290 lines
9.8 KiB
TeX
290 lines
9.8 KiB
TeX
% vim: set ts=2 sw=2 et tw=80:
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\documentclass[12pt,a4paper]{article}
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\usepackage[utf8]{inputenc} \usepackage[margin=2cm]{geometry}
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\usepackage{amstext} \usepackage{amsmath} \usepackage{array}
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\newcommand{\lra}{\Leftrightarrow}
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\title{Howework 3 -- Introduction to Computational Science}
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\author{Claudio Maggioni}
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\begin{document} \maketitle
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\section*{Question 1}
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\[i=1 \hspace{1cm} l_1 = \begin{bmatrix}1\\1\\5\\-3\\\end{bmatrix} \hspace{1cm} u_1 = \begin{bmatrix}
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2 & 1 & 1 & -2 \\\end{bmatrix}\]
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\[A_2 = \begin{bmatrix}
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2 & 1 & 1 &-2 \\ 2 & 2 & -2 & -1 \\ 10 & 4 & 23 & -8 \\
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-6 & -2 & 4 & 6 \\\end{bmatrix} - \begin{bmatrix}
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2 & 1 & 1 & -2 \\ 2 &1& 1 &-2 \\ 10 & 5 & 5 & -10 \\ -6 & -3 & -3 & 6 \\\end{bmatrix} = \begin{bmatrix}\\& 1 & -3 & 1 \\ & -1 & 18 & 2 \\ & 1 & 7 & 0 \end{bmatrix}\]
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\[i = 2 \hspace{1cm} l_2 = \begin{bmatrix}\\1 \\-1 \\1\\\end{bmatrix} \hspace{1cm} u_2 = \begin{bmatrix} & 1& -3& 1\\\end{bmatrix}\]
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\[A_3 = \begin{bmatrix}\\& 1 & -3 & 1 \\ & -1 & 18 & 2 \\ & 1 & 7 & 0 \end{bmatrix} - \begin{bmatrix}\\&1 & -3 & 1 \\ & -1 & 3 & -1 \\ & 1 & -3 & 1 \\\end{bmatrix} = \begin{bmatrix}
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\\
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\\
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&&15&3\\&&10&-1\\\end{bmatrix}\]
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\[i = 3 \hspace{1cm} l_3 = \begin{bmatrix}\\\\1\\2/3\\\end{bmatrix} \hspace{1cm} u_3 = \begin{bmatrix}&& 15 & 3 \\ \end{bmatrix}\]
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\[A_4 = \begin{bmatrix}
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\\
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\\
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&&15&3\\&&10&-1\\\end{bmatrix} - \begin{bmatrix}\\\\&&15 & 3 \\&&10 & 2 \\\end{bmatrix} = \begin{bmatrix}\\\\\\&&&-3\\\end{bmatrix}\]
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\[i = 4 \hspace{1cm} l_4 = \begin{bmatrix}\\\\\\1\\\end{bmatrix} \hspace{1cm} u_4 = \begin{bmatrix}&&& -3 \\ \end{bmatrix}\]
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\[L = \begin{bmatrix}1\\1&1\\5&-1&1\\-3 &1 &2/3&1\\\end{bmatrix} \hspace{1cm} U = \begin{bmatrix}2 &1 &1&-2\\&1 &-3&1\\&&15&3\\&&&-3\\\end{bmatrix}\]
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\[Ly = B \Rightarrow \begin{bmatrix}1\\1&1\\5&-1&1\\-3 &1 &2/3&1\\\end{bmatrix} \begin{bmatrix}y_1\\y_2\\y_3\\y_4\\\end{bmatrix} = \begin{bmatrix}-1\\-3\\36\\18\\\end{bmatrix}
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\]
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\[y_1 = -1\]
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\[y_2 = -3 - (-1) \cdot 1 = -2\]
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\[y_3 = 36 - 1 \cdot 2 - (-1) \cdot 5 = 39\]
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\[y_4 = 18 - \frac{2}{3} \cdot 39 - (-2) - (-3) \cdot (-1) = -9\]
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\[Ux = Y \Rightarrow \begin{bmatrix}2 &1 &1&-2\\&1 &-3&1\\&&15&3\\&&&-3\\\end{bmatrix} \begin{bmatrix}x_1\\x_2\\x_3\\x_4\\\end{bmatrix} = \begin{bmatrix}-1\\-2\\39\\-9\\\end{bmatrix}
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\]
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\[x_4 = 3\]
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\[x_3 = \frac{39 - 3 - 3}{15} = 2\]
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\[x_2 = \frac{-2 -3 - (-3) \cdot 2}{1} = 1\]
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\[x_1 = \frac{ -1 - (-2) \cdot 3 - 1 \cdot 2 - 1 \cdot 1}{2} = 1\]
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\[x = \begin{bmatrix}1\\1\\2\\3\\\end{bmatrix}\]
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\section*{Question 2}
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\[i = 1 \hspace{1cm} k = 4 \hspace{1cm} \begin{bmatrix}4&2&3&1\\\end{bmatrix}\]
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\[
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l_1 =
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\begin{bmatrix}
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1/8 \\
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1/4 \\
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1/2 \\
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1 \\
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\end{bmatrix}
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\hspace{1cm}
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u_1 = \begin{bmatrix} 32 & 24 & 10 & 11\end{bmatrix}
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\hspace{1cm}
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\]\[
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A_2 = \begin{bmatrix}4 &3 &2& 1\\ 8& 8& 5& 2\\ 16& 12& 10& 5\\ 32& 24& 20 &11 \\\end{bmatrix} - \begin{bmatrix}
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4 & 3 & 5/2 & 11/8 \\
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8 & 6 & 5 & 11/4 \\
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16 & 12 & 10 & 11/2 \\
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32 & 24 & 20 & 11 \\
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\end{bmatrix} =
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\begin{bmatrix}
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0 & 0 & -1/2 & -3/8 \\
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0 & 2 & 0 & -3/4 \\
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0 & 0 & 0 & -1/2 \\
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0 & 0 & 0 & 0 \\
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\end{bmatrix}
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\]
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\[i = 2 \hspace{1cm} k = 2 \hspace{1cm} p = \begin{bmatrix}4&2&3&1\\\end{bmatrix}\]
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\[
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l_2 =\begin{bmatrix}
