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Table of Contents
One more example
Solve the following linear system:
Solution
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
We know that we can apply EROs to any augmented matrix into REF.
Suppose the system has $n$ equations and $m$ variables, and let $k$ be the number of non-zero rows in REF. Also suppose the system is consistent: then the REF has no row of the form $[0~0~0~\dots~1]$.
- $k\le n$, because there are only $n$ rows in the whole matrix
- $k$ is precisely the number of leading variables. So $k$ is no bigger $m$, the total number of variables; in symbols, we have $k\le m$.
- All the other variables are free variables, so $$ \text{$m-k$ is the number of free variables.} $$
What does this tell us about the set of solutions? For example, how many solutions are there?
Observation 1: free variables and the number of solutions
For consistent systems, this shows that:
- either $k=m$;
- so $m-k=0$
- there are no free variables
- the system has one solution and no more
- We say it has a unique solution.
- or $k<m$
- so $m-k>0$
- there is at least one free variable
- so the system has infinitely many solutions (one for each value of each free variable)
- The number of free variables, $m-k$, is called the dimension of the solution set.
Observation 2: systems with fewer equations than variables
For consistent systems where $n<m$ (fewer equations than variables):
- $k\le n < m$, so $k<m$.
- So there is at least one free variable.
- So in this situation we always have infinitely many solutions.
Chapter 2: The algebra of matrices
Definition
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
- 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.
- $\begin{bmatrix}3\\2\\4\\0\\-1\end{bmatrix}$ 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 the row matrix and the column matrix above have the same entries, they have a different “shape”, or “size”, so we must think of them has being different matrices. Let's give the definitions to make this precise.
Definition
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.
Definition
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 $1\times 5$.
- $\begin{bmatrix}1\\2\end{bmatrix}\ne\begin{bmatrix}1 &0\\2&0\end{bmatrix}$ since these matrices are not the same size.
- $\begin{bmatrix}1&0\\0&1\end{bmatrix}\ne \begin{bmatrix}1&0\\1&0\end{bmatrix}$ because even though they have the same size, 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, so we get four equations:\begin{align*}3x&=1\\7y+2&=2z\\8z-3&=\sqrt2\\w^2&=9\end{align*}