Call Now to Set Up Tutoring:

(888) 888-0446

Linear Algebra : Least Squares

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 #61 : Linear Algebra

Let $y=(1 \ 2 \ 3 \ 4 \ 5 \10 )$ , and $x=(0 \ 1 \ 2 \3 \4 \5)$ , find the least squares solution for a linear line.

Possible Answers:

$x=\begin{bmatrix} -\frac{5}{21}\\ \\ \frac{11}{7} \end{bmatrix}$

$x=\begin{bmatrix} -\frac{7}{21}\\ \\ \frac{1}{7} \end{bmatrix}$

No solutions exist

$x=\begin{bmatrix} \frac{5}{21}\\ \\ \frac{11}{7} \end{bmatrix}$

$x=\begin{bmatrix} 0\\ \\ 11 \end{bmatrix}$

Correct answer:

$x=\begin{bmatrix} \frac{5}{21}\\ \\ \frac{11}{7} \end{bmatrix}$

Explanation:

The equation for least squares solution for a linear fit looks as follows.

$y=\theta_1+\theta_2x_i$

Recall the formula for method of least squares.

Ax=b

$x=(A^T\cdot A)^{-1}A^T\cdot b$

Remember when setting up the A matrix, that we have to fill one column full of ones.

$A=\begin{bmatrix} 1 & 0 \\ 1 & 1\\ 1 & 2\\ 1 & 3\\ 1& 4\\ 1& 5 \end{bmatrix}$

$B=\begin{bmatrix} 1 \\ 2\\ 3\\ 4\\ 5\\10 \end{bmatrix}$

To make things simpler, lets make $C=A^T\cdot A$ , and $D=A^T\cdot b$

$C=\begin{bmatrix} 1 & 1&1 &1 &1 &1 \\ 0 & 1& 2& 3& 4& 5 \end{bmatrix}.\begin{bmatrix} 1&0 \\ 1& 1\\ 1& 2\\ 1& 3\\ 1& 4\\ 1& 5 \end{bmatrix}$

$C=\begin{bmatrix} 6 & 15 \\ 15 & 55 \end{bmatrix}$

Now we need to solve for the inverse, we can do this simply by doing the following. We flip the sign on the off diagonal, and change the spots on the main diagonal, then we multiply by $\frac{1}{\det(C)}$ .

$\det(C)=55(6)-(15)(15)=105$

$C^{-1}=\frac{1}{105}\begin{bmatrix}55 & -15 \\ -15 & 6 \end{bmatrix}$

$D=\begin{bmatrix} 1&1 &1 &1 &1 & 1\\ 0&1 &2 &3 &4 &5 \end{bmatrix} . \begin{bmatrix} 1\\ 2\\ 3\\ 4\\ 5\\ 10\\ \end{bmatrix}$

$D=\begin{bmatrix} 25\\ 90 \end{bmatrix}$

$x=C^{-1}\cdot D=\frac{1}{105}\begin{bmatrix} 55& -15\\ -15& 6 \end{bmatrix} \cdot \begin{bmatrix} 25\\ 90 \end{bmatrix}=\begin{bmatrix} \frac{5}{21}\\ \\ \frac{11}{7} \end{bmatrix}$

Report an Error

Example Question #2 : Least Squares

$\begin{align*}&\text{Approximate }\begin{bmatrix}x\\y\end{bmatrix}\\&\text{for the system: }\begin{bmatrix}-12&9\\13&14\\-5&-16\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-100\\-5\\14\end{bmatrix}\end{align*}$

Possible Answers:

$\begin{bmatrix}12.80\\4.37\end{bmatrix}$

$\begin{bmatrix}0.80\\-7.63\end{bmatrix}$

$\begin{bmatrix}4.80\\-3.63\end{bmatrix}$

$\begin{bmatrix}3.80\\-4.63\end{bmatrix}$

Correct answer:

$\begin{bmatrix}4.80\\-3.63\end{bmatrix}$

Explanation:

