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lecture_5

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lecture_5 [2015/02/04 12:19] – [Gaussian elimination] rupertlecture_5 [2017/02/07 10:11] (current) – [Gaussian elimination] rupert
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 Use Gaussian elimination to solve the linear system Use Gaussian elimination to solve the linear system
-\begin{align*} 2x+y+3x+4w&=27\\ x+2y+3z+2w&=30\\x+y+3z+w&=25\end{align*}+\begin{align*} 2x+y+3z+4w&=27\\ x+2y+3z+2w&=30\\x+y+3z+w&=25\end{align*}
  
 == Solution 1 == == Solution 1 ==
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 \go{1&1&3&1&25}{0&1&0&1&5}{0&-1&-3&2&-23} \go{1&1&3&1&25}{0&1&0&1&5}{0&-1&-3&2&-23}
 \ar{R3\to R3+R2} \ar{R3\to R3+R2}
-\go{1&1&3&1&25}{0&1&0&1&5}{0&0&-3&3&18}+\go{1&1&3&1&25}{0&1&0&1&5}{0&0&-3&3&-18}
 \ar{R3\to-\tfrac13R3} \ar{R3\to-\tfrac13R3}
 \go{1&1&3&1&25}{0&1&0&1&5}{0&0&1&-1&6} \go{1&1&3&1&25}{0&1&0&1&5}{0&0&1&-1&6}
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 So  So 
 $$ \begin{bmatrix}x\\y\\z\\w\end{bmatrix}=\begin{bmatrix} 2\\5\\6\\0\end{bmatrix}+t\begin{bmatrix}-3\\-1\\1\\1\end{bmatrix},\quad t\in\mathbb{R}.$$ $$ \begin{bmatrix}x\\y\\z\\w\end{bmatrix}=\begin{bmatrix} 2\\5\\6\\0\end{bmatrix}+t\begin{bmatrix}-3\\-1\\1\\1\end{bmatrix},\quad t\in\mathbb{R}.$$
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lecture_5.1423052346.txt.gz · Last modified: by rupert
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