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--- lecture_4_slides [2016/02/03 15:32] – rupert
+++ lecture_4_slides [2017/02/01 17:19] (current) – [Example] rupert
@@ Line 67: / Line 67: @@
 ===== Row echelon form and reduced row echelon form =====
-==== Row echelon form (REF) ====
+==== Definition: zero row ====
-/*
-=== Definition ===
 {{page>zero row}}
-=== Definition ===
+==== Definition: leading entry ====
 {{page>leading entry}}
-*/
+==== Row echelon form (REF) ====
 A matrix is in **row echelon form**, or **REF**, if:
@@ Line 98: / Line 95: @@
   * e.g. $\left[\begin{smallmatrix} {\color{blue}1}&{\color{red}2}&{\color{red}3}&4&5\\0&{\color{blue}1}&{\color{red}2}&3&4\\0&0&{\color{blue}1}&2&3\end{smallmatrix}\right]$ and $\left[\begin{smallmatrix} {\color{blue}1}&0&{\color{red}3}&4&5\\0&{\color{blue}1}&0&3&4\\0&0&{\color{blue}1}&2&3\end{smallmatrix}\right]$ are in REF but not RREF
   * e.g. $\left[\begin{smallmatrix} {\color{blue}1}&0&0&4&5\\0&{\color{blue}1}&0&3&4\\0&0&{\color{blue}1}&2&3\end{smallmatrix}\right]$ is in RREF.
+  * How about $\left[\begin{smallmatrix} {\color{blue}1}&9&0&4&5\\0&0&{\color{blue}1}&0&3\\0&0&0&{\color{blue}1}&2\end{smallmatrix}\right]$, $\left[\begin{smallmatrix} {\color{blue}1}&9&0&0&5\\0&0&{\color{blue}1}&0&3\\0&0&0&{\color{blue}1}&2\end{smallmatrix}\right]$, $\left[\begin{smallmatrix} {\color{blue}1}&0&0&0&5\\0&0&{\color{blue}1}&0&3\\0&0&0&{\color{blue}1}&2\end{smallmatrix}\right]$?
 ==== Example ====
@@ Line 138: / Line 135: @@
-$\newcommand{\sm}{\left[\begin{smallmatrix}}\newcommand{\esm}{\end{smallmatrix}\right]} \sm 1&2&3&0&0&8\\0&0&1&1&1&5\\0&0&0&1&3&4\esm$
+Solve the linear system for $\newcommand{\sm}{\left[\begin{smallmatrix}}\newcommand{\esm}{\end{smallmatrix}\right]} \sm 1&2&3&0&0&8\\0&0&1&1&1&5\\0&0&0&1&3&4\esm$.
-is in REF.
-  * Using variables $x_1,x_2,x_3,x_4,x_5$:
-    * $x_1$, $x_3$ and $x_4$ are leading: columns have a leading entry
-    * $x_2$ and $x_5$ are free: columns don't have a leading entry
-  - $x_2$, $x_5$ free, so set $x_2=s$ and $x_5=t$ where $s,t\in \mathbb{R}$
+  * This is in REF, so can use the method just described.
-  - Working from the bottom:
+  * Free variables: $x_2$, $x_5$. Set $x_2=s$, $x_5=t$.
+  * Working from bottom:
     * $ x_4+3x_5=4\implies x_4=4-3x_5=4-3t$
-    * $ x_3+x_4+x_5=5\implies x_3=5-x_4-x_5=5-(4-3t)-t=1+2t$
+    * $ x_3+x_4+x_5=5\implies x_3=1+2t$
-    * $ x_1+2x_2+3x_3=8\implies x_1=8-2x_2-3x_3=8-2s-3(1+2t)=5-2s-6t.$
+    * $ x_1+2x_2+3x_3=8\implies x_1=5-2s-6t.$
   * So $ \left[\begin{smallmatrix} x_1\\x_2\\x_3\\x_4 \\x_5\end{smallmatrix}\right]= \left[\begin{smallmatrix} 5\\0\\1\\4\\0\end{smallmatrix}\right] +s\left[\begin{smallmatrix} -2\\1\\0\\0\\0\end{smallmatrix}\right] +t\left[\begin{smallmatrix} -6\\0\\2\\-3\\1\end{smallmatrix}\right],\quad s,t\in \mathbb{R}.$
-  * (The solution set is a $2$-dimensional subset of $5$-dimensional space).
+  * 2 free vars, 5 vars in all
+  * Solution set is $2$-dimensional, in $5$-dimensional space $\mathbb{R}^5$.