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0\\1\\0\\0\\
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\end{bmatrix}
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\hspace{1cm}
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u_2 =\begin{bmatrix} 0 & 2 & 0 & -3/4 \end{bmatrix}\]
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\[
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A_3 = \begin{bmatrix}
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0 & 0 & -1/2 & -3/8 \\
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0 & 2 & 0 & -3/4 \\
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0 & 0 & 0 & -1/2 \\
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0 & 0 & 0 & 0 \\
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\end{bmatrix} -
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\begin{bmatrix}
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0 & 0 & 0 & 0 \\
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0 & 2 & 0 & -3/4 \\
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0 & 0 & 0 & 0 \\
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0 & 0 & 0 & 0 \\
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\end{bmatrix}
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=
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\begin{bmatrix}
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0 & 0 & -1/2 & -3/8 \\
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0 & 0 & 0 & 0 \\
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0 & 0 & 0 & -1/2 \\
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0 & 0 & 0 & 0 \\
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\end{bmatrix}
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\]
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\[ i = 3 \hspace{1cm} k = 4 \hspace{1cm} p = \begin{bmatrix}
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4&2&1&3\\
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\end{bmatrix}
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\]
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\[
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l_3 = \begin{bmatrix}
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1 \\0\\0\\0\\
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\end{bmatrix}
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\hspace{1cm}
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u_3 = \begin{bmatrix} 0 & 0& -1/2 & -3/8\\\end{bmatrix}
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\]\[
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A_4 = \begin{bmatrix}
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0 & 0 & -1/2 & -3/8 \\
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0 & 0 & 0 & 0 \\
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0 & 0 & 0 & -1/2 \\
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0 & 0 & 0 & 0 \\
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\end{bmatrix} -
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\begin{bmatrix}
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0 & 0 & -1/2& -3/8\\
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0 & 0 & 0 & 0 \\
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0 & 0 & 0 & 0 \\
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0 & 0 & 0 & 0 \\
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\end{bmatrix} =
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\begin{bmatrix}
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0 & 0 & 0 & 0 \\
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0 & 0 & 0 & 0 \\
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0 & 0 & 0 & -1/2 \\
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0 & 0 & 0 & 0 \\
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\end{bmatrix}
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\]
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\[ i =4 \hspace{1cm} k = 4 \hspace{1cm} p = \begin{bmatrix}
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4&2&1&3\\
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\end{bmatrix}\]
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\[l_4 = \begin{bmatrix} 0 \\0\\1\\0\\\end{bmatrix}
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u_4 = \begin{bmatrix}0&0&0&-1/2\end{bmatrix}
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\]
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\[
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P =
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\begin{bmatrix}
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0 & 0 & 0 & 1 \\
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0 & 1 & 0 & 0 \\
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1 & 0 & 0 & 0 \\
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0 & 0 & 1 & 0 \\
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\end{bmatrix}
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\]
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\[
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L = P *
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\begin{bmatrix}
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1/8 & 0 & 1 & 0 \\
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1/4 & 1 & 0 & 0 \\
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1/2 & 0 & 0 & 1 \\
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1 & 0 & 0 & 0 \\
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\end{bmatrix}
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=
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\begin{bmatrix}
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1 & 0 & 0 & 0 \\
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1/4 & 1 & 0 & 0 \\
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1/8 & 0 & 1 & 0 \\
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1/2 & 0 & 0 & 1 \\
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\end{bmatrix}
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\]
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\[
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U =
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\begin{bmatrix}
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32 & 24 & 20 & 11 \\
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0 & 2 & 0 & -3/4 \\
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0 & 0 & -1/2 & -3/8 \\
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0 & 0 & 0 & -1/2 \\
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\end{bmatrix}
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\]
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\section*{Question 4}
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\[