$\begin{align*}&\text{When the rows of a matrix outnumber the columns in an equation,}\\&\text{there are a greater number of equations than variables. In order}\\&\text{to approximate the values of these variables, a least-squares}\\&\text{approach is appropriate. This is done by multiplying each side}\\&\text{of the equation by the transpose of the equation matrix, and}\\&\text{solving the resulting equation:}\\&Ax=B\\&A^{T}Ax=A^{T}B\\&(A^{T}A)^{-1}(A^{T}A)x=(A^{T}A)^{-1}A^{T}B\\&x=(A^{T}A)^{-1}A^{T}B\\&\text{Using this, we can approximate our values:}\\&\begin{bmatrix}-12&13&-5\\9&14&-16\end{bmatrix}\begin{bmatrix}-12&9\\13&14\\-5&-16\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-12&13&-5\\9&14&-16\end{bmatrix}\begin{bmatrix}-100\\-5\\14\end{bmatrix}\\&\begin{bmatrix}338&154\\154&533\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}1065\\-1194\end{bmatrix}\\&\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}4.80\\-3.63\end{bmatrix}\end{align*}$

Report an Error

Example Question #1 : Least Squares

$\begin{align*}&\text{Approximate the values of }\begin{bmatrix}x\\y\end{bmatrix}\\&\text{for the function: }\begin{bmatrix}15&15\\-7&-12\\-3&9\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-120\\-4\\-180\end{bmatrix}\end{align*}$

Possible Answers:

$\begin{bmatrix}3.29\\-17.69\end{bmatrix}$

$\begin{bmatrix}10.29\\-10.69\end{bmatrix}$

$\begin{bmatrix}16.29\\-4.69\end{bmatrix}$

$\begin{bmatrix}8.29\\-12.69\end{bmatrix}$

Correct answer:

$\begin{bmatrix}8.29\\-12.69\end{bmatrix}$

Explanation:

$\begin{align*}&\text{When the rows of a matrix outnumber the columns in an equation,}\\&\text{there are a greater number of equations than variables. In order}\\&\text{to approximate the values of these variables, a least-squares}\\&\text{approach is appropriate. This is done by multiplying each side}\\&\text{of the equation by the transpose of the equation matrix, and}\\&\text{solving the resulting equation:}\\&Ax=B\\&A^{T}Ax=A^{T}B\\&(A^{T}A)^{-1}(A^{T}A)x=(A^{T}A)^{-1}A^{T}B\\&x=(A^{T}A)^{-1}A^{T}B\\&\text{Using this, we can approximate our values:}\\&\begin{bmatrix}15&-7&-3\\15&-12&9\end{bmatrix}\begin{bmatrix}15&15\\-7&-12\\-3&9\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}15&-7&-3\\15&-12&9\end{bmatrix}\begin{bmatrix}-120\\-4\\-180\end{bmatrix}\\&\begin{bmatrix}283&282\\282&450\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-1232\\-3372\end{bmatrix}\\&\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}8.29\\-12.69\end{bmatrix}\end{align*}$

Report an Error

Example Question #4 : Least Squares

$\begin{align*}&\text{Approximate }\begin{bmatrix}x\\y\end{bmatrix}\\&\text{for the system: }\begin{bmatrix}-13&13\\9&9\\11&-11\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-109\\-168\\46\end{bmatrix}\end{align*}$

Possible Answers:

$\begin{bmatrix}-12.02\\-18.65\end{bmatrix}$

$\begin{bmatrix}-6.02\\-12.65\end{bmatrix}$

$\begin{bmatrix}-15.02\\-21.65\end{bmatrix}$

$\begin{bmatrix}-3.02\\-9.65\end{bmatrix}$

Correct answer:

$\begin{bmatrix}-6.02\\-12.65\end{bmatrix}$

Explanation:

$\begin{align*}&\text{When the rows of a matrix outnumber the columns in an equation,}\\&\text{there are a greater number of equations than variables. In order}\\&\text{to approximate the values of these variables, a least-squares}\\&\text{approach is appropriate. This is done by multiplying each side}\\&\text{of the equation by the transpose of the equation matrix, and}\\&\text{solving the resulting equation:}\\&Ax=B\\&A^{T}Ax=A^{T}B\\&(A^{T}A)^{-1}(A^{T}A)x=(A^{T}A)^{-1}A^{T}B\\&x=(A^{T}A)^{-1}A^{T}B\\&\text{Using this, we can approximate our values:}\\&\begin{bmatrix}-13&9&11\\13&9&-11\end{bmatrix}\begin{bmatrix}-13&13\\9&9\\11&-11\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-13&9&11\\13&9&-11\end{bmatrix}\begin{bmatrix}-109\\-168\\46\end{bmatrix}\\&\begin{bmatrix}371&-209\\-209&371\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}411\\-3435\end{bmatrix}\\&\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-6.02\\-12.65\end{bmatrix}\end{align*}$

