Differences

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

--- lecture_6 [2015/02/05 12:18] – 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 25: / Line 66: @@
 \go{1&0&0&3}{0&1&0&-1}{0&0&1&2}
 \end{align*}
-The solution is $x=3$, $y=-1$, $z=2$. So there is a unique solution: just one opint in $\mathbb{R}^3$, namely $(3,-1,2)$.
+The solution is $x=3$, $y=-1$, $z=2$. So there is a unique solution: just one point in $\mathbb{R}^3$, namely $(3,-1,2)$.
-=== Example 2 ===
+=== Example 3 ===
 Solve the linear system
-\begin{align*}3x+y-2z+4w&=5\\x+z+w&=2\\4x+2y-6z+6w=0\end{align*}
+\begin{align*}3x+y-2z+4w&=5\\x+z+w&=2\\4x+2y-6z+6w&=0\end{align*}
 == Solution ==
@@ Line 41: / Line 82: @@
 \go{1&0&1&1&2}{3&1&-2&4&5}{4&2&-6&6&0}
 \ar{R2\to R2-3R1\text{ and }R3\to R3-4R1}
-\go{1&0&1&1&2}{&0&1&-5&1&-1}{0&2&-10&2&-8}
+\go{1&0&1&1&2}{0&1&-5&1&-1}{0&2&-10&2&-8}
 \ar{R3\to R3-2R2}
 \go{1&0&1&1&2}{0&1&-5&1&-1}{0&0&0&0&-6}
@@ Line 53: / Line 94: @@
 {{page>inconsistent}}
-=== Example 3 ===
-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
-\[ f(x)=1.5x^2-5.5x+7.\]
 === Example 4 ===
@@ Line 111: / Line 114: @@
 \end{align*}
-If $k-4=0$, then this matrix is in REF:
+If $k-4=0$, then this matrix is in [[REF]]:
-\[\go{1&1&1&1}{0&1&2&-4}{0&0&0&0}\] In this situation, $z$ is a free
+\[\go{1&1&1&1}{0&1&2&-4}{0&0&0&0}\] In this situation, $z$ is a [[free variable]] (since there's no [[leading entry]] in the third column).  For
-variable (since there's no leading entry in the third column).  For
 each value of $z$ we get a different solution, so if $k-4=0$,
 or equivalently, if $k=4$, then there are infinitely many different
@@ Line 131: / Line 133: @@
 $k=4$.
+/*
+==== One more example ====
+Solve the following [[linear system]]:
+\begin{align*} x+y+z&=2\\2x-z&=0\\x+3y+4z&=6\\3x+y&=2\end{align*}
+== Solution ==
+We remark that there are more equations than unknowns... but this isn't a problem! We proceed as usual:
+\begin{align*}
+\def\go#1#2#3#4{\begin{bmatrix}#1\\#2\\#3\\#4\end{bmatrix}}
+\def\ar#1{\\[6pt]\xrightarrow{#1}&}
+&\go{1&1&1&2}{2&0&-1&0}{1&3&4&6}{3&1&0&2}
+\ar{R2\to R2-2R1,\ R3\to R3-R1\text{ and }R4\to R4-3R1}
+\go{1&1&1&2}{0&-2&-3&-4}{0&2&3&4}{0&-2&-3&-4}
+\ar{R3\to R3+R2\text{ and }R4\to R4-R2}
+\go{1&1&1&2}{0&1&1.5&2}{0&0&0&0}{0&0&0&0}
+\ar{R1\to R1-R2}
+\go{1&0&-0.5&0}{0&1&1.5&2}{0&0&0&0}{0&0&0&0}
+\end{align*}
+Now $z$ is a free variable, say $z=t$ where $t\in \mathbb{R}$, and from row 2 we get $y=2-1.5t$ and row 1 gives $x=0.5t$, so the solution is
+\[ \begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\2\\0\end{bmatrix}+t\begin{bmatrix}0.5\\-1.5\\1\end{bmatrix},\quad t\in\mathbb{R}.\]
+*/
+===== Observations about Gaussian elimination =====
+{{page>gaussian elimination remarks}}
+====== Chapter 2: The algebra of matrices ======
+=== Definition ===
+{{page>matrix}}
+{{page>(i,j) entry}}
+=== Example ===
+If $B=\begin{bmatrix} 99&3&5\\7&-20&14\end{bmatrix}$, then $B$ is a $2\times 3$ matrix, and the $(1,1)$ entry of $B$ is $b_{11}=99$, the $(1,3)$ entry of $B$ is $b_{13}=5$, the $(2,1)$ entry is $b_{21}=7$, etc.