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--- lecture_6 [2015/02/05 11:57] – rupert
+++ lecture_6 [2017/02/07 10:14] (current) – rupert
@@ Line 2: / Line 2: @@
 === 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 4 ===
+For which value(s) of $k$ does the following linear system have
+infinitely many solutions?
+\begin{gather*}x+y+z=1\\x-z=5\\2x+3y+kz=-2\end{gather*}
+== Solution ==
+\begin{align*}
+&\go{1&1&1&1}{1&0&-1&5}{2&3&k&-2}
+\ar{R2\to R2-R1\text{ and }R3\to R3-2R1}
+\go{1&1&1&1}{0&-1&-2&4}{0&1&k-2&-4}
+\ar{R2\to -R2}
+\go{1&1&1&1}{0&1&2&-4}{0&1&k-2&-4}
+\ar{R3\to R3-R2}
+\go{1&1&1&1}{0&1&2&-4}{0&0&k-4&0}
+\end{align*}
+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 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
+solutions.
+If $k-4\ne0$, then we can divide the third row by $k-4$ to get the
+REF: \[\go{1&1&1&1}{0&1&2&-4}{0&0&1&0}\] In this situtation, there are
+no free variables since $x$, $y$ and $z$ are all leading variables. So
+if $k-4\ne 0$, or equivalently if $k\ne 4$, then there are no free
+variables so there is not an infinite number of solutions. (The only
+possibilities are that there is is a unique solution or that the
+system is inconsistent; and in this case you can check that there is a
+unique solution, although we don't need to know this to answer the
+question).
+In conclusion, the system has infinitely many solutions if and only if
+$k=4$.
+/*
+==== 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.