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Linear Algebra : Matrix Calculus

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Example Questions

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:

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

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Example Question #72 : Matrix Calculus

$\begin{align*}&\text{Using the method of least-squares, approximate x and y values for: }\\&\begin{bmatrix}14&-9\\-10&-4\\-8&-6\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-12\\-198\\-179\end{bmatrix}\end{align*}$

Possible Answers:

$\begin{bmatrix}11.91\\18.96\end{bmatrix}$

$\begin{bmatrix}10.91\\17.96\end{bmatrix}$

$\begin{bmatrix}17.91\\24.96\end{bmatrix}$

$\begin{bmatrix}14.91\\21.96\end{bmatrix}$

Correct answer:

$\begin{bmatrix}10.91\\17.96\end{bmatrix}$

Explanation:

$\begin{align*}&\text{When the rows of a matrix outnumber the columns in an equation,}\\&\text{there are a greater number of equations than variables. In order}\\&\text{to approximate the values of these variables, a least-squares}\\&\text{approach is appropriate. This is done by multiplying each side}\\&\text{of the equation by the transpose of the equation matrix, and}\\&\text{solving the resulting equation:}\\&Ax=B\\&A^{T}Ax=A^{T}B\\&(A^{T}A)^{-1}(A^{T}A)x=(A^{T}A)^{-1}A^{T}B\\&x=(A^{T}A)^{-1}A^{T}B\\&\text{Using this, we can approximate our values:}\\&\begin{bmatrix}14&-10&-8\\-9&-4&-6\end{bmatrix}\begin{bmatrix}14&-9\\-10&-4\\-8&-6\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}14&-10&-8\\-9&-4&-6\end{bmatrix}\begin{bmatrix}-12\\-198\\-179\end{bmatrix}\\&\begin{bmatrix}360&-38\\-38&133\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}3244\\1974\end{bmatrix}\\&\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}10.91\\17.96\end{bmatrix}\end{align*}$

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Example Question #73 : Matrix Calculus

$\begin{align*}&\text{Approximate }\begin{bmatrix}x\\y\end{bmatrix}\\&\text{for the system: }\begin{bmatrix}11&4\\-8&0\\8&20\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-42\\-66\\-36\end{bmatrix}\end{align*}$

Possible Answers:

$\begin{bmatrix}-7.57\\-11.84\end{bmatrix}$

$\begin{bmatrix}1.43\\-2.84\end{bmatrix}$

$\begin{bmatrix}-1.57\\-5.84\end{bmatrix}$

$\begin{bmatrix}5.43\\1.16\end{bmatrix}$

Correct answer:

$\begin{bmatrix}1.43\\-2.84\end{bmatrix}$

Explanation:

$\begin{align*}&\text{When the rows of a matrix outnumber the columns in an equation,}\\&\text{there are a greater number of equations than variables. In order}\\&\text{to approximate the values of these variables, a least-squares}\\&\text{approach is appropriate. This is done by multiplying each side}\\&\text{of the equation by the transpose of the equation matrix, and}\\&\text{solving the resulting equation:}\\&Ax=B\\&A^{T}Ax=A^{T}B\\&(A^{T}A)^{-1}(A^{T}A)x=(A^{T}A)^{-1}A^{T}B\\&x=(A^{T}A)^{-1}A^{T}B\\&\text{Using this, we can approximate our values:}\\&\begin{bmatrix}11&-8&8\\4&0&20\end{bmatrix}\begin{bmatrix}11&4\\-8&0\\8&20\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}11&-8&8\\4&0&20\end{bmatrix}\begin{bmatrix}-42\\-66\\-36\end{bmatrix}\\&\begin{bmatrix}249&204\\204&416\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-222\\-888\end{bmatrix}\\&\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}1.43\\-2.84\end{bmatrix}\end{align*}$

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Example Question #71 : Matrix Calculus

$\begin{align*}&\text{Using the method of least-squares, approximate x and y values for: }\\&\begin{bmatrix}15&-4\\19&-15\\-2&19\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-195\\-163\\15\end{bmatrix}\end{align*}$

Possible Answers:

$\begin{bmatrix}-7.03\\2.82\end{bmatrix}$

$\begin{bmatrix}-13.03\\-3.18\end{bmatrix}$

$\begin{bmatrix}-18.03\\-8.18\end{bmatrix}$

$\begin{bmatrix}-11.03\\-1.18\end{bmatrix}$

Correct answer:

$\begin{bmatrix}-11.03\\-1.18\end{bmatrix}$

Explanation:

