==== 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\\
4a+2b+c=2\\
9a+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
\begin{align*}2x+4y-2z&=1\\x+y+z&=4\\2x+5y+z&=3\end{align*}
by transforming the [[augmented matrix]] into [[reduced row echelon form]].
== Solution ==

\begin{align*} 
\def\go#1#2#3{\begin{bmatrix}#1\\#2\\#3\end{bmatrix}}
\def\ar#1{\\[6pt]\xrightarrow{#1}&}
&\go{3&4&-2&1}{1&1&1&4}{2&5&1&3}
\ar{R1\leftrightarrow R2}
\go{1&1&1&4}{3&4&-2&1}{2&5&1&3}
\ar{R2\to R2-3R1\text{ and }R3\to R3-2R1}
\go{1&1&1&4}{0&1&-5&-11}{0&3&-1&-5}
\ar{R3\to R3-3R1}
\go{1&1&1&4}{0&1&-5&-11}{0&0&14&28}
\ar{R3\to \tfrac1{14}R3}
\go{1&1&1&4}{0&1&-5&-11}{0&0&1&2}
\ar{R1\to R1-R3\text{ and }R2\to R2+5R3}
\go{1&1&0&2}{0&1&0&-1}{0&0&1&2}
\ar{R1\to R1-R2}
\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 point in $\mathbb{R}^3$, namely $(3,-1,2)$.


=== Example 3 ===

Solve the linear system
\begin{align*}3x+y-2z+4w&=5\\x+z+w&=2\\4x+2y-6z+6w&=0\end{align*}

== Solution ==


\begin{align*} 
&\go{3&1&-2&4&5}{1&0&1&1&2}{4&2&-6&6&0}
\ar{R1\leftrightarrow R2}
\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}
\ar{R3\to R3-2R2}
\go{1&0&1&1&2}{0&1&-5&1&-1}{0&0&0&0&-6}
\ar{R3\to -\tfrac16 R3}
\go{1&0&1&1&2}{0&1&-5&1&-1}{0&0&0&0&1}
\end{align*}

This is in [[REF]]. The last row corresponds to the equation 
\[ 0=1\]
which clearly has no solution! We conclude that this system has no solutions, and hence the original linear system has no solutions either.

{{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.