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--- lecture_5_slides [2016/02/08 18:42] – 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 21: / Line 34: @@
 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 90: / Line 114: @@
 \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
-$$ \left[\begin{smallmatrix}x\\y\\z\\w\end{smallmatrix}\right]=\left[\begin{smallmatrix} 2\\5\\6\\0\end{smallmatrix}\right]+t\begin{smallmatrix}-3\\-1\\1\\1\end{bmatrix},\quad t\in\mathbb{R}.$$
+==== ====
+$\go{1&0&0&3&2}{0&1&0&1&5}{0&0&1&-1&6}$ is in RREF.
+  * $w=t$ is a free variable
+  * $z-w=6\implies z=6+w=6+t$
+  * $ y+w=5\implies y=5-w=5-t$
+  * $ x+3w=2\implies x=2-3w=2-3t$
+  * $\left[\begin{smallmatrix}x\\y\\z\\w\end{smallmatrix}\right]=\left[\begin{smallmatrix}2\\5\\6\\0\end{smallmatrix}\right]+t\left[\begin{smallmatrix}-3\\-1\\1\\1\end{smallmatrix}\right],\quad t\in\mathbb{R}.$
+==== Example ====
+If $ f(x)=ax^2+bx+c$ and $f(1)=3$, $f(2)=2$ and $f(3)=4$, find $f(x)$.
+  * $f(1)=3\implies a+b+c=3$
+  * $f(2)=2\implies 4a+2b+c=2$
+  * $f(3)=4\implies 9a+3b+c=4$
+  * $\begin{gather*}  a+b+c=3\\4a+2b+c=2\\9a+3b+c=4\end{gather*}$
+  * Solve using RREF.
+==== ====
+\begin{align*}
+\go{1&1&1&3}{4&2&1&2}{9&3&1&4}
+\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}
+\go{1&1&1&3}{0&1&\tfrac32&5}{0&-6&-8&-23}
+\ar{R3\to R3+6R2}
+\go{1&1&1&3}{0&1&\tfrac32&5}{0&0&1&7}
+\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 $f(x)=1.5x^2-5.5x+7$.