Report an Error

Example Question #1 : Least Squares

$\begin{align*}&\text{Approximate the values of }\begin{bmatrix}x\\y\end{bmatrix}\\&\text{for the function: }\begin{bmatrix}20&-19\\-9&-18\\14&-10\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-8\\11\\37\end{bmatrix}\end{align*}$

Possible Answers:

$\begin{bmatrix}0.13\\-0.47\end{bmatrix}$

$\begin{bmatrix}5.13\\4.53\end{bmatrix}$

$\begin{bmatrix}-7.87\\-8.47\end{bmatrix}$

$\begin{bmatrix}-0.87\\-1.47\end{bmatrix}$

Correct answer:

$\begin{bmatrix}0.13\\-0.47\end{bmatrix}$

Explanation:

$\begin{align*}&\text{When the rows of a matrix outnumber the columns in an equation,}\\&\text{there are a greater number of equations than variables. In order}\\&\text{to approximate the values of these variables, a least-squares}\\&\text{approach is appropriate. This is done by multiplying each side}\\&\text{of the equation by the transpose of the equation matrix, and}\\&\text{solving the resulting equation:}\\&Ax=B\\&A^{T}Ax=A^{T}B\\&(A^{T}A)^{-1}(A^{T}A)x=(A^{T}A)^{-1}A^{T}B\\&x=(A^{T}A)^{-1}A^{T}B\\&\text{Using this, we can approximate our values:}\\&\begin{bmatrix}20&-9&14\\-19&-18&-10\end{bmatrix}\begin{bmatrix}20&-19\\-9&-18\\14&-10\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}20&-9&14\\-19&-18&-10\end{bmatrix}\begin{bmatrix}-8\\11\\37\end{bmatrix}\\&\begin{bmatrix}677&-358\\-358&785\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}259\\-416\end{bmatrix}\\&\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}0.13\\-0.47\end{bmatrix}\end{align*}$

Report an Error

Example Question #3 : Least Squares

$\begin{align*}&\text{Approximate }\begin{bmatrix}x\\y\end{bmatrix}\\&\text{for the system: }\begin{bmatrix}14&-16\\5&-18\\5&-14\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-92\\-165\\-168\end{bmatrix}\end{align*}$

Possible Answers:

$\begin{bmatrix}10.31\\15.37\end{bmatrix}$

$\begin{bmatrix}-1.69\\3.37\end{bmatrix}$

$\begin{bmatrix}7.31\\12.37\end{bmatrix}$

$\begin{bmatrix}12.31\\17.37\end{bmatrix}$

Correct answer:

$\begin{bmatrix}7.31\\12.37\end{bmatrix}$

Explanation:

$\begin{align*}&\text{When the rows of a matrix outnumber the columns in an equation,}\\&\text{there are a greater number of equations than variables. In order}\\&\text{to approximate the values of these variables, a least-squares}\\&\text{approach is appropriate. This is done by multiplying each side}\\&\text{of the equation by the transpose of the equation matrix, and}\\&\text{solving the resulting equation:}\\&Ax=B\\&A^{T}Ax=A^{T}B\\&(A^{T}A)^{-1}(A^{T}A)x=(A^{T}A)^{-1}A^{T}B\\&x=(A^{T}A)^{-1}A^{T}B\\&\text{Using this, we can approximate our values:}\\&\begin{bmatrix}14&5&5\\-16&-18&-14\end{bmatrix}\begin{bmatrix}14&-16\\5&-18\\5&-14\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}14&5&5\\-16&-18&-14\end{bmatrix}\begin{bmatrix}-92\\-165\\-168\end{bmatrix}\\&\begin{bmatrix}246&-384\\-384&776\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-2953\\6794\end{bmatrix}\\&\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}7.31\\12.37\end{bmatrix}\end{align*}$

Report an Error

Example Question #7 : Least Squares

$\begin{align*}&\text{Using the method of least-squares, approximate x and y values for: }\\&\begin{bmatrix}0&16\\-2&-18\\3&5\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-73\\158\\-35\end{bmatrix}\end{align*}$

Possible Answers:

$\begin{bmatrix}0.18\\1.74\end{bmatrix}$

$\begin{bmatrix}-4.82\\-3.26\end{bmatrix}$

$\begin{bmatrix}-7.82\\-6.26\end{bmatrix}$

$\begin{bmatrix}-1.82\\-0.26\end{bmatrix}$

Correct answer:

$\begin{bmatrix}-7.82\\-6.26\end{bmatrix}$

Explanation:

