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Linear Algebra : Least Squares

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Example Question #1 : Least Squares

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

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

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

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 $\frac{1}{\det(C)}$ .

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

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

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

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

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

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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:

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

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

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

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

Correct answer:

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

Explanation:

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

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Example Question #3 : 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:

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

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

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

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

Correct answer:

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

Explanation:

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

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Example Question #1 : 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:

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

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Example Question #5 : 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:

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

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

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

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

Correct answer:

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

Explanation:

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

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Example Question #6 : 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:

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

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

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

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

Correct answer:

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

Explanation:

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

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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:

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

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

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

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

Correct answer:

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

Explanation:

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

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Example Question #1 : 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:

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

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

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

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

Correct answer:

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

Explanation:

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

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Example Question #9 : 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:

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

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

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

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

Correct answer:

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

Explanation:

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

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Example Question #1 : Least Squares

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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Linear Algebra : Least Squares

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