$\begin{align*}&\text{When the rows of a matrix outnumber the columns in an equation,}\\&\text{there are a greater number of equations than variables. In order}\\&\text{to approximate the values of these variables, a least-squares}\\&\text{approach is appropriate. This is done by multiplying each side}\\&\text{of the equation by the transpose of the equation matrix, and}\\&\text{solving the resulting equation:}\\&Ax=B\\&A^{T}Ax=A^{T}B\\&(A^{T}A)^{-1}(A^{T}A)x=(A^{T}A)^{-1}A^{T}B\\&x=(A^{T}A)^{-1}A^{T}B\\&\text{Using this, we can approximate our values:}\\&\begin{bmatrix}15&19&-2\\-4&-15&19\end{bmatrix}\begin{bmatrix}15&-4\\19&-15\\-2&19\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}15&19&-2\\-4&-15&19\end{bmatrix}\begin{bmatrix}-195\\-163\\15\end{bmatrix}\\&\begin{bmatrix}590&-383\\-383&602\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-6052\\3510\end{bmatrix}\\&\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-11.03\\-1.18\end{bmatrix}\end{align*}$

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Example Question #74 : Matrix Calculus

It is recommended that you use a calculator with matrix arithmetic capability for this question.

Give the equation of the least squares regression line for the following data:

(0, 128) , (10, 194 ) , (20, 253) , (30, 330) .

Round your coefficients to three decimal digits, if applicable.

Possible Answers:

y=6.733x + 128

y=0.15x -18.928

y=6.8x + 123

y = 6.65 x + 126.5

y=0.146x -17.867

Correct answer:

y = 6.65 x + 126.5

Explanation:

Form the matrices $X = \begin{bmatrix} 1 & x_{1} \\ 1 & x_{2} \\1 & x_{3} \\1 & x_{4} \end{bmatrix}$ and $Y = \begin{bmatrix} y_{1} \\ y _{2} \\ y_{3} \\ y_{4}\end{bmatrix}$

using the abscissas and ordinates of the four points:

$X = \begin{bmatrix} 1 & 0 \\ 1 & 10 \\1 & 20 \\1 & 30 \end{bmatrix}$ and $Y = \begin{bmatrix} 128 \\194\\ 253 \\ 330\end{bmatrix}$

The least squares regression line is the line of the equation y = mx + b ,

where $A = \begin{bmatrix} b \\ m \end{bmatrix}$ can be found using the equation

$A =( X^{T}X)^{-1} X^{T}Y$ .

This can be calculated as follows:

$X^{T}X =\begin{bmatrix} 4 & 60 \\ 60 & 1,400 \end{bmatrix}$

$(X^{T}X )^{-1}=\begin{bmatrix} 0.7 & -0.03 \\ -0.03&0.002\end{bmatrix}$

$X^{T}Y = \begin{bmatrix} 905 \\ 16,900 \end{bmatrix}$

$A =( X^{T}X)^{-1} X^{T}Y = \begin{bmatrix}126.5 \\ 6.65 \end{bmatrix}$

The least squares regression line is the line of the equation

y = 6.65 x + 126.5 .

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Example Question #1 : Gradients Of The Determinant

Which of the following expressions is one for the gradient of the determinant of an $n \times n$ matrix ?

Possible Answers:

$(adj(A))^{-1}$

$\mathbf{x}^{-1}A\mathbf{x}, (\|\mathbf{x}\|=1)$

None of the other answers

(adj(A))^T

Correct answer:

(adj(A))^T

Explanation:

The expression for the determinant of using co-factor expansion (along any row) is

$|A| = \sum_{i=1}^n(-1)^{i+j}A_{ij}M_{ij}$

In order to find the gradient of the determinant, we take the partial derivative of the determinant expression with respect to some entry $A_{kl}$ in our matrix, yielding $(-1)^{k+l}M_{kl} = (adj(A))_{kl} = (adj(A))^T$ .

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Example Question #71 : Matrix Calculus

True or False, the Constrained Extremum Theorem only applies to skew-symmetric matrices.

Possible Answers:

False

True

Correct answer:

False

Explanation:

It only applies to symmetric matrices, not skew-symmetric ones. The Constrained Extremum Theorem concerns the maximum and minimum values of the quadratic form $\mathbf{x}^TA\mathbf{x}$ when $\|\mathbf{x}\|=1$ .

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Example Question #1 : Eigenvalues As Optimization

The maximum value of a quadratic form $\mathbf{x}^TA\mathbf{x}$ ( is an $n \times n$ symmetric matrix, $\|\mathbf{x}\|=1$ ) corresponds to which eigenvalue of ?

Possible Answers:

The eigenvalue with the greatest multiplicity

None of the other answers

The largest eigenvalue

The second largest eigenvalue

The smallest eigenvalue

Correct answer:

The largest eigenvalue

Explanation:

This is the statement of the Constrained Extremum Theorem. Likewise, the minimum value of the quadratic form corresponds to the smallest eigenvalue of .

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Linear Algebra : Matrix Calculus

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

Example Question #71 : Matrix Calculus

Example Question #72 : Matrix Calculus

Example Question #73 : Matrix Calculus

Example Question #71 : Matrix Calculus

Example Question #74 : Matrix Calculus

Example Question #1 : Gradients Of The Determinant

Example Question #71 : Matrix Calculus

Example Question #1 : Eigenvalues As Optimization

All Linear Algebra Resources

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