$\begin{align*}&\text{When the rows of a matrix outnumber the columns in an equation,}\\&\text{there are a greater number of equations than variables. In order}\\&\text{to approximate the values of these variables, a least-squares}\\&\text{approach is appropriate. This is done by multiplying each side}\\&\text{of the equation by the transpose of the equation matrix, and}\\&\text{solving the resulting equation:}\\&Ax=B\\&A^{T}Ax=A^{T}B\\&(A^{T}A)^{-1}(A^{T}A)x=(A^{T}A)^{-1}A^{T}B\\&x=(A^{T}A)^{-1}A^{T}B\\&\text{Using this, we can approximate our values:}\\&\begin{bmatrix}0&-2&3\\16&-18&5\end{bmatrix}\begin{bmatrix}0&16\\-2&-18\\3&5\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}0&-2&3\\16&-18&5\end{bmatrix}\begin{bmatrix}-73\\158\\-35\end{bmatrix}\\&\begin{bmatrix}13&51\\51&605\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-421\\-4187\end{bmatrix}\\&\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-7.82\\-6.26\end{bmatrix}\end{align*}$

Report an Error

Example Question #8 : Least Squares

$\begin{align*}&\text{Using the method of least-squares, approximate x and y values for: }\\&\begin{bmatrix}1&-17\\-6&1\\1&-13\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}188\\149\\19\end{bmatrix}\end{align*}$

Possible Answers:

$\begin{bmatrix}-26.87\\-9.28\end{bmatrix}$

$\begin{bmatrix}-35.87\\-18.28\end{bmatrix}$

$\begin{bmatrix}-32.87\\-15.28\end{bmatrix}$

$\begin{bmatrix}-28.87\\-11.28\end{bmatrix}$

Correct answer:

$\begin{bmatrix}-26.87\\-9.28\end{bmatrix}$

Explanation:

$\begin{align*}&\text{When the rows of a matrix outnumber the columns in an equation,}\\&\text{there are a greater number of equations than variables. In order}\\&\text{to approximate the values of these variables, a least-squares}\\&\text{approach is appropriate. This is done by multiplying each side}\\&\text{of the equation by the transpose of the equation matrix, and}\\&\text{solving the resulting equation:}\\&Ax=B\\&A^{T}Ax=A^{T}B\\&(A^{T}A)^{-1}(A^{T}A)x=(A^{T}A)^{-1}A^{T}B\\&x=(A^{T}A)^{-1}A^{T}B\\&\text{Using this, we can approximate our values:}\\&\begin{bmatrix}1&-6&1\\-17&1&-13\end{bmatrix}\begin{bmatrix}1&-17\\-6&1\\1&-13\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}1&-6&1\\-17&1&-13\end{bmatrix}\begin{bmatrix}188\\149\\19\end{bmatrix}\\&\begin{bmatrix}38&-36\\-36&459\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-687\\-3294\end{bmatrix}\\&\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-26.87\\-9.28\end{bmatrix}\end{align*}$

Report an Error

Example Question #2 : Least Squares

$\begin{align*}&\text{Approximate the values of }\begin{bmatrix}x\\y\end{bmatrix}\\&\text{for the function: }\begin{bmatrix}14&-8\\-4&-5\\-9&-2\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-161\\188\\30\end{bmatrix}\end{align*}$

Possible Answers:

$\begin{bmatrix}-12.01\\-6.26\end{bmatrix}$

$\begin{bmatrix}-21.01\\-15.26\end{bmatrix}$

$\begin{bmatrix}-13.01\\-7.26\end{bmatrix}$

$\begin{bmatrix}-7.01\\-1.26\end{bmatrix}$

Correct answer:

$\begin{bmatrix}-13.01\\-7.26\end{bmatrix}$

Explanation:

$\begin{align*}&\text{When the rows of a matrix outnumber the columns in an equation,}\\&\text{there are a greater number of equations than variables. In order}\\&\text{to approximate the values of these variables, a least-squares}\\&\text{approach is appropriate. This is done by multiplying each side}\\&\text{of the equation by the transpose of the equation matrix, and}\\&\text{solving the resulting equation:}\\&Ax=B\\&A^{T}Ax=A^{T}B\\&(A^{T}A)^{-1}(A^{T}A)x=(A^{T}A)^{-1}A^{T}B\\&x=(A^{T}A)^{-1}A^{T}B\\&\text{Using this, we can approximate our values:}\\&\begin{bmatrix}14&-4&-9\\-8&-5&-2\end{bmatrix}\begin{bmatrix}14&-8\\-4&-5\\-9&-2\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}14&-4&-9\\-8&-5&-2\end{bmatrix}\begin{bmatrix}-161\\188\\30\end{bmatrix}\\&\begin{bmatrix}293&-74\\-74&93\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-3276\\288\end{bmatrix}\\&\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-13.01\\-7.26\end{bmatrix}\end{align*}$

