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Merge branch 'master' of tea.maggioni.xyz:maggicl/ICS

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Claudio Maggioni 2020-04-21 16:05:08 +02:00
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\begin{document} \maketitle \begin{document} \maketitle
\section*{Question 1} \section*{Question 1}
to be transcribed \[i=1 \hspace{1cm} l_1 = \begin{bmatrix}1\\1\\5\\-3\\\end{bmatrix} \hspace{1cm} u_1 = \begin{bmatrix}
2 & 1 & 1 & -2 \\\end{bmatrix}\]
\[A_2 = \begin{bmatrix}
2 & 1 & 1 &-2 \\ 2 & 2 & -2 & -1 \\ 10 & 4 & 23 & -8 \\
-6 & -2 & 4 & 6 \\\end{bmatrix} - \begin{bmatrix}
2 & 1 & 1 & -2 \\ 2 &1& 1 &-2 \\ 10 & 5 & 5 & -10 \\ -6 & -3 & -3 & 6 \\\end{bmatrix} = \begin{bmatrix}\\& 1 & -3 & 1 \\ & -1 & 18 & 2 \\ & 1 & 7 & 0 \end{bmatrix}\]
\[i = 2 \hspace{1cm} l_2 = \begin{bmatrix}\\1 \\-1 \\1\\\end{bmatrix} \hspace{1cm} u_2 = \begin{bmatrix} & 1& -3& 1\\\end{bmatrix}\]
\[A_3 = \begin{bmatrix}\\& 1 & -3 & 1 \\ & -1 & 18 & 2 \\ & 1 & 7 & 0 \end{bmatrix} - \begin{bmatrix}\\&1 & -3 & 1 \\ & -1 & 3 & -1 \\ & 1 & -3 & 1 \\\end{bmatrix} = \begin{bmatrix}
\\
\\
&&15&3\\&&10&-1\\\end{bmatrix}\]
\[i = 3 \hspace{1cm} l_3 = \begin{bmatrix}\\\\1\\2/3\\\end{bmatrix} \hspace{1cm} u_3 = \begin{bmatrix}&& 15 & 3 \\ \end{bmatrix}\]
\[A_4 = \begin{bmatrix}
\\
\\
&&15&3\\&&10&-1\\\end{bmatrix} - \begin{bmatrix}\\\\&&15 & 3 \\&&10 & 2 \\\end{bmatrix} = \begin{bmatrix}\\\\\\&&&-3\\\end{bmatrix}\]
\[i = 4 \hspace{1cm} l_4 = \begin{bmatrix}\\\\\\1\\\end{bmatrix} \hspace{1cm} u_4 = \begin{bmatrix}&&& -3 \\ \end{bmatrix}\]
\[L = \begin{bmatrix}1\\1&1\\5&-1&1\\-3 &1 &2/3&1\\\end{bmatrix} \hspace{1cm} U = \begin{bmatrix}2 &1 &1&-2\\&1 &-3&1\\&&15&3\\&&&-3\\\end{bmatrix}\]
\[Ly = B \Rightarrow \begin{bmatrix}1\\1&1\\5&-1&1\\-3 &1 &2/3&1\\\end{bmatrix} \begin{bmatrix}y_1\\y_2\\y_3\\y_4\\\end{bmatrix} = \begin{bmatrix}-1\\-3\\36\\18\\\end{bmatrix}
\]
\[y_1 = -1\]
\[y_2 = -3 - (-1) \cdot 1 = -2\]
\[y_3 = 36 - 1 \cdot 2 - (-1) \cdot 5 = 39\]
\[y_4 = 18 - \frac{2}{3} \cdot 39 - (-2) - (-3) \cdot (-1) = -9\]
\[Ux = Y \Rightarrow \begin{bmatrix}2 &1 &1&-2\\&1 &-3&1\\&&15&3\\&&&-3\\\end{bmatrix} \begin{bmatrix}x_1\\x_2\\x_3\\x_4\\\end{bmatrix} = \begin{bmatrix}-1\\-2\\39\\-9\\\end{bmatrix}
\]
\[x_4 = 3\]
\[x_3 = \frac{39 - 3 - 3}{15} = 2\]
\[x_2 = \frac{-2 -3 - (-3) \cdot 2}{1} = 1\]
\[x_1 = \frac{ -1 - (-2) \cdot 3 - 1 \cdot 2 - 1 \cdot 1}{2} = 1\]
\[x = \begin{bmatrix}1\\1\\2\\3\\\end{bmatrix}\]
\section*{Question 2} \section*{Question 2}
\[i = 1 \hspace{1cm} k = 4 \hspace{1cm} \begin{bmatrix}4&2&3&1\\\end{bmatrix}\] \[i = 1 \hspace{1cm} k = 4 \hspace{1cm} \begin{bmatrix}4&2&3&1\\\end{bmatrix}\]