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Linear Algebra : Linear Algebra

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Example Questions

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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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Example Question #71 : Linear Algebra

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

Possible Answers:

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

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

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

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

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 #11 : Least Squares

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

Possible Answers:

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

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

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

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

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 #73 : Linear Algebra

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

Possible Answers:

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

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

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

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

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 #71 : 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 = 6.65 x + 126.5

y=6.8x + 123

y=6.733x + 128

y=0.146x -17.867

y=0.15x -18.928

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:

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

(adj(A))^T

None of the other answers

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

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 #76 : Linear Algebra

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

Possible Answers:

True

False

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 #71 : Linear Algebra

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 second largest eigenvalue

The eigenvalue with the greatest multiplicity

The smallest eigenvalue

The largest eigenvalue

None of the other answers

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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Example Question #78 : Linear Algebra

Which of the following matrices is a scalar multiple of the identity matrix?

$\begin{pmatrix} 1&2 \\ 3&4 \end{pmatrix}$ , $\begin{pmatrix} e&0 \\ 0&e \end{pmatrix}$ , $\begin{pmatrix} 0& 2\\ 2&0 \end{pmatrix}$

Possible Answers:

$\begin{pmatrix} 0& 2\\ 2&0 \end{pmatrix}$

$\begin{pmatrix} e&0 \\ 0&e \end{pmatrix}$

$\begin{pmatrix} 1&2 \\ 3&4 \end{pmatrix}$

Correct answer:

$\begin{pmatrix} e&0 \\ 0&e \end{pmatrix}$

Explanation:

The x identity matrix is

$\begin{pmatrix} 1&0 \\ 0&1 \end{pmatrix}$

For this problem we see that

$\begin{pmatrix} e&0 \\ 0&e \end{pmatrix}=\begin{pmatrix} e*1&e*0 \\ e*0&e*1 \end{pmatrix}=e\begin{pmatrix} 1&0 \\ 0&1 \end{pmatrix}$

And so

$\begin{pmatrix} e&0 \\ 0&e \end{pmatrix}$ is a scalar multiple of the identity matrix.

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Example Question #1 : The Identity Matrix And Diagonal Matrices

Which of the following is true concerning diagonal matrices?

Possible Answers:

The zero matrix (of any size) is not a diagonal matrix.

The determinant of any diagonal matrix is .

The product of two diagonal matrices (in either order) is always another diagonal matrix.

The trace of any diagonal matrix is equal to its determinant.

All of the other answers are false.

Correct answer:

The product of two diagonal matrices (in either order) is always another diagonal matrix.

Explanation:

You can verify this directly by proving it, or by multiplying a few examples on your calculator.

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Linear Algebra : Linear Algebra

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

Example Question #1 : Least Squares

Example Question #71 : Linear Algebra

Example Question #11 : Least Squares

Example Question #73 : Linear Algebra

Example Question #71 : Matrix Calculus

Example Question #1 : Gradients Of The Determinant

Example Question #76 : Linear Algebra

Example Question #71 : Linear Algebra

Example Question #78 : Linear Algebra

Example Question #1 : The Identity Matrix And Diagonal Matrices

All Linear Algebra Resources

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