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A_1 = \begin{bmatrix}
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1 & 4 & 8 & 3 \\
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4 & 20 & 40 & 28 \\
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8 & 40 & 89 & 71 \\
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3 & 28 & 71 & 114 \\
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\end{bmatrix}
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\]
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\[ i = 1 \hspace{1cm} l_1 = \begin{bmatrix}1\\4\\8\\3\\\end{bmatrix}\]
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\[
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A_2 = \begin{bmatrix}
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1 & 4 & 8 & 3 \\
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4 & 20 & 40 & 28 \\
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8 & 40 & 89 & 71 \\
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3 & 28 & 71 & 114 \\
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\end{bmatrix}
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-
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\begin{bmatrix}
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1 & 4 & 8 & 3\\
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4 & 16 & 32 & 12\\
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8 & 32 & 64 & 24\\
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3 & 12 & 24 & 9\\
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\end{bmatrix}
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=
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\begin{bmatrix}
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\\
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& 4 & 8 & 16\\
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& 8 & 25 & 47 \\
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& 16 & 47 & 105 \\
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\end{bmatrix}
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\]
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\[i=2 \hspace{1cm} l_2 = \begin{bmatrix}\\2\\4\\8\\\end{bmatrix}\]
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\[A_3 =
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\begin{bmatrix}
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\\
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& 4 & 8 & 16\\
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& 8 & 25 & 47 \\
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& 16 & 47 & 105 \\
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\end{bmatrix}
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-
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\begin{bmatrix}
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\\
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& 4 & 8 & 16 \\
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& 8 & 16 & 32 \\
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& 16 & 32 & 64 \\
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\end{bmatrix}
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=
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\begin{bmatrix}\\\\&& 9 & 15 \\&&15 & 41\\\end{bmatrix}
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\]
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\[i = 3 \hspace{1cm} l_3 = \begin{bmatrix}\\\\3\\5\\\end{bmatrix}\]
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\[A_4 =
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\begin{bmatrix}\\\\&& 9 & 15 \\&&15 & 41\\\end{bmatrix}
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-
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\begin{bmatrix}
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\\
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\\
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&&9&15\\
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&& 15 & 25\\
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\end{bmatrix}
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=
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\begin{bmatrix}
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\\\\\\&&&16\\
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\end{bmatrix}
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\]
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\[i = 4 \hspace{1cm} l_4 = \begin{bmatrix}\\\\\\4\\\end{bmatrix}\]
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\[L = \begin{bmatrix}1\\4&2\\8&4&3\\3&8&5&4\\\end{bmatrix}\]
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\section*{Question 5}
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\subsection*{Point a)}
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First of all, to show that $A_{1,1}$ is symmetric, we say:
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\[\begin{bmatrix}A_{1,1} & A_{1,2}\\A_{2,1} & A_{2,2}\\\end{bmatrix} = A = A^T = \begin{bmatrix}A_{1,1}^T & A_{2,1}\\A_{1,2} & A_{2,2}^T\\\end{bmatrix}\]
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Therefore we can say that
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$$A_{1,1} = A_{1,1}^T$$
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and thus $A_{1,1}$ is shown to be symmetric.
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Then, since $A$ is positive definite, $x^TAx > 0$ for any vector $x \in R^n$. We choose a vector $x$ such that $x_k = 0$ for any $p < k \leq n$.
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Then:
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$$ x^TAx = \sum_{j=1}^n x_j \cdot \sum_{i=1}^n x_i a_{i,j} = \sum_{j=1}^p x_j \cdot \sum_{i=1}^n x_i a_{i,j} +
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\sum_{j=p+1}^n 0 \cdot \sum_{i=1}^n 0 a_{i,j} =
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\sum_{j=1}^p x_j \cdot (\sum_{i=i}^p x_i a_{i,j} +
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\sum_{i=p+1}^n 0 a_{i,j}) =$$
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$$=\sum_{j=1}^p x_j \cdot \sum_{i=i}^p x_i a_{i,j} =
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\begin{bmatrix}x_1 & x_2 & \ldots & x_n\end{bmatrix}
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A_{1,1}\begin{bmatrix}x_1 & x_2 & \ldots & x_n\end{bmatrix}^T$$
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Then:
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$$\begin{bmatrix}x_1 & x_2 & \ldots & x_n\end{bmatrix}A_{1,1}\begin{bmatrix}x_1 & x_2 & \ldots & x_n\end{bmatrix}^T>0$$
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which is the definition of positive definiteness for $A_{1,1}$.
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\end{document}
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