Differences

This shows you the differences between two versions of the page.

--- lecture_5 [2015/02/04 12:14] – [Gaussian elimination] rupert
+++ lecture_5 [2017/02/07 10:11] (current) – [Gaussian elimination] rupert
@@ Line 41: / Line 41: @@
 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 ==
+== Solution 1 ==
+We put the augmented matrix into REF:
 \begin{align*}
 \def\go#1#2#3{\begin{bmatrix}#1\\#2\\#3\end{bmatrix}}
@@ Line 54: / Line 55: @@
 \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}
+\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}
@@ Line 64: / Line 65: @@
 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
+$$ \begin{bmatrix}x\\y\\z\\w\end{bmatrix}=\begin{bmatrix} 2\\5\\6\\0\end{bmatrix}+t\begin{bmatrix}-3\\-1\\1\\1\end{bmatrix},\quad t\in\mathbb{R}.$$