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--- lecture_5_slides [2016/02/08 18:49] – rupert
+++ lecture_5_slides [2017/02/07 10:15] (current) – rupert
@@ Line 1: / Line 1: @@
 ~~REVEAL~~
+===== The story so far... =====
+  * Systems of linear equations
+  * The augmented matrix
+  * Elementary Row Operations (EROs)
+  * Row Echelon Form (REF)
+  * Reduced Row Echelon Form (RREF)
+===== Today =====
+  - How to solve a system in REF or RREF
+  - How to find REF or RREF for a system
 ===== Solving a system in REF or RREF =====
@@ Line 19: / Line 32: @@
-Solve the linear system for $\newcommand{\sm}{\left[\begin{smallmatrix}}\newcommand{\esm}{\end{smallmatrix}\right]} \sm 1&2&3&0&0&8\\0&0&1&1&1&5\\0&0&0&1&4\esm$.
+Solve the linear system for $\newcommand{\sm}{\left[\begin{smallmatrix}}\newcommand{\esm}{\end{smallmatrix}\right]} \sm 1&2&3&0&0&8\\0&0&1&1&1&5\\0&0&0&1&3&4\esm$.
+  * Observe: it's in REF
   * Free variables: $x_2$, $x_5$. Set $x_2=s$, $x_5=t$.
     * $ x_4+3x_5=4\implies x_4=4-3x_5=4-3t$
@@ Line 27: / Line 41: @@
   * $ \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}.$
   * (A $2$-dimensional set in $5$-dimensional space $\mathbb{R}^5$).
+====== Gaussian elimination ======
+A systematic way to put an augmented matrix into REF or RREF using EROs
+  * REF=row echelon form
+  * RREF=reduced row echelon form
+  * ERO=elementary row operation
+    - reordering rows
+    - scaling a row
+    - combining two different rows
 ===== Gaussian elimination 1 =====
@@ Line 49: / Line 73: @@
 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*}
 ==== ====
@@ Line 101: / Line 125: @@
 ==== Example ====
+If $ f(x)=ax^2+bx+c$ and $f(1)=3$, $f(2)=2$ and $f(3)=4$, find $f(x)$.
-A function $f(x)$ has the form \[ f(x)=ax^2+bx+c\] where $a,b,c$ are
+  * $f(1)=3\implies a+b+c=3$
-constants. Given that $f(1)=3$, $f(2)=2$ and $f(3)=4$, find $f(x)$.
+  * $f(2)=2\implies 4a+2b+c=2$
+  * $f(3)=4\implies 9a+3b+c=4$
+  * $\begin{gather*}  a+b+c=3\\4a+2b+c=2\\9a+3b+c=4\end{gather*}$
+  * Solve using RREF.
 ==== ====
-  * $f(1)=3\implies a\cdot 1^2+b\cdot 1 +c = 3\implies a+b+c=3$
-  * f(2)=2\implies a\cdot 2^2+b\cdot 2 +c = 3\implies 4a+2b+c=2$
-  * $f(3)=4\implies a\cdot 3^2+b\cdot 3 +c = 3\implies 9a+3b+c=4$
-  * We get a system of three linear equations in the variables $a,b,c$:\begin{gather*}  a+b+c=3\\4a+2b+c=2\\9a+3b+c=4\end{gather*}$
-  * Let's reduce the augmented matrix for this system to RREF.
-==== ====
 \begin{align*}
-&\go{1&1&1&3}{4&2&1&2}{9&3&1&4}
+\go{1&1&1&3}{4&2&1&2}{9&3&1&4}
-\ar{R2\to R2-4R1\text{ and }R3\to R3-9R1}
+\xrightarrow{R2\to R2-4R1\text{ and }R3\to R3-9R1}&
 \go{1&1&1&3}{0&-2&-3&-10}{0&-6&-8&-23}
 \ar{R2\to -\tfrac12 R2}
@@ Line 123: / Line 141: @@
 \ar{R3\to R3+6R2}
 \go{1&1&1&3}{0&1&\tfrac32&5}{0&0&1&7}
-\ar{R1\to R1-R3\text{ and }R2\to R2-\tfrac32R3}
+\end{align*}
+  * So far: in REF!
+==== ====
+\begin{align*}
+\go{1&1&1&3}{0&1&\tfrac32&5}{0&0&1&7}
+\xrightarrow{R1\to R1-R3\text{ and }R2\to R2-\tfrac32R3}&
 \go{1&1&0&-4}{0&1&0&-5.5}{0&0&1&7}
 \ar{R1\to R1-R2}
 \go{1&0&0&1.5}{0&1&0&-5.5}{0&0&1&7}
 \end{align*}
-So $a=1.5$, $b=-5.5$ and $c=7$; so
+  * So $a=1.5$, $b=-5.5$ and $c=7$
-\[ f(x)=1.5x^2-5.5x+7.\]
+  * So $f(x)=1.5x^2-5.5x+7$.