Matrix Calculus - Linear Algebra

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Question

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

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Answer

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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Question

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 ?

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Answer

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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Question

$\begin{align}&\text{Find the matrix }Hf \text{ for }f(x,y,z)=13y^{2}tan(2x)tan(4z)\end{align}$

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Answer

$\begin{align}&\text{In asking for H, the question asks for the Hessian.}\&\text{The Hessian of a function is a matrix of second derivatives taken with}\&\text{respect to its constituent variables. The form is as follows:}\&\begin{bmatrix}\frac{d^2f}{d_{x_1}^2}&...&\frac{d^2f}{d_{x_1}d_{x_n}}\...&...&...\\frac{d^2f}{d_{x_n}d_{x_1}}&...&\frac{d^2f}{d_{x_n}^2}\end{bmatrix}\&\text{Considering our function: }f(x,y,z)=13y^{2}tan(2x)tan(4z)\&\text{And utilizing derivative rules:}\&(f\circ g )'=(f'\circ g )\cdot g'\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\&Hf=\begin{bmatrix}52y^{2}tan(2x)tan(4z)\cdot (2tan(2x)^{2} + 2)&26ytan(4z)\cdot (2tan(2x)^{2} + 2)&13y^{2}\cdot (2tan(2x)^{2} + 2)\cdot (4tan(4z)^{2} + 4)\26ytan(4z)\cdot (2tan(2x)^{2} + 2)&26tan(2x)tan(4z)&26ytan(2x)\cdot (4tan(4z)^{2} + 4)\13y^{2}\cdot (2tan(2x)^{2} + 2)\cdot (4tan(4z)^{2} + 4)&26ytan(2x)\cdot (4tan(4z)^{2} + 4)&104y^{2}tan(2x)tan(4z)\cdot (4tan(4z)^{2} + 4)\end{bmatrix}\end{align}$

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Question

$\begin{align}&\text{Determine the Hessian, }H\text{, of the function}\&f(x,y,z)=-cos(6x)tan(5y)sin(5z)\end{align}$

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Answer

$\begin{align}&\text{The Hessian of a function is a matrix of second derivatives taken with}\&\text{respect to its constituent variables. The form is as follows:}\&\begin{bmatrix}\frac{d^2f}{d_{x_1}^2}&...&\frac{d^2f}{d_{x_1}d_{x_n}}\...&...&...\\frac{d^2f}{d_{x_n}d_{x_1}}&...&\frac{d^2f}{d_{x_n}^2}\end{bmatrix}\&\text{Considering our function: }f(x,y,z)=-cos(6x)tan(5y)sin(5z)\&\text{And utilizing derivative rules:}\&(f\circ g )'=(f'\circ g )\cdot g'\&d[cos(u)]=-sin(u)du\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\&d[sin(u)]=cos(u)du\&Hf=\begin{bmatrix}36cos(6x)tan(5y)sin(5z)&6sin(6x)sin(5z)\cdot (5tan(5y)^{2} + 5)&30cos(5z)sin(6x)tan(5y)\6sin(6x)sin(5z)\cdot (5tan(5y)^{2} + 5)&-10cos(6x)tan(5y)sin(5z)\cdot (5tan(5y)^{2} + 5)&-5cos(6x)cos(5z)\cdot (5tan(5y)^{2} + 5)\30cos(5z)sin(6x)tan(5y)&-5cos(6x)cos(5z)\cdot (5tan(5y)^{2} + 5)&25cos(6x)tan(5y)sin(5z)\end{bmatrix}\end{align}$

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Question

$\begin{align}&\text{Find the matrix }Hf \text{ for }f(x,y)=- 8ytan(5x) - 13\cdot 5^{(2y)}x^{2}\end{align}$