Report an Error

Example Question #71 : Matrix Calculus

$\begin{align*}&\text{Approximate the values of }\begin{bmatrix}x\\y\end{bmatrix}\\&\text{for the function: }\begin{bmatrix}-13&12\\-12&1\\-2&19\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-103\\-178\\-16\end{bmatrix}\end{align*}$

Possible Answers:

$\begin{bmatrix}12.04\\1.51\end{bmatrix}$

$\begin{bmatrix}5.04\\-5.49\end{bmatrix}$

$\begin{bmatrix}10.04\\-0.49\end{bmatrix}$

$\begin{bmatrix}17.04\\6.51\end{bmatrix}$

Correct answer:

$\begin{bmatrix}12.04\\1.51\end{bmatrix}$

Explanation:

$\begin{align*}&\text{When the rows of a matrix outnumber the columns in an equation,}\\&\text{there are a greater number of equations than variables. In order}\\&\text{to approximate the values of these variables, a least-squares}\\&\text{approach is appropriate. This is done by multiplying each side}\\&\text{of the equation by the transpose of the equation matrix, and}\\&\text{solving the resulting equation:}\\&Ax=B\\&A^{T}Ax=A^{T}B\\&(A^{T}A)^{-1}(A^{T}A)x=(A^{T}A)^{-1}A^{T}B\\&x=(A^{T}A)^{-1}A^{T}B\\&\text{Using this, we can approximate our values:}\\&\begin{bmatrix}-13&-12&-2\\12&1&19\end{bmatrix}\begin{bmatrix}-13&12\\-12&1\\-2&19\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-13&-12&-2\\12&1&19\end{bmatrix}\begin{bmatrix}-103\\-178\\-16\end{bmatrix}\\&\begin{bmatrix}317&-206\\-206&506\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}3507\\-1718\end{bmatrix}\\&\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}12.04\\1.51\end{bmatrix}\end{align*}$

Report an Error

View Linear Algebra Tutors

Foysal
Certified Tutor

Towson University, Bachelor, Physics . Virginia Commonwealth University, Master of Science, Health Physics.

View Linear Algebra Tutors

Scott
Certified Tutor

Florida State University, Bachelor of Science, Applied Mathematics. Florida State University, Bachelor of Science, Astrophysi...

View Linear Algebra Tutors

Yosef
Certified Tutor

Cooper Union for the Advancement of Science and Art, Bachelor of Engineering, Mechanical Engineering. Rensselaer Polytechnic ...

All Linear Algebra Resources

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

Popular Subjects

SAT Tutors in Denver, SAT Tutors in Washington DC, English Tutors in Chicago, Biology Tutors in Denver, Biology Tutors in Seattle, Reading Tutors in San Francisco-Bay Area, SAT Tutors in San Francisco-Bay Area, Biology Tutors in San Diego, English Tutors in Seattle, GRE Tutors in Chicago

Popular Courses & Classes

ACT Courses & Classes in Boston, ACT Courses & Classes in Washington DC, LSAT Courses & Classes in Dallas Fort Worth, ISEE Courses & Classes in San Francisco-Bay Area, SAT Courses & Classes in Chicago, GMAT Courses & Classes in Philadelphia, LSAT Courses & Classes in Phoenix, SAT Courses & Classes in Los Angeles, ISEE Courses & Classes in Washington DC, Spanish Courses & Classes in Seattle

Popular Test Prep

SAT Test Prep in Los Angeles, GRE Test Prep in Houston, MCAT Test Prep in New York City, LSAT Test Prep in San Diego, SAT Test Prep in Miami, ACT Test Prep in Washington DC, GRE Test Prep in San Francisco-Bay Area, ISEE Test Prep in Washington DC, SSAT Test Prep in Denver, 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 : Least Squares

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

Example Question #61 : Linear Algebra

Example Question #2 : Least Squares

Example Question #1 : Least Squares

Example Question #4 : Least Squares

Example Question #1 : Least Squares

Example Question #3 : Least Squares

Example Question #7 : Least Squares

Example Question #8 : Least Squares

Example Question #2 : Least Squares

Example Question #71 : Matrix Calculus

All Linear Algebra Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: