find the nontrivial solutions for the system of equations: }\\&\begin{bmatrix}8&-13&12\\1&2&-12\\\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\\end{bmatrix}\end{align*}$

Possible Answers:

$\begin{align*}&x=-\frac{(132z)}{145}\\&y=-\frac{(27z)}{29}\end{align*}$

$\begin{align*}&x=\frac{(132z)}{29}\\&y=-\frac{(108z)}{29}\end{align*}$

$\begin{align*}&x=-\frac{(44z)}{87}\\&y=-\frac{(864z)}{29}\end{align*}$

$\begin{align*}&x=-\frac{(396z)}{29}\\&y=-\frac{(324z)}{29}\end{align*}$

Correct answer:

$\begin{align*}&x=\frac{(132z)}{29}\\&y=-\frac{(108z)}{29}\end{align*}$

Explanation:

$\begin{align*}&\text{When faced with a system of equations like the}\\&\text{one represented by }\\&\begin{bmatrix}8&-13&12\\1&2&-12\\\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\\end{bmatrix}\\&\text{We can manipulate coefficients to eliminate}\\&\text{variables in each row. That is to say, to reduce it to echelon form.}\\&\text{Reducing our matrix }\begin{bmatrix}8&-13&12\\1&2&-12\\\end{bmatrix}\\&\text{We find the reduced echelon form: }\begin{bmatrix}1&0&-\frac{132}{29}\\0&1&-\frac{108}{29}\\\end{bmatrix}\\&\text{Since each row is equal to zero, we can therefore relate variables to find}\\&\text{nontrivial solutions:}\\&x=\frac{(132z)}{29}\\&y=\frac{(108z)}{29}\end{align*}$

Report an Error

Example Question #86 : Linear Equations

$\begin{align*}&\text{Find nonzero solutions for }\begin{bmatrix}x\\y\\z\end{bmatrix}\\&\text{for the system: }\begin{bmatrix}-19&-13&18\\18&-2&13\\\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\\end{bmatrix}\end{align*}$

Possible Answers:

$\begin{align*}&x=-\frac{(133z)}{272}\\&y=-\frac{(571z)}{272}\end{align*}$

$\begin{align*}&x=\frac{(133z)}{34}\\&y=-\frac{(571z)}{2176}\end{align*}$

$\begin{align*}&x=\frac{(665z)}{272}\\&y=-\frac{(571z)}{1632}\end{align*}$

$\begin{align*}&x=\frac{(133z)}{136}\\&y=-\frac{(571z)}{544}\end{align*}$

Correct answer:

$\begin{align*}&x=-\frac{(133z)}{272}\\&y=-\frac{(571z)}{272}\end{align*}$

Explanation:

$\begin{align*}&\text{When faced with a system of equations like the}\\&\text{one represented by }\\&\begin{bmatrix}-19&-13&18\\18&-2&13\\\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\\end{bmatrix}\\&\text{We can manipulate coefficients to eliminate}\\&\text{variables in each row. That is to say, to reduce it to echelon form.}\\&\text{Reducing our matrix }\begin{bmatrix}-19&-13&18\\18&-2&13\\\end{bmatrix}\\&\text{We find the reduced echelon form: }\begin{bmatrix}1&0&\frac{133}{272}\\0&1&-\frac{571}{272}\\\end{bmatrix}\\&\text{Since each row is equal to zero, we can therefore relate variables to find}\\&\text{nontrivial solutions:}\\&x=-\frac{(133z)}{272}\\&y=\frac{(571z)}{272}\end{align*}$

Report an Error

Example Question #87 : Linear Equations

$\begin{align*}&\text{Find the nontrivial solutions for the system of equations: }\\&\begin{bmatrix}8&15&-1\\-2&13&13\\\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\\end{bmatrix}\end{align*}$

Possible Answers:

$\begin{align*}&x=-\frac{(208z)}{67}\\&y=\frac{(51z)}{134}\end{align*}$

$\begin{align*}&x=-\frac{(936z)}{67}\\&y=\frac{(17z)}{201}\end{align*}$

$\begin{align*}&x=-\frac{(624z)}{67}\\&y=\frac{(204z)}{67}\end{align*}$

$\begin{align*}&x=\frac{(104z)}{67}\\&y=\frac{(51z)}{67}\end{align*}$

Correct answer:

$\begin{align*}&x=\frac{(104z)}{67}\\&y=\frac{(51z)}{67}\end{align*}$

Explanation:

$\begin{align*}&\text{When faced with a system of equations like the}\\&\text{one represented by }\\&\begin{bmatrix}8&15&-1\\-2&13&13\\\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\\end{bmatrix}\\&\text{We can manipulate coefficients to eliminate}\\&\text{variables in each row. That is to say, to reduce it to echelon form.}\\&\text{Reducing our matrix }\begin{bmatrix}8&15&-1\\-2&13&13\\\end{bmatrix}\\&\text{We find the reduced echelon form: }\begin{bmatrix}1&0&-\frac{104}{67}\\0&1&\frac{51}{67}\\\end{bmatrix}\\&\text{Since each row is equal to zero, we can therefore relate variables to find}\\&\text{nontrivial solutions:}\\&x=\frac{(104z)}{67}\\&y=-\frac{(51z)}{67}\end{align*}$

Report an Error

Example Question #88 : Linear Equations

$\begin{align*}&\text{Find nontrivial solutions of }\begin{bmatrix}x\\y\\z\end{bmatrix}\\&\text{for the function: }\begin{bmatrix}11&16&-15\\12&-13&-19\\\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\\end{bmatrix}\end{align*}$

Possible Answers:

$\begin{align*}&x=\frac{(499z)}{335}\\&y=\frac{(29z)}{335}\end{align*}$

$\begin{align*}&x=-\frac{(499z)}{67}\\&y=\frac{(29z)}{1340}\end{align*}$

$\begin{align*}&x=-\frac{(499z)}{670}\\&y=\frac{(29z)}{670}\end{align*}$

$\begin{align*}&x=-\frac{(499z)}{2345}\\&y=\frac{(29z)}{3015}\end{align*}$

Correct answer:

$\begin{align*}&x=\frac{(499z)}{335}\\&y=\frac{(29z)}{335}\end{align*}$

Explanation:

$\begin{align*}&\text{When faced with a system of equations like the}\\&\text{one represented by }\\&\begin{bmatrix}11&16&-15\\12&-13&-19\\\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\\end{bmatrix}\\&\text{We can manipulate coefficients to eliminate}\\&\text{variables in each row. That is to say, to reduce it to echelon form.}\\&\text{Reducing our matrix }\begin{bmatrix}11&16&-15\\12&-13&-19\\\end{bmatrix}\\&\text{We find the reduced echelon form: }\begin{bmatrix}1&0&-\frac{499}{335}\\0&1&\frac{29}{335}\\\end{bmatrix}\\&\text{Since each row is equal to zero, we can therefore relate variables to find}\\&\text{nontrivial solutions:}\\&x=\frac{(499z)}{335}\\&y=-\frac{(29z)}{335}\end{align*}$

Report an Error

Example Question #89 : Linear Equations

$\begin{align*}&\text{Find the nontrivial solutions for the system of equations: }\\&\begin{bmatrix}-6&-16&2\\-3&1&16\\\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\\end{bmatrix}\end{align*}$

Possible Answers:

$\begin{align*}&x=-\frac{(215z)}{9}\\&y=\frac{(25z)}{3}\end{align*}$

$\begin{align*}&x=\frac{(43z)}{9}\\&y=\frac{(5z)}{3}\end{align*}$

$\begin{align*}&x=-\frac{(86z)}{9}\\&y=5z\end{align*}$

$\begin{align*}&x=-\frac{(344z)}{9}\\&y=\frac{(5z)}{24}\end{align*}$

Correct answer:

$\begin{align*}&x=\frac{(43z)}{9}\\&y=\frac{(5z)}{3}\end{align*}$

Explanation:

$\begin{align*}&\text{When faced with a system of equations like the}\\&\text{one represented by }\\&\begin{bmatrix}-6&-16&2\\-3&1&16\\\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\\end{bmatrix}\\&\text{We can manipulate coefficients to eliminate}\\&\text{variables in each row. That is to say, to reduce it to echelon form.}\\&\text{Reducing our matrix }\begin{bmatrix}-6&-16&2\\-3&1&16\\\end{bmatrix}\\&\text{We find the reduced echelon form: }\begin{bmatrix}1&0&-\frac{43}{9}\\0&1&\frac{5}{3}\\\end{bmatrix}\\&\text{Since each row is equal to zero, we can therefore relate variables to find}\\&\text{nontrivial solutions:}\\&x=\frac{(43z)}{9}\\&y=-\frac{(5z)}{3}\end{align*}$

Report an Error

Example Question #90 : Linear Equations

$\begin{align*}&\text{Find nonzero solutions for }\begin{bmatrix}x\\y\\z\end{bmatrix}\\&\text{for the system: }\begin{bmatrix}-5&-1&13\\-8&13&3\\\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\\end{bmatrix}\end{align*}$

Possible Answers:

$\begin{align*}&x=-\frac{(43z)}{73}\\&y=-\frac{(445z)}{73}\end{align*}$

$\begin{align*}&x=-\frac{(516z)}{73}\\&y=-\frac{(178z)}{73}\end{align*}$

$\begin{align*}&x=\frac{(172z)}{73}\\&y=-\frac{(89z)}{73}\end{align*}$

$\begin{align*}&x=-\frac{(1204z)}{73}\\&y=-\frac{(712z)}{73}\end{align*}$

Correct answer:

$\begin{align*}&x=\frac{(172z)}{73}\\&y=-\frac{(89z)}{73}\end{align*}$

Explanation:

$\begin{align*}&\text{When faced with a system of equations like the}\\&\text{one represented by }\\&\begin{bmatrix}-5&-1&13\\-8&13&3\\\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\\end{bmatrix}\\&\text{We can manipulate coefficients to eliminate}\\&\text{variables in each row. That is to say, to reduce it to echelon form.}\\&\text{Reducing our matrix }\begin{bmatrix}-5&-1&13\\-8&13&3\\\end{bmatrix}\\&\text{We find the reduced echelon form: }\begin{bmatrix}1&0&-\frac{172}{73}\\0&1&-\frac{89}{73}\\\end{bmatrix}\\&\text{Since each row is equal to zero, we can therefore relate variables to find}\\&\text{nontrivial solutions:}\\&x=\frac{(172z)}{73}\\&y=\frac{(89z)}{73}\end{align*}$

Report an Error

View Linear Algebra Tutors

Godswill
Certified Tutor

Rivers State University, Bachelor of Engineering, Mechanical Engineering. Texas Southern University, Master of Science, Compu...

View Linear Algebra Tutors

Ehab
Certified Tutor

Military Technical College, Bachelor of Engineering, Telecommunications Engineering. Military Technical College, Master of Sc...

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

MCAT Tutors in Phoenix, MCAT Tutors in Seattle, SAT Tutors in Denver, ISEE Tutors in San Diego, LSAT Tutors in San Diego, Spanish Tutors in San Francisco-Bay Area, Computer Science Tutors in Denver, Spanish Tutors in Atlanta, GMAT Tutors in Dallas Fort Worth, SSAT Tutors in Chicago

Popular Courses & Classes

GMAT Courses & Classes in Chicago, ISEE Courses & Classes in Boston, LSAT Courses & Classes in San Diego, MCAT Courses & Classes in Houston, GMAT Courses & Classes in Atlanta, SSAT Courses & Classes in Dallas Fort Worth, SAT Courses & Classes in New York City, ISEE Courses & Classes in Denver, ISEE Courses & Classes in Phoenix, MCAT Courses & Classes in New York City

Popular Test Prep

GRE Test Prep in San Francisco-Bay Area, SAT Test Prep in Houston, ACT Test Prep in Phoenix, MCAT Test Prep in San Francisco-Bay Area, ISEE Test Prep in Miami, SAT Test Prep in Miami, GRE Test Prep in Houston, LSAT Test Prep in Atlanta, GRE Test Prep in Miami, LSAT 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 : Linear Equations

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

Example Question #81 : Linear Equations

Example Question #82 : Linear Equations

Example Question #83 : Linear Equations

Example Question #84 : Linear Equations

Example Question #85 : Linear Equations

Example Question #86 : Linear Equations

Example Question #87 : Linear Equations

Example Question #88 : Linear Equations

Example Question #89 : Linear Equations

Example Question #90 : Linear Equations

All Linear Algebra Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: