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--- lecture_5 [2016/02/09 10:27] – rupert
+++ lecture_5 [2017/02/07 10:11] (current) – [Gaussian elimination] rupert
@@ Line 55: / 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 87: / Line 87: @@
 $$ \begin{bmatrix}x\\y\\z\\w\end{bmatrix}=\begin{bmatrix} 2\\5\\6\\0\end{bmatrix}+t\begin{bmatrix}-3\\-1\\1\\1\end{bmatrix},\quad t\in\mathbb{R}.$$
-=== Example ===
-A function $f(x)$ has the form \[ f(x)=ax^2+bx+c\] where $a,b,c$ are
-constants. Given that $f(1)=3$, $f(2)=2$ and $f(3)=4$, find $f(x)$.
-== Solution ==
-\begin{gather*} 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
-\end{gather*}
-We get a system of three linear equations in the variables $a,b,c$:
-\begin{gather*}
- a+b+c=3\\
-a+2b+c=2\\
-a+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}
-\ar{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}
-\ar{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
-\[ f(x)=1.5x^2-5.5x+7.\]