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Table of Contents
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, 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<html><br /></html> #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:<html><br /></html>always have #vars ≤ #eqs
Chapter 2: The algebra of matrices
An $n\times m$ matrix is a grid of numbers with $n$ rows and $m$ columns: \[ A=\begin{bmatrix}a_{11}&a_{12}&\dots&a_{1m}\\a_{21}&a_{22}&\dots&a_{2m}\\\vdots&\vdots&&\vdots\\a_{n1}&a_{n2}&\dots&a_{nm}\end{bmatrix}\]
The $(i,j)$ entry of a matrix $A$ is $a_{ij}$, the number in row $i$ and column $j$ of $A$.
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
Two matrices $A$ and $B$ have the same size if they have the same number of rows, and they have the same number of columns.
If two matrices do not have the same size, we say they have different sizes.
Equality of matrices
Two matrices $A$ and $B$ are said to be equal if both of the following conditions hold:
- $A$ and $B$ have the same size; and
- every entry of $A$ is equal to the corresponding entry of $B$; in other words, for every $(i,j)$ so that $A$ and $B$ have an $(i,j)$ entry, we have $a_{ij}=b_{ij}$.
When $A$ and $B$ are equal matrices, we write $A=B$. Otherwise, we write $A\ne B$.
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*}