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Answer

$\begin{align}&\text{In asking for H, the question asks for the Hessian.}\&\text{The Hessian of a function is a matrix of second derivatives taken with}\&\text{respect to its constituent variables. The form is as follows:}\&\begin{bmatrix}\frac{d^2f}{d_{x_1}^2}&...&\frac{d^2f}{d_{x_1}d_{x_n}}\...&...&...\\frac{d^2f}{d_{x_n}d_{x_1}}&...&\frac{d^2f}{d_{x_n}^2}\end{bmatrix}\&\text{Considering our function: }f(x,y)=- 8ytan(5x) - 13\cdot 5^{(2y)}x^{2}\&\text{And utilizing derivative rules:}\&(f\circ g )'=(f'\circ g )\cdot g'\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\&d[a^u]=a^uduln(a)\&Hf=\begin{bmatrix}- 26\cdot 5^{(2y)} - 80ytan(5x)\cdot (5tan(5x)^{2} + 5)&- 40tan(5x)^{2} - 52\cdot 5^{(2y)}xln(5) - 40\- 40tan(5x)^{2} - 52\cdot 5^{(2y)}xln(5) - 40&-52\cdot 5^{(2y)}x^{2}ln(5)^{2}\end{bmatrix}\end{align}$

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Question

$\begin{align}&\text{Find the matrix }Hf \text{ for }f(x,y,z)=13y^{2}tan(2x)tan(4z)\end{align}$

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Answer

$\begin{align}&\text{In asking for H, the question asks for the Hessian.}\&\text{The Hessian of a function is a matrix of second derivatives taken with}\&\text{respect to its constituent variables. The form is as follows:}\&\begin{bmatrix}\frac{d^2f}{d_{x_1}^2}&...&\frac{d^2f}{d_{x_1}d_{x_n}}\...&...&...\\frac{d^2f}{d_{x_n}d_{x_1}}&...&\frac{d^2f}{d_{x_n}^2}\end{bmatrix}\&\text{Considering our function: }f(x,y,z)=13y^{2}tan(2x)tan(4z)\&\text{And utilizing derivative rules:}\&(f\circ g )'=(f'\circ g )\cdot g'\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\&Hf=\begin{bmatrix}52y^{2}tan(2x)tan(4z)\cdot (2tan(2x)^{2} + 2)&26ytan(4z)\cdot (2tan(2x)^{2} + 2)&13y^{2}\cdot (2tan(2x)^{2} + 2)\cdot (4tan(4z)^{2} + 4)\26ytan(4z)\cdot (2tan(2x)^{2} + 2)&26tan(2x)tan(4z)&26ytan(2x)\cdot (4tan(4z)^{2} + 4)\13y^{2}\cdot (2tan(2x)^{2} + 2)\cdot (4tan(4z)^{2} + 4)&26ytan(2x)\cdot (4tan(4z)^{2} + 4)&104y^{2}tan(2x)tan(4z)\cdot (4tan(4z)^{2} + 4)\end{bmatrix}\end{align}$

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Question

$\begin{align}&\text{Determine the gradient, }\nabla\text{, of the function}\&f(x,y,z)=-cos(6x)tan(5y)sin(5z)\end{align}$

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Answer

$\begin{align}&\text{The gradient of a function is a vector of first derivatives taken with}\&\text{respect to its constituent variables. The form is as follows:}\&\begin{bmatrix}\frac{d^f}{d_{x_1}}\ \frac{d^f}{d_{x_2}}\...\\frac{df}{d_{x_n}}\end{bmatrix}\&\text{Considering our function: }f(x,y,z)=-cos(6x)tan(5y)sin(5z)\&\text{And utilizing derivative rules:}\&(f\circ g )'=(f'\circ g )\cdot g'\&d[cos(u)]=-sin(u)du\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\&d[sin(u)]=cos(u)du\&\nabla f=\begin{bmatrix}6sin(6x)tan(5y)sin(5z)\-cos(6x)sin(5z)\cdot (5tan(5y)^{2} + 5)\-5cos(6x)cos(5z)tan(5y)\end{bmatrix}\end{align}$

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Question

$\begin{align}&\text{Find the vector }\nabla f \text{ for }f(x,y)=- 8ytan(5x) - 13\cdot 5^{(2y)}x^{2}\end{align}$

