Differences

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

--- lecture_6 [2016/02/11 11:19] – rupert
+++ lecture_6 [2017/02/07 10:14] (current) – rupert
@@ Line 1: / Line 1: @@
- ==== Examples ====
+==== Examples ====
 === Example 1 ===
+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*}
+\def\go#1#2#3{\begin{bmatrix}#1\\#2\\#3\end{bmatrix}}
+\def\ar#1{\\[6pt]\xrightarrow{#1}&}
+&\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
+\[ f(x)=1.5x^2-5.5x+7.\]
+=== Example 2 ===
 Solve the linear system
@@ Line 28: / Line 69: @@
-=== Example 2 ===
+=== Example 3 ===
 Solve the linear system
@@ Line 54: / Line 95: @@
 {{page>inconsistent}}
-=== Example 3 ===
+=== Example 4 ===
 For which value(s) of $k$ does the following linear system have