~~REVEAL~~
===== Solving linear systems. Examples; how many solutions? =====
==== More examples ====
==== Example 1 ====
If $ f(x)=ax^2+bx+c$ and $f(1)=3$, $f(2)=2$ and $f(3)=4$, find $f(x)$.
* $f(1)=3\implies a+b+c=3$
* $f(2)=2\implies 4a+2b+c=2$
* $f(3)=4\implies 9a+3b+c=4$
* $\begin{gather*} a+b+c=3\\4a+2b+c=2\\9a+3b+c=4\end{gather*}$
* Solve using RREF.
==== ====
\begin{align*}\def\go#1#2#3{\left[\begin{smallmatrix}#1\\#2\\#3\end{smallmatrix}\right]}
\def\ar#1{\\\xrightarrow{#1}&}
\go{1&1&1&3}{4&2&1&2}{9&3&1&4}
\xrightarrow{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}
\end{align*}
* So far: in REF!
==== ====
\begin{align*}
\go{1&1&1&3}{0&1&\tfrac32&5}{0&0&1&7}
\xrightarrow{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*}3x+4y-2z&=1\\x+y+z&=4\\2x+5y+z&=3\end{align*}
by transforming the [[augmented matrix]] into [[reduced row echelon form]].
==== ====
\begin{align*}
\def\go#1#2#3{\left[\begin{smallmatrix}#1\\#2\\#3\end{smallmatrix}\right]}
\def\ar#1{\\\xrightarrow{#1}&}
\go{3&4&-2&1}{1&1&1&4}{2&5&1&3}
\xrightarrow{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}
\end{align*}
* in REF, keep going for RREF...
==== ====
\begin{align*}
\go{1&1&1&4}{0&1&-5&-11}{0&0&1&2}
\xrightarrow{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*}
* $x=3$, $y=-1$, $z=2$
* There is a **unique solution**: $(3,-1,2)$.
* No free variables.
==== 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*}
\go{3&1&-2&4&5}{1&0&1&1&2}{4&2&-6&6&0}
\xrightarrow{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*}
==== ====
\[ \go{1&0&1&1&2}{0&1&-5&1&-1}{0&0&0&0&1}\]
* This is in [[REF]].
* The last row corresponds to the equation $0=1$
* This has no solution!
* So the original linear system has no solutions.
* Call the system **inconsistent** (no solutions).
* To detect this: put in REF and find a row $[0~0~\dots~0~1]$.
==== 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*}
==== ====
\begin{align*}
\go{1&1&1&1}{1&0&-1&5}{2&3&k&-2}
\xrightarrow{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*}
==== ====
\[\go{1&1&1&1}{0&1&2&-4}{0&0&k-4&0}\]
* If $k=4$:
* get $\go{1&1&1&1}{0&1&2&-4}{0&0&0&0}$, in REF
* $z$ [[free variable|free]], so infinitely many solutions
* If $k\ne4$ then $k-4\ne0$.
* $R3\to \tfrac1{k-4}R3$: $\go{1&1&1&1}{0&1&2&-4}{0&0&1&0}$, in REF
* no free vars, so not infinitely many solutions
* So infinitely many solutions $\iff k=4$.
==== Observations ====
For a system of linear equations with
#vars variables, #eqs equations:
* If #vars > #eqs, at least one var is free (see REF!)
- either system is inconsistent....
- ....or it has infinitely many solutions, one for each value of the free vars.
- The **dimension** of the set of solutions is the number of free variables.
* If there's a **unique** solution:
always have #vars ≤ #eqs
====== Chapter 2: The algebra of matrices ======
==== ====
{{page>matrix}}
* {{page>(i,j) entry}}
==== Examples ====
* $B=\begin{bmatrix} 99&3&5\\7&-20&14\end{bmatrix}$ is a $2\times 3$ matrix
* the $(1,1)$ entry of $B$ is $b_{11}=99$
* the $(1,3)$ entry of $B$ is $b_{13}=5$
* the $(2,1)$ entry of $B$ is $b_{21}=7$
* etc.
* $(3,2)$ entry of $B$?
* undefined!
==== =====
* $\left[\begin{smallmatrix}3\\2\\4\\0\\-1\end{smallmatrix}\right]$ is a $5\times 1$ matrix.
* A matrix like this with one column is called a **column vector**.
* $\begin{bmatrix}3&2&4&0&-1\end{bmatrix}$ is a $1\times 5$ matrix.
* A matrix like this with one row is called a **row vector**.
* Even though these have the same entries, they have a different "shape", or "size" and they are different matrices.
==== Size of a matrix ====
{{page>same size}}
==== Equality of matrices ====
{{page>equal matrices}}
==== Examples ====
* $\begin{bmatrix}3\\2\\4\\0\\-1\end{bmatrix}\ne \begin{bmatrix}3&2&4&0&-1\end{bmatrix}$, since these matrices have different sizes: the first is $5\times 1$ but the second is $1\times 5$.
==== ====
* $\begin{bmatrix}1\\2\end{bmatrix}\ne\begin{bmatrix}1 &0\\2&0\end{bmatrix}$
* not the same size.
* $\begin{bmatrix}1&0\\0&1\end{bmatrix}\ne \begin{bmatrix}1&0\\1&1\end{bmatrix}$
* same size but the $(2,1)$ entries are different.
==== ====
* If $\begin{bmatrix}3x&7y+2\\8z-3&w^2\end{bmatrix}=\begin{bmatrix}1&2z\\\sqrt2&9\end{bmatrix}$ then we know that all the corresponding entries are equal
* We get four equations:\begin{align*}3x&=1\\7y+2&=2z\\8z-3&=\sqrt2\\w^2&=9\end{align*}