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Answer

$\begin{align}&\text{In asking for }\nabla\text{, the question asks for the gradient.}\&\text{The gradient of a function is a vector of first derivatives taken with}\&\text{respect to its constituent variables. The form is as follows:}\&\begin{bmatrix}\frac{d^f}{d_{x_1}}\\frac{d^f}{d_{x_2}}\...\\frac{df}{d_{x_n}}\end{bmatrix}\&\text{Considering our function: }f(x,y)=- 8ytan(5x) - 13\cdot 5^{(2y)}x^{2}\&\text{And utilizing derivative rules:}\&(f\circ g )'=(f'\circ g )\cdot g'\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\&d[a^u]=a^uduln(a)\&\nabla f=\begin{bmatrix}- 26\cdot 5^{(2y)}x - 8y\cdot (5tan(5x)^{2} + 5)\- 8tan(5x) - 26\cdot 5^{(2y)}x^{2}ln(5)\end{bmatrix}\end{align}$

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Question

$\begin{align}&\text{Calculate the gradient of the function}\&f(x,y)=-7\cdot 2^{(5y)}x\end{align}$

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Answer

$\begin{align}&\text{The gradient of a function is a vector of first derivatives taken with}\&\text{respect to its constituent variables. The form is as follows:}\&\begin{bmatrix}\frac{d^f}{d_{x_1}}\\frac{d^f}{d_{x_2}}\...\\frac{df}{d_{x_n}}\end{bmatrix}\&\text{Considering our function: }f(x,y)=-7\cdot 2^{(5y)}x\&\text{And utilizing derivative rules:}\&(f\circ g )'=(f'\circ g )\cdot g'\&d[a^u]=a^uduln(a)\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\&\nabla f=\begin{bmatrix}-7\cdot 2^{(5y)}\-35\cdot 2^{(5y)}xln(2)\end{bmatrix}\end{align}$

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Question

$\begin{align}&\text{Find the vector }\nabla f \text{ for }f(x,y,z)=19z^{2}cos(2y)e^{(2x)} - 17\cdot 5^{(2y)}ln(4x)e^{(5z)}\end{align}$

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Answer

$\begin{align}&\text{In asking for }\nabla\text{, the question asks for the gradient.}\&\text{The gradient of a function is a vector of first derivatives taken with}\&\text{respect to its constituent variables. The form is as follows:}\&\begin{bmatrix}\frac{d^f}{d_{x_1}}\\frac{d^f}{d_{x_2}}\...\\frac{df}{d_{x_n}}\end{bmatrix}\&\text{Considering our function: }f(x,y,z)=19z^{2}cos(2y)e^{(2x)} - 17\cdot 5^{(2y)}ln(4x)e^{(5z)}\&\text{And utilizing derivative rules:}\&(f\circ g )'=(f'\circ g )\cdot g'\&d[e^u]=e^udu\&d[cos(u)]=-sin(u)du\&d[ln(u)]=\frac{du}{u}\&d[a^u]=a^uduln(a)\&\nabla f=\begin{bmatrix}38z^{2}cos(2y)e^{(2x)} -\frac{ (17\cdot 5^{(2y)}e^{(5z)})}{x}\- 38z^{2}sin(2y)e^{(2x)} - 34\cdot 5^{(2y)}ln(4x)e^{(5z)}ln(5)\38zcos(2y)e^{(2x)} - 85\cdot 5^{(2y)}ln(4x)e^{(5z)}\end{bmatrix}\end{align}$

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Question

$\begin{align}&\text{Find the vector }\nabla f \text{ for }f(x,y,z)=-16xyln(z)\end{align}$

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Answer

$\begin{align}&\text{In asking for }\nabla\text{, the question asks for the gradient.}\&\text{The gradient of a function is a vector of first derivatives taken with}\&\text{respect to its constituent variables. The form is as follows:}\&\begin{bmatrix}\frac{d^f}{d_{x_1}}\\frac{d^f}{d_{x_2}}\...\\frac{df}{d_{x_n}}\end{bmatrix}\&\text{Considering our function: }f(x,y,z)=-16xyln(z)\&\text{And utilizing derivative rules:}\&(f\circ g )'=(f'\circ g )\cdot g'\&d[ln(u)]=\frac{du}{u}\&\nabla f=\begin{bmatrix}-16yln(z)\-16xln(z)\-\frac{(16xy)}{z}\end{bmatrix}\end{align}$

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Question

$\begin{align}&\text{Determine the gradient, }\nabla\text{, of the function}\&f(x,y)=10sin(2x)sin(6y)\end{align}$

