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lecture_5 [2015/02/04 12:04] rupertlecture_5 [2017/02/07 10:11] (current) – [Gaussian elimination] rupert
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 {{page>gaussian elimination algorithm}} {{page>gaussian elimination algorithm}}
 +
 +=== Example ===
 +
 +Use Gaussian elimination to solve the linear system
 +\begin{align*} 2x+y+3z+4w&=27\\ x+2y+3z+2w&=30\\x+y+3z+w&=25\end{align*}
 +
 +== Solution 1 ==
 +
 +We put the augmented matrix into REF:
 +\begin{align*} 
 +\def\go#1#2#3{\begin{bmatrix}#1\\#2\\#3\end{bmatrix}}
 +\def\ar#1{\\[6pt]\xrightarrow{#1}&}
 +&\go{2&1&3&4&27}{1&2&3&2&30}{1&1&3&1&25}
 +\ar{\text{reorder rows (to avoid division)}}
 +\go{1&1&3&1&25}{1&2&3&2&30}{2&1&3&4&27}
 +\ar{R2\to R2-R1\text{ and }R3\to R3-2R2}
 +\go{1&1&3&1&25}{0&1&0&1&5}{0&-1&-3&2&-23}
 +\ar{R3\to R3+R2}
 +\go{1&1&3&1&25}{0&1&0&1&5}{0&0&-3&3&-18}
 +\ar{R3\to-\tfrac13R3}
 +\go{1&1&3&1&25}{0&1&0&1&5}{0&0&1&-1&6}
 +\end{align*}
 +This is in REF. There are leading entries in the columns for $x,y,z$ but not for $w$, so $w=t$ is a free variable (where $t\in\mathbb{R}$). Now
 +$$ z-w=6\implies z=6+w=6+t$$
 +$$ y+w=5\implies y=5-w=5-t$$
 +$$ x+y+3z+w=25\implies x=25-y-3z-w=25-(5-t)-3(6+t)-t=2-3t.$$
 +So 
 +$$ \begin{bmatrix}x\\y\\z\\w\end{bmatrix}=\begin{bmatrix} 2\\5\\6\\0\end{bmatrix}+t\begin{bmatrix}-3\\-1\\1\\1\end{bmatrix},\quad t\in\mathbb{R}.$$
 +
 +== Solution 2 ==
 +
 +We put the augmented matrix into RREF. 
 +\begin{align*} 
 +\def\go#1#2#3{\begin{bmatrix}#1\\#2\\#3\end{bmatrix}}
 +\def\ar#1{\\[6pt]\xrightarrow{#1}&}
 +&\go{2&1&3&4&27}{1&2&3&2&30}{1&1&3&1&25}
 +\ar{\text{do everything as above}}
 +\go{1&1&3&1&25}{0&1&0&1&5}{0&0&1&-1&6}
 +\ar{R1\to R1-3R3}
 +\go{1&1&0&4&7}{0&1&0&1&5}{0&0&1&-1&6}
 +\ar{R1\to R1-R2}
 +\go{1&0&0&3&2}{0&1&0&1&5}{0&0&1&-1&6}
 +\end{align*}
 +This is in RREF. There are leading entries in the columns for $x,y,z$ but not for $w$, so $w=t$ is a free variable (where $t\in\mathbb{R}$). Now
 +$$ z-w=6\implies z=6+w=6+t$$
 +$$ y+w=5\implies y=5-w=5-t$$
 +$$ x+3w=2\implies x=2-3w=2-3t$$
 +So 
 +$$ \begin{bmatrix}x\\y\\z\\w\end{bmatrix}=\begin{bmatrix} 2\\5\\6\\0\end{bmatrix}+t\begin{bmatrix}-3\\-1\\1\\1\end{bmatrix},\quad t\in\mathbb{R}.$$
 +
 +
lecture_5.1423051456.txt.gz · Last modified: by rupert

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