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Answer

$\begin{align}&\text{The gradient of a function is a vector of first derivatives taken with}\&\text{respect to its constituent variables. The form is as follows:}\&\begin{bmatrix}\frac{d^f}{d_{x_1}}\\frac{d^f}{d_{x_2}}\...\\frac{df}{d_{x_n}}\end{bmatrix}\&\text{Considering our function: }f(x,y)=10sin(2x)sin(6y)\&\text{And utilizing derivative rules:}\&(f\circ g )'=(f'\circ g )\cdot g'\&d[sin(u)]=cos(u)du\&\nabla f=\begin{bmatrix}20cos(2x)sin(6y)\60cos(6y)sin(2x)\end{bmatrix}\end{align}$

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Question

$\begin{align}&\text{Calculate the Hessian matrix of the function}\&f(x,y)=2cos(2x)cos(5y) + 6\cdot 2^{(6x)}tan(4y)\end{align}$

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Answer

$\begin{align}&\text{The Hessian of a function is a matrix of second derivatives taken with}\&\text{respect to its constituent variables. The form is as follows:}\&\begin{bmatrix}\frac{d^2f}{d_{x_1}^2}&...&\frac{d^2f}{d_{x_1}d_{x_n}}\...&...&...\\frac{d^2f}{d_{x_n}d_{x_1}}&...&\frac{d^2f}{d_{x_n}^2}\end{bmatrix}\&\text{Considering our function: }f(x,y)=2cos(2x)cos(5y) + 6\cdot 2^{(6x)}tan(4y)\&\text{And utilizing derivative rules:}\&(f\circ g )'=(f'\circ g )\cdot g'\&d[cos(u)]=-sin(u)du\&d[a^u]=a^uduln(a)\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\&\&Hf=\begin{bmatrix}216\cdot 2^{(6x)}tan(4y)ln(2)^{2} - 8cos(2x)cos(5y)&20sin(2x)sin(5y) + 36\cdot 2^{(6x)}ln(2)\cdot (4tan(4y)^{2} + 4)\20sin(2x)sin(5y) + 36\cdot 2^{(6x)}ln(2)\cdot (4tan(4y)^{2} + 4)&48\cdot 2^{(6x)}tan(4y)\cdot (4tan(4y)^{2} + 4) - 50cos(2x)cos(5y)\end{bmatrix}\end{align}$

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Question

$\begin{align}&\text{Determine the Hessian, }H\text{, of the function}\&f(x,y,z)=9ln(2x)ln(2z)e^{(y)}\end{align}$

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Answer

$\begin{align}&\text{The Hessian of a function is a matrix of second derivatives taken with}\&\text{respect to its constituent variables. The form is as follows:}\&\begin{bmatrix}\frac{d^2f}{d_{x_1}^2}&...&\frac{d^2f}{d_{x_1}d_{x_n}}\...&...&...\\frac{d^2f}{d_{x_n}d_{x_1}}&...&\frac{d^2f}{d_{x_n}^2}\end{bmatrix}\&\text{Considering our function: }f(x,y,z)=9ln(2x)ln(2z)e^{(y)}\&\text{And utilizing derivative rules:}\&(f\circ g )'=(f'\circ g )\cdot g'\&d[ln(u)]=\frac{du}{u}\&d[e^u]=e^udu\&Hf=\begin{bmatrix}-\frac{(9ln(2z)e^{(y)})}{x^{2}}&\frac{(9ln(2z)e^{(y)})}{x}&\frac{(9e^{(y)})}{(xz)}\\frac{(9ln(2z)e^{(y)})}{x}&9ln(2x)ln(2z)e^{(y)}&\frac{(9ln(2x)e^{(y)})}{z}\\frac{(9e^{(y)})}{(xz)}&\frac{(9ln(2x)e^{(y)})}{z}&-\frac{(9ln(2x)e^{(y)})}{z^{2}}\end{bmatrix}\end{align}$

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Question

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

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Answer

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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Question

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.

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Answer

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.

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

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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Question

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

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Answer

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

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Question

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

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Answer

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

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Question

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

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Answer

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

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Question

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

